July 2024, Week 2: July 8th - July 14th

The central concept of this solution is to leverage a deque (double-ended queue) to simulate the circular elimination game efficiently. This approach directly models the game's mechanics by rotating players and removing them as per the rules.

• Circular Nature Simulation: The deque allows for efficient rotation of players, mimicking the circular arrangement described in the problem.

• Player Elimination: By using the deque's popleft() method, we can easily remove players from the game when they are counted out.

• Efficient Rotation: The deque's ability to perform operations at both ends makes it ideal for simulating the counting process without the need for complex index calculations.

1. Initialization:

   active_players = deque(range(n))

   We create a deque containing integers from 0 to n-1, representing the players. Using 0-based indexing simplifies the simulation process.

2. Game Simulation Loop:

   while len(active_players) > 1:

   This loop continues until only one player remains, simulating the entire game process.

3. Player Rotation:

   for _ in range(k - 1): active_players.append(active_players.popleft())

   We rotate k-1 players from the front to the back of the deque. This simulates the counting process without actually removing any players yet. We use k-1 because the kth player will be removed in the next step.

4. Player Elimination:

   active_players.popleft()

   After the rotation, the player at the front of the deque is the kth player in the count. We remove this player using popleft(), simulating their elimination from the game.

5. Result Calculation/Return:

   return active_players[0] + 1

   Once the loop ends, only one player remains in the deque. We return this player's number, adding 1 to convert from 0-based to 1-based indexing as required by the problem statement.

Time Complexity:

• , where is the number of players and is the counting interval. This is because we perform eliminations (as we start with players and end with 1), and for each elimination, we rotate the deque times.

Space Complexity:

• , where is the number of players. This is because we store all players in the deque initially. The space used remains constant throughout the execution of the algorithm, even as players are eliminated.

Input:

Step-by-Step Walkthrough:

1. Initialization:

   • Create a deque called active_players with values [0, 1, 2, 3, 4, 5], representing the players in 0-indexed form.

2. Main Loop (Elimination Process):

   • Iteration 1:

     • Current active players: [0, 1, 2, 3, 4, 5]

     • We need to remove the fifth player (k=5)

     • Rotate the deque 4 times (k-1):

       • Move 0 to the end: [1, 2, 3, 4, 5, 0]

       • Move 1 to the end: [2, 3, 4, 5, 0, 1]

       • Move 2 to the end: [3, 4, 5, 0, 1, 2]

       • Move 3 to the end: [4, 5, 0, 1, 2, 3]

     • Remove the front player (4)

     • Resulting deque: [5, 0, 1, 2, 3]

   • Iteration 2:

     • Current active players: [5, 0, 1, 2, 3]

     • Rotate 4 times:

       • Final state after rotation: [3, 5, 0, 1, 2]

     • Remove the front player (3)

     • Resulting deque: [5, 0, 1, 2]

   • Iteration 3:

     • Current active players: [5, 0, 1, 2]

     • Rotate 4 times:

       • Final state after rotation: [5, 0, 1, 2]

     • Remove the front player (5)

     • Resulting deque: [0, 1, 2]

   • Iteration 4:

     • Current active players: [0, 1, 2]

     • Rotate 4 times:

       • Final state after rotation: [1, 2, 0]

     • Remove the front player (1)

     • Resulting deque: [2, 0]

   • Iteration 5:

     • Current active players: [2, 0]

     • Rotate 4 times:

       • Final state after rotation: [2, 0]

     • Remove the front player (2)

     • Resulting deque: [0]

3. Loop Termination:

   • The loop terminates when only one player remains in the active_players deque.

   • Final state of active_players: [0]

4. Visual Aids:

   The following image represents the simulation process for the example input n = 6, k = 5 where the red elements represent the players that will be removed in each round. The visualization shows the rotation of the deque and the elimination of players until only one player remains.

   

   Iteration Summary:

   Round	Active Players	Removed Player
   1	[5, 0, 1, 2, 3]	4
   2	[5, 0, 1, 2]	3
   3	[0, 1, 2]	5
   4	[2, 0]	1
   5	[0]	2

5. Result Calculation/Final Steps:

   • The last remaining player in active_players is 0 (0-indexed).

   • Convert to 1-indexed by adding 1: 0 + 1 = 1

   • Return 1 as the final result.

The winner of the game is player 1 (in 1-indexed form).

The central concept of this solution is to leverage the recursive nature of the Josephus problem to calculate the winner's position efficiently. Instead of simulating the game step-by-step, it uses a mathematical relationship between the survivor's position in games with and players.

• Recursive Formulation: The solution is built on the idea that the survivor's position in a game with players can be derived from the survivor's position in a game with players.

• Position Transformation: The key mathematical relationship used is (survivor_position + k) % n, which adjusts the survivor's position from an ( )-player game to an -player game.

• Bottom-up Calculation: The solution starts from the base case of one player and builds up to players through recursive calls.

1. Helper Function Definition:

   def simulate_game(players: int, step: int) -> int:

   This inner function is defined to handle the recursive calculations. It takes the current number of players and the step size as parameters.

2. Base Case:

   if players == 1: return 0

   When only one player remains, they are at position 0 (0-based indexing). This serves as the stopping condition for the recursion.

3. Recursive Case:

   survivor_position = simulate_game(players - 1, step)

   We recursively calculate the survivor's position for a game with one less player. This builds the solution from the bottom up.

4. Position Adjustment:

   return (survivor_position + step) % players

   We adjust the survivor's position from the (n-1)-player game to the n-player game. Adding step moves the position forward, and the modulo operation wraps it around the circle if necessary.

5. Main Function Call and Result Conversion:

   return simulate_game(n, k) + 1

   We call the helper function with the initial number of players and step size. The result is then converted to 1-based indexing by adding 1.

Time Complexity:

• , where is the number of players. This is because the function makes exactly recursive calls, each performing constant time operations (addition and modulo). Despite being recursive, the linear nature of the recursion leads to a linear time complexity.

Space Complexity:

• , where is the number of players. This space complexity is due to the recursion stack. In the worst case, the recursion will go levels deep before reaching the base case, resulting in stack frames being stored simultaneously.

Input:

Step-by-Step Walkthrough:

1. Initialization:

   • The function findTheWinner2(n=6, k=5) is called.

   • It immediately calls the helper function simulate_game(6, 5).

2. Main Recursive Process:

   • Recursive Call: simulate_game(6, 5)

     • Not a base case, so it makes a recursive call to simulate_game(5, 5)

   • Recursive Call: simulate_game(5, 5)

     • Not a base case, so it makes a recursive call to simulate_game(4, 5)

   • Recursive Call: simulate_game(4, 5)

     • Not a base case, so it makes a recursive call to simulate_game(3, 5)

   • Recursive Call: simulate_game(3, 5)

     • Not a base case, so it makes a recursive call to simulate_game(2, 5)

   • Recursive Call: simulate_game(2, 5)

     • Not a base case, so it makes a recursive call to simulate_game(1, 5)

   • Recursive Call: simulate_game(1, 5)

     • Base case reached: only one player remains

     • Returns position 0

3. Unwinding the Recursion:

   • simulate_game(2, 5):

     • Calculates new position: (0 + 5) % 2 = 1

     • Returns 1

   • simulate_game(3, 5):

     • Calculates new position: (1 + 5) % 3 = 0

     • Returns 0

   • simulate_game(4, 5):

     • Calculates new position: (0 + 5) % 4 = 1

     • Returns 1

   • simulate_game(5, 5):

     • Calculates new position: (1 + 5) % 5 = 1

     • Returns 1

   • simulate_game(6, 5):

     • Calculates new position: (1 + 5) % 6 = 0

     • Returns 0

4. Visual Aids:

   The following image represents the recursive process for the example input n = 6, k = 5. Each row represents a recursive call, with the top row being the base case. The green elements represent the survivor's position at each step, and the arrows show how the position is calculated and passed up through the recursive calls.

   

   Recursive Calls Summary:

   Players	Step	Case	Survivor Position
   6	5	Recursive case	0
   5	5	Recursive case	1
   4	5	Recursive case	1
   3	5	Recursive case	0
   2	5	Recursive case	1
   1	5	Base case	0

5. Result Calculation/Final Steps:

   • The recursive function returns 0 (0-indexed result).

   • The main function adds 1 to convert to 1-indexed: 0 + 1 = 1

   • Return 1 as the final result.

The winner of the game is player 1 (in 1-indexed form).

The central concept of this solution is to solve the Josephus problem iteratively, building up the solution from smaller game sizes to the target size. It leverages the mathematical relationship between survivor positions in games of consecutive sizes.

• Iterative Build-up: The solution starts with a trivial game of one player and iteratively calculates the survivor's position for games of increasing sizes up to n players.

• Position Transformation: Similar to the recursive approach, it uses the formula (survivor_position + k) % circle_size to adjust the survivor's position as the game size increases.

• In-place Calculation: Unlike the simulation approach, this method doesn't maintain a list of players, instead working directly with the survivor's position.

1. Initialization:

   survivor_position = 0

   We start with a survivor position of 0, which represents the winner's position in a trivial 1-player game.

2. Iterative Calculation Loop:

   for circle_size in range(2, n + 1):

   This loop iterates from 2 to , calculating the survivor's position for each game size. We start from 2 because the 1-player case is trivially handled by the initialization.

3. Position Adjustment:

   survivor_position = (survivor_position + k) % circle_size

   For each circle size, we adjust the survivor's position from the previous game size. Adding simulates the counting process, and the modulo operation wraps the position around the current circle size if necessary.

4. Result Conversion and Return:

   return survivor_position + 1

   After the loop completes, survivor_position holds the winner's position for the n-player game in 0-based indexing. We add 1 to convert to 1-based indexing as required by the problem statement.

Time Complexity:

• , where is the number of players. This is because the function iterates from 2 to , performing a constant number of operations (addition and modulo) in each iteration. Despite solving the problem for all game sizes from 2 to , the linear nature of the iteration results in a linear time complexity.

Space Complexity:

• , or constant space. This solution uses only a fixed amount of additional space (the survivor_position variable) regardless of the input size. It doesn't use any data structures that grow with the input, nor does it use recursion that would create a variable-sized call stack.

Input:

Step-by-Step Walkthrough:

1. Initialization:

   • Set survivor_position = 0, representing the initial position of the survivor in a hypothetical 1-player game.

2. Main Loop (Iterative Calculation):

   • Iteration 1 (circle size 2):

     • Current state: survivor_position = 0, circle_size = 2

     • Calculation: new_position = (0 + 5) % 2 = 1

     • Update survivor_position to 1

   • Iteration 2 (circle size 3):

     • Current state: survivor_position = 1, circle_size = 3

     • Calculation: new_position = (1 + 5) % 3 = 0

     • Update survivor_position to 0

   • Iteration 3 (circle size 4):

     • Current state: survivor_position = 0, circle_size = 4

     • Calculation: new_position = (0 + 5) % 4 = 1

     • Update survivor_position to 1

   • Iteration 4 (circle size 5):

     • Current state: survivor_position = 1, circle_size = 5

     • Calculation: new_position = (1 + 5) % 5 = 1

     • survivor_position remains 1

   • Iteration 5 (circle size 6):

     • Current state: survivor_position = 1, circle_size = 6

     • Calculation: new_position = (1 + 5) % 6 = 0

     • Update survivor_position to 0

3. Loop Termination:

   • The loop terminates when circle_size reaches n (6 in this case).

   • Final state of survivor_position: 0

4. Visual Aids:

   The following image represents the iterative process for the example input n = 6, k = 5. Each row represents a circle of increasing size, with the green element representing the survivor's position at each step. The arrows and labels show how the position is calculated from one step to the next.

   

   Iteration Summary:

   Circle Size	Survivor Position
   2	1
   3	0
   4	1
   5	1
   6	0

5. Result Calculation/Final Steps:

   • The final survivor_position is 0 (0-indexed result).

   • Convert to 1-indexed by adding 1: 0 + 1 = 1

   • Return 1 as the final result.

The winner of the game is player 1 (in 1-indexed form).

The central concept of this solution is to simulate the chef's order processing in real-time, leveraging a linear scan through the customer list to calculate waiting times efficiently. This approach mimics the actual process of a single chef handling orders as they come in.

• Time Tracking: The solution maintains a running order_finish_time to keep track of when the chef will be available next.

• Greedy Processing: Orders are processed greedily in the sequence they arrive, without any optimization or reordering.

• Cumulative Wait Time: The total waiting time is accumulated as each order is processed, allowing for a simple average calculation at the end.

1. Initialization:

   order_finish_time, total_customer_wait_time = 0, 0

   The function initializes two key variables: order_finish_time to track when the chef will next be available, and total_customer_wait_time to accumulate the total waiting time across all customers.

2. Iterating Through Customers:

   for arrival_time, prep_time in customers:

   This loop processes each customer's order sequentially, unpacking their arrival time and preparation time. This approach aligns with the problem statement that orders are prepared in the given input order.

3. Calculating Order Finish Time:

   order_finish_time = max(arrival_time, order_finish_time) + prep_time

   This line is crucial as it determines when the current order will be finished. It uses max() to handle two scenarios:

   • If the chef is idle (order_finish_time < arrival_time), cooking starts at arrival_time.

   • If the chef is busy (order_finish_time > arrival_time), cooking starts when the previous order finishes. The preparation time is then added to this start time to get the new order_finish_time.

4. Calculating and Accumulating Wait Time:

   total_customer_wait_time += order_finish_time - arrival_time

   The waiting time for each customer is the difference between when their order is finished and when they arrived. This is added to the running total of wait times.

5. Calculating Average Wait Time:

   average_wait_time = total_customer_wait_time / len(customers) return average_wait_time

   After processing all customers, the function calculates the average wait time by dividing the total wait time by the number of customers, then returns this value.

Time Complexity:

• , where is the number of customers. This is because the function iterates through the customer list exactly once, performing constant-time operations for each customer.

Space Complexity:

• , or constant space. This is because the function uses only a fixed amount of additional space (order_finish_time and total_customer_wait_time) regardless of the input size. It doesn't create any data structures that grow with the input.

Input:

Step-by-Step Walkthrough:

1. Initialization:

   • Initialize order_finish_time = 0 (representing when the chef will be available next)

   • Initialize total_customer_wait_time = 0 (to accumulate total waiting time for all customers)

2. Main Loop (Processing each customer):

   • Iteration 1 (Customer 1):

     • Current state: order_finish_time = 0, total_customer_wait_time = 0

     • Customer arrives at time 5 with order prep time 2

     • Calculate new order_finish_time:

       • max(5, 0) + 2 = 7 (chef starts at customer's arrival time)

     • Calculate customer wait time: 7 - 5 = 2

     • Update total_customer_wait_time: 0 + 2 = 2

     • Update order_finish_time to 7

   • Iteration 2 (Customer 2):

     • Current state: order_finish_time = 7, total_customer_wait_time = 2

     • Customer arrives at time 5 with order prep time 4

     • Calculate new order_finish_time:

       • max(5, 7) + 4 = 11 (chef starts after finishing the previous order)

     • Calculate customer wait time: 11 - 5 = 6

     • Update total_customer_wait_time: 2 + 6 = 8

     • Update order_finish_time to 11

   • Iteration 3 (Customer 3):

     • Current state: order_finish_time = 11, total_customer_wait_time = 8

     • Customer arrives at time 10 with order prep time 3

     • Calculate new order_finish_time:

       • max(10, 11) + 3 = 14 (chef starts after finishing the previous order)

     • Calculate customer wait time: 14 - 10 = 4

     • Update total_customer_wait_time: 8 + 4 = 12

     • Update order_finish_time to 14

   • Iteration 4 (Customer 4):

     • Current state: order_finish_time = 14, total_customer_wait_time = 12

     • Customer arrives at time 20 with order prep time 1

     • Calculate new order_finish_time:

       • max(20, 14) + 1 = 21 (chef starts at customer's arrival time)

     • Calculate customer wait time: 21 - 20 = 1

     • Update total_customer_wait_time: 12 + 1 = 13

     • Update order_finish_time to 21

3. Visual Aid:

   Customer	Arrival Time	Prep Time	Order Finish Time	Wait Time	Total Wait Time
   1	5	2	7	2	2
   2	5	4	11	6	8
   3	10	3	14	4	12
   4	20	1	21	1	13

4. Result Calculation:

   • Calculate average waiting time: total_customer_wait_time / number_of_customers = 13 / 4 = 3.25

The function returns 3.25, which is the average waiting time for all customers.

The central concept of this solution is to leverage a single integer variable to represent the current folder depth. This approach simplifies the problem of tracking folder navigation into basic arithmetic operations.

• Depth Representation: The main folder is represented by depth 0, and each subfolder increases the depth by 1.

• Navigational Operations: Moving into a subfolder increases depth, moving up decreases it, and staying in the same folder doesn't change the depth.

• Boundary Handling: The solution prevents going above the main folder by using the max() function.

1. Initialization:

   folder_depth = 0

   We initialize folder_depth to 0, representing the starting position in the main folder. This variable will keep track of our current depth throughout the navigation process.

2. Iterating Through Operations:

   for operation in logs:

   We iterate through each operation in the logs list. This allows us to process each folder change sequentially, simulating the user's navigation.

3. Handling "Stay in the Same Folder" Operation:

   if operation == "./": continue

   When we encounter "./", we simply continue to the next operation without changing folder_depth. This efficiently handles the "stay in the same folder" operation without unnecessary computations.

4. Handling "Move Up" Operation:

   if operation == "../": folder_depth = max(0, folder_depth - 1)

   For "../", we decrease folder_depth by 1, but use max() to ensure it never goes below 0. This elegantly handles moving up from the main folder without requiring an additional conditional check.

5. Handling "Move to Subfolder" Operation:

   else: folder_depth += 1

   For any other operation (which represents moving into a subfolder), we increment folder_depth by 1. This assumes all other operations are valid "x/" commands as per the problem statement.

6. Result Calculation/Return:

   return folder_depth

   After processing all operations, folder_depth represents the current depth from the main folder. This is exactly the number of "../" operations needed to return to the main folder, so we return it directly.

Time Complexity:

• , where is the number of operations in the logs list. This is because we iterate through each operation exactly once, performing constant-time operations for each.

Space Complexity:

• , as we only use a single integer variable folder_depth regardless of the input size. This solution doesn't require any additional data structures that grow with the input size.

Input:

Step-by-Step Walkthrough:

1. Initialization:

   • We start by initializing folder_depth = 0, representing that we're in the main folder.

2. Main Loop (Processing each operation):

   • Iteration 1:

     • Operation: 'd1/'

     • This is a move to child folder operation.

     • Action: Increment folder_depth from 0 to 1.

     • We are now one level deep in the folder structure.

   • Iteration 2:

     • Operation: 'd2/'

     • Another move to child folder operation.

     • Action: Increment folder_depth from 1 to 2.

     • We are now two levels deep.

   • Iteration 3:

     • Operation: './'

     • This operation means stay in the same folder.

     • Action: No change to folder_depth, it remains 2.

   • Iteration 4:

     • Operation: 'd3/'

     • Another move to child folder operation.

     • Action: Increment folder_depth from 2 to 3.

     • We are now three levels deep.

   • Iteration 5:

     • Operation: '../'

     • This is a move to parent folder operation.

     • Action: Decrement folder_depth, but ensure it doesn't go below 0.

     • Calculation: max(0, 3 - 1) = 2

     • folder_depth is updated from 3 to 2.

   • Iteration 6:

     • Operation: 'd31/'

     • Another move to child folder operation.

     • Action: Increment folder_depth from 2 to 3.

     • We end up three levels deep again.

3. Visual Aid:

   Operation #	Command	Resulting Depth
   1	d1/	1
   2	d2/	2
   3	./	2
   4	d3/	3
   5	../	2
   6	d31/	3

4. Result Calculation:

   • After processing all operations, our final folder_depth is 3.

   • This means we are three levels deep in the folder structure.

   • To return to the main folder, we need to perform three '../' operations.

The function returns 3, which represents the minimum number of '../' operations needed to return to the main folder from our current position.

The central concept of this solution is to leverage a stack data structure to simulate the folder hierarchy. This approach provides a more intuitive representation of the file system structure.

• Stack Representation: Each element in the stack represents a subfolder, with the bottom of the stack being the main folder.

• Navigational Operations: Moving into a subfolder pushes it onto the stack, moving up pops the top folder, and staying in the same folder does nothing to the stack.

• Depth Calculation: The depth of the current folder is directly represented by the size of the stack.

1. Initialization:

   folder_stack = []

   We initialize an empty list folder_stack to represent our folder structure. The empty stack corresponds to being in the main folder.

2. Iterating Through Operations:

   for operation in logs:

   We iterate through each operation in the logs list, processing them sequentially to simulate the user's navigation.

3. Handling "Stay in the Same Folder" Operation:

   if operation == "./": continue

   When we encounter "./", we simply continue to the next operation without modifying the stack. This efficiently handles the "stay in the same folder" operation without unnecessary computations.

4. Handling "Move Up" Operation:

   elif operation == "../": if folder_stack: folder_stack.pop()

   For "../", we check if the stack is not empty (ensuring we're not in the main folder) and then pop the top element. This simulates moving up one level in the folder hierarchy. The check prevents attempts to move above the main folder.

5. Handling "Move to Subfolder" Operation:

   else: folder_stack.append(operation)

   For any other operation (which represents moving into a subfolder), we append it to the stack. This simulates entering a new subfolder by adding it to our current path.

6. Result Calculation/Return:

   return len(folder_stack)

   After processing all operations, the length of folder_stack represents the current depth from the main folder. This is exactly the number of "../" operations needed to return to the main folder, so we return it directly.

Time Complexity:

• , where is the number of operations in the logs list. This is because we iterate through each operation exactly once, and each operation (push, pop, or continue) is performed in constant time.

Space Complexity:

• in the worst case, where is the number of operations in the logs list. This occurs when all operations are "move to subfolder" operations, causing the stack to grow to the size of the input. In the best case (all operations are "../" or "./"), the space complexity would be .

Step-by-Step Walkthrough:

1. Initialization:

   • We start by initializing folder_stack = [], an empty list representing that we're in the main folder.

2. Main Loop (Processing each operation):

   • Iteration 1:

     • Operation: 'd1/'

     • This is a move to child folder operation.

     • Action: Append 'd1/' to folder_stack.

     • Current stack: ['d1/']

   • Iteration 2:

     • Operation: 'd2/'

     • Another move to child folder operation.

     • Action: Append 'd2/' to folder_stack.

     • Current stack: ['d1/', 'd2/']

   • Iteration 3:

     • Operation: './'

     • This operation means stay in the same folder.

     • Action: No change to folder_stack.

     • Current stack remains: ['d1/', 'd2/']

   • Iteration 4:

     • Operation: 'd3/'

     • Another move to child folder operation.

     • Action: Append 'd3/' to folder_stack.

     • Current stack: ['d1/', 'd2/', 'd3/']

   • Iteration 5:

     • Operation: '../'

     • This is a move to parent folder operation.

     • Action: Remove the last item from folder_stack (pop 'd3/').

     • Current stack: ['d1/', 'd2/']

   • Iteration 6:

     • Operation: 'd31/'

     • Another move to child folder operation.

     • Action: Append 'd31/' to folder_stack.

     • Final stack: ['d1/', 'd2/', 'd31/']

3. Visual Aid:

   The following image illustrates the stack representation after each operation in the example:

   

   Iteration Summary:

   Operation #	Command	Stack Depth	Current Stack
   1	d1/	1	['d1/']
   2	d2/	2	['d1/', 'd2/']
   3	./	2	['d1/', 'd2/']
   4	d3/	3	['d1/', 'd2/', 'd3/']
   5	../	2	['d1/', 'd2/']
   6	d31/	3	['d1/', 'd2/', 'd31/']

4. Result Calculation:

   • After processing all operations, our final folder_stack is ['d1/', 'd2/', 'd31/'].

   • The length of this stack is 3.

   • This means we are three levels deep in the folder structure.

   • To return to the main folder, we need to perform three '../' operations, one for each folder in the stack.

The function returns 3, which represents the minimum number of '../' operations needed to return to the main folder from our current position. This is equivalent to the length of the folder_stack at the end of processing all operations.

The central concept of this solution is to leverage a stack data structure to handle nested parentheses and perform in-place reversals. This approach allows us to process the string from left to right, reversing substrings within parentheses as we encounter them.

• Stack Usage: The stack is used to store characters and handle nested parentheses efficiently.

• Immediate Reversal: Substrings within parentheses are reversed as soon as a closing parenthesis is encountered, ensuring proper handling of nested structures.

• In-place Processing: The solution modifies the stack contents directly, avoiding the need for additional data structures for each nested level.

1. Initialization:

   stack = []

   We initialize an empty list stack to serve as our stack data structure. This stack will store characters from the input string and handle the reversal process.

2. Iterating Through the String:

   for char in s:

   We iterate through each character in the input string s. This allows us to process the string from left to right, handling each character individually.

3. Handling Closing Parenthesis:

   if char == ')': reversed_substring = [] while stack and stack[-1] != '(': reversed_substring.append(stack.pop())

   When we encounter a closing parenthesis, we start the reversal process. We create a temporary list reversed_substring to store the characters we'll reverse. We then pop characters from the stack and append them to reversed_substring until we find an opening parenthesis. This effectively reverses the order of characters within the parentheses.

4. Removing Opening Parenthesis:

   if stack: stack.pop()

   After extracting the substring to be reversed, we remove the opening parenthesis from the stack. This step ensures that the parentheses themselves are not included in the final result.

5. Adding Reversed Substring Back to Stack:

   stack.extend(reversed_substring)

   We add the reversed substring back to the stack. This step completes the reversal process for the current pair of parentheses.

6. Handling Other Characters:

   else: stack.append(char)

   For any character that is not a closing parenthesis (including opening parentheses and regular characters), we simply append it to the stack.

7. Result Calculation/Return:

   return "".join(stack)

   After processing all characters, the stack contains the final result with all parentheses removed and substrings appropriately reversed. We join the characters in the stack to form the final string and return it.

Time Complexity:

• , where is the length of the input string. This is because in the worst case (deeply nested parentheses), we might need to reverse and re-add almost the entire string for each closing parenthesis. For example, in a string like "(((...)))", each closing parenthesis might require reversing almost the entire preceding substring.

Space Complexity:

• , where is the length of the input string. This is because in the worst case (no parentheses or all opening parentheses), the stack might need to store all characters of the input string. The reversed_substring list is temporary and never grows larger than the stack, so it doesn't affect the overall space complexity.

Input:

Step-by-Step Walkthrough:

1. Initialization:

   • An empty stack stack = [] is created to store characters and handle parentheses.

2. Main Loop (Iterating through the input string):

   • Iteration 1 ('('):

     • The opening parenthesis is appended to the stack.

     • Current stack: ['(']

   • Iterations 2-3 ('e', 'l'):

     • Characters 'e' and 'l' are appended to the stack.

     • Current stack: ['(', 'e', 'l']

   • Iterations 4-8 ('(', 'a', '(', 'p', 'm'):

     • These characters, including nested parentheses, are appended to the stack.

     • Current stack: ['(', 'e', 'l', '(', 'a', '(', 'p', 'm']

   • Iteration 9 (')'):

     • Closing parenthesis encountered. The algorithm starts reversing the innermost substring.

     • 'm' and 'p' are popped from the stack and stored in reversed_substring.

     • The opening parenthesis '(' is removed.

     • reversed_substring ['m', 'p'] is added back to the stack.

     • Current stack: ['(', 'e', 'l', '(', 'a', 'm', 'p']

   • Iteration 10 (')'):

     • Another closing parenthesis. The algorithm reverses the next substring.

     • 'p', 'm', and 'a' are popped and stored in reversed_substring.

     • The opening parenthesis '(' is removed.

     • reversed_substring ['p', 'm', 'a'] is added back to the stack.

     • Current stack: ['(', 'e', 'l', 'p', 'm', 'a']

   • Iterations 11-12 ('x', 'e'):

     • Characters 'x' and 'e' are appended to the stack.

     • Current stack: ['(', 'e', 'l', 'p', 'm', 'a', 'x', 'e']

   • Iteration 13 (')'):

     • Final closing parenthesis. The algorithm reverses the entire remaining substring.

     • All characters except the first '(' are popped and stored in reversed_substring.

     • The opening parenthesis '(' is removed.

     • reversed_substring ['e', 'x', 'a', 'm', 'p', 'l', 'e'] becomes the final stack content.

3. Loop Termination:

   • The loop terminates after processing all characters in the input string.

   • Final stack: ['e', 'x', 'a', 'm', 'p', 'l', 'e']

4. Visual Aid:

   Here's a summary of the stack's state after each iteration:

   Step	Character	Stack Content
   1	(	(
   2	e	(e
   3	l	(el
   4	(	(el(
   5	a	(el(a
   6	(	(el(a(
   7	p	(el(a(p
   8	m	(el(a(pm
   9	)	(el(amp
   10	)	(elpma
   11	x	(elpmax
   12	e	(elpmaxe
   13	)	example

   Here's an image illustrating the stack's state after each iteration, including the reversal process where:

   1. Light blue cells represent characters added to the stack normally.

   2. Red cells indicate opening parentheses that will be removed.

   3. Green cells show characters that are about to be reversed.

   4. Dark blue cells represent reversed substrings that have been added back to the stack.

   

   The visualization demonstrates how the algorithm handles nested parentheses:

   • It first processes the innermost pair (pm), reversing it to 'mp'.

   • Then it handles the next level (amp), reversing it to 'pma'.

   • Finally, it reverses the entire remaining string, resulting in 'example'.

5. Result Calculation/Final Steps:

   • The algorithm joins all characters in the final stack to form the result string.

   • Final Result: "example"

The central concept of this solution is to leverage a two-pass approach combined with direction reversal to handle nested parentheses efficiently. This method avoids explicit string reversals by cleverly using directional traversal.

• Parentheses Pairing: The first pass pairs up matching parentheses, storing their indices for quick access.

• Directional Traversal: The second pass builds the result by traversing the string, changing the direction at parentheses to implicitly reverse substrings.

• Index Jumping: Instead of pushing and popping from a stack, this approach uses index jumps to navigate between matching parentheses.

1. Initialization:

   parentheses_pairs = {} opening_parentheses = []

   We initialize an empty dictionary parentheses_pairs to store the indices of matching parentheses, and a list opening_parentheses to keep track of opening parentheses indices during the first pass.

2. First Pass - Pairing Parentheses:

   for index, char in enumerate(s): if char == '(': opening_parentheses.append(index) elif char == ')': if opening_parentheses: opening_index = opening_parentheses.pop() parentheses_pairs[index] = opening_index parentheses_pairs[opening_index] = index

   We iterate through the string, using a stack-like approach to pair up matching parentheses. When we encounter an opening parenthesis, we add its index to opening_parentheses. For a closing parenthesis, we pop the last opening parenthesis index and create a two-way mapping in parentheses_pairs. This allows us to quickly jump between matching parentheses in the second pass.

3. Second Pass Initialization:

   result = [] current_index = 0 direction = 1 # 1 for forward, -1 for backward

   We initialize an empty list result to build our output string, current_index to keep track of our position in the string, and direction to determine whether we're moving forward or backward through the string.

4. Second Pass - Building the Result:

   while current_index < len(s): if s[current_index] in '()': current_index = parentheses_pairs[current_index] direction = -direction else: result.append(s[current_index]) current_index += direction

   We traverse the string, changing direction when we hit a parenthesis. If the current character is a parenthesis, we jump to its matching pair and reverse direction. Otherwise, we add the character to our result. This clever technique implicitly reverses substrings within parentheses without actually reversing any part of the string.

5. Result Calculation/Return:

   return ''.join(result)

   After processing all characters, we join the characters in the result list to form the final string and return it.

Time Complexity:

• , where is the length of the input string. This is because we make two passes through the string, and in each pass, we process each character exactly once. Even though we change the direction in the second pass, we still visit each character no more than twice in total.

Space Complexity:

• , where is the length of the input string. This is because in the worst case (where the string consists entirely of paired parentheses), the parentheses_pairs dictionary would store pairs of indices. The result list will also store up to characters. The opening_parentheses list in the first pass could also store up to indices in the worst case, but it's cleared before the second pass begins.

Input:

Step-by-Step Walkthrough:

1. Initialization:

   • Two empty data structures are created:

     • parentheses_pairs = {}: A dictionary to store matching parentheses indices.

     • opening_parentheses = []: A list to keep track of opening parentheses indices.

2. First Pass: Pairing Parentheses

   • Iteration 1 ('('):

     • Opening parenthesis at index 0 is added to opening_parentheses.

     • Current opening_parentheses: [0]

   • Iterations 2-3 ('e', 'l'):

     • Non-parenthesis characters are skipped.

   • Iteration 4 ('('):

     • Opening parenthesis at index 3 is added to opening_parentheses.

     • Current opening_parentheses: [0, 3]

   • Iteration 5 ('a'):

     • Non-parenthesis character is skipped.

   • Iteration 6 ('('):

     • Opening parenthesis at index 5 is added to opening_parentheses.

     • Current opening_parentheses: [0, 3, 5]

   • Iterations 7-8 ('p', 'm'):

     • Non-parenthesis characters are skipped.

   • Iteration 9 (')'):

     • Closing parenthesis found. Paired with the last opening parenthesis.

     • Popped index 5 from opening_parentheses.

     • Added pair to parentheses_pairs: {8: 5, 5: 8}

   • Iteration 10 (')'):

     • Another closing parenthesis. Paired with the next opening parenthesis.

     • Popped index 3 from opening_parentheses.

     • Updated parentheses_pairs: {8: 5, 5: 8, 9: 3, 3: 9}

   • Iterations 11-12 ('x', 'e'):

     • Non-parenthesis characters are skipped.

   • Iteration 13 (')'):

     • Final closing parenthesis. Paired with the first opening parenthesis.

     • Popped index 0 from opening_parentheses.

     • Final parentheses_pairs: {8: 5, 5: 8, 9: 3, 3: 9, 12: 0, 0: 12}

   • Final Pairing:

     Index	Matching Index
     8	5
     5	8
     9	3
     3	9
     12	0
     0	12

3. Second Pass: Building Result

   • Initialize result = [], current_index = 0, direction = 1

   • Step 1:

     • At index 0 ('('): Jump to matching parenthesis at index 12, reverse direction.

     • Updated: result = "", current_index = 11, direction = -1

   • Steps 2-3:

     • Process 'e' and 'x' backwards, appending to result.

     • After step 3: result = "ex", current_index = 9, direction = -1

   • Step 4:

     • At index 9 (')'): Jump to matching parenthesis at index 3, reverse direction.

     • Updated: result = "ex", current_index = 4, direction = 1

   • Step 5:

     • Process 'a' forward, appending to result.

     • Updated: result = "exa", current_index = 5, direction = 1

   • Step 6:

     • At index 5 ('('): Jump to matching parenthesis at index 8, reverse direction.

     • Updated: result = "exa", current_index = 7, direction = -1

   • Steps 7-8:

     • Process 'm' and 'p' backwards, appending to result.

     • After step 8: result = "examp", current_index = 5, direction = -1

   • Steps 9-10:

     • Navigate through parentheses, changing directions.

     • After step 10: result = "examp", current_index = 2, direction = -1

   • Steps 11-12:

     • Process 'l' and 'e' backwards, appending to result.

     • After step 12: result = "example", current_index = 0, direction = -1

   • Step 13:

     • At index 0 ('('): Jump to matching parenthesis at index 12, reverse direction.

     • Loop terminates as current_index = 13 is out of string bounds.

4. Visual Aid:

   The attached image serves a dual-purpose in explaining the reverseParentheses2 algorithm:

   First Pass: Parentheses Pairing

   • The color-coded parentheses demonstrate how the algorithm identifies and pairs matching brackets:

     • Green cells (indices 0 and 12) represent the outermost pair of parentheses.

     • Yellow cells (indices 3 and 9) show the second level of nested parentheses.

     • Blue cells (indices 5 and 8) indicate the innermost pair of parentheses.

   • This color-coding visualizes the creation of the parentheses_pairs dictionary, with each color representing a key-value pair.

   

   Second Pass: Result Building

   • The same image can be used to trace the algorithm's path as it builds the result:

     1. Start at index 0 (green '('): Jump to its pair at index 12, reverse direction to -1.

     2. Moving backwards, append 'e' (index 11) and 'x' (index 10) to the result.

     3. Reach yellow ')' at index 9: Jump to its pair at index 3, reverse direction to +1.

     4. Moving forward, append 'a' (index 4).

     5. Encounter blue '(' at index 5: Jump to its pair at index 8, reverse direction to -1.

     6. Moving backwards, append 'm' (index 7) and 'p' (index 6).

     7. Hit blue '(' at index 5 again: Jump back to index 8, reverse to +1.

     8. Immediately encounter yellow ')' at index 9: Jump to index 3, reverse to -1.

     9. Moving backwards, append 'l' (index 2) and 'e' (index 1).

     10. Reach green '(' at index 0: Jump to index 12, reverse to +1.

     11. Terminate as index 13 is out of bounds.

5. Result Calculation/Final Steps:

   • The algorithm has built the result string character by character during the second pass.

   • Final Result: "example"

The central concept of this solution is to leverage a stack-based approach combined with a greedy strategy to maximize point gain from substring removals. The solution processes the string twice, prioritizing the higher-value substring removal first.

• Greedy Prioritization: The algorithm first determines which substring ('ab' or 'ba') yields higher points and processes that substring first. This ensures that the more valuable removals are prioritized.

• Stack-based Removal: For each substring type, the function uses a stack to efficiently identify and remove matching substrings in a single pass through the string.

• Two-Pass Approach: The string is processed twice, once for each substring type, ensuring all possible removals are made, even if removing one type of substring reveals opportunities for the other type.

1. Initialization and Prioritization:

   high_value_seq, low_value_seq = ('ab', x), ('ba', y) if x < y: high_value_seq, low_value_seq = low_value_seq, high_value_seq total_points = 0

   The code starts by determining which substring has higher value and sets up tuples to represent these sequences along with their point values. This prioritization is crucial for the greedy approach.

2. Iterative Processing:

   for (first_char, second_char), points in [high_value_seq, low_value_seq]:

   The function iterates twice, first for the high-value sequence and then for the low-value sequence. This ensures that we always prioritize the more valuable removals while still capturing all possible point gains.

3. Stack-based Removal:

   stack = [] for char in s: if stack and stack[-1] == first_char and char == second_char: stack.pop() total_points += points else: stack.append(char)

   For each character in the string, the algorithm checks if it forms a target substring with the top of the stack. If so, it removes both characters and adds points. Otherwise, it adds the character to the stack. This efficient approach allows for substring removal in a single pass.

4. Preparing for the Next Iteration:

   s = stack

   After processing one substring type, the remaining characters become the input for the next iteration. This step is crucial as it allows the algorithm to find new opportunities for removal that might have been created by the previous pass.

5. Result Calculation/Return:

   return total_points

   The function returns the total points accumulated from all substring removals, representing the maximum points that can be gained.

Time Complexity:

• , where is the length of the input string. This is because the function processes the string twice in the worst case, once for each substring type. Each character is processed at most twice, resulting in a linear time complexity.

Space Complexity:

• , where is the length of the input string. In the worst case, where no removals are possible, the stack would store all characters of the input string. Therefore, the space required grows linearly with the input size.

Input:

Step-by-Step Walkthrough:

1. Initialization:

   • The function starts by determining which substring has a higher value:

     • high_value_seq = ('ba', 5) (since y > x)

     • low_value_seq = ('ab', 4)

   • total_points is initialized to 0

2. Main Loop (Processing high-value sequence 'ba'):

   • Iteration 1-6:

     • Characters 'c', 'd', 'b', 'c', 'b', 'b' are pushed onto the stack.

     • Stack after six iterations: ['c', 'd', 'b', 'c', 'b', 'b']

   • Iteration 7:

     • Current character: 'a'

     • Forms 'ba' with top of stack

     • Pop 'b' from stack, add 5 points

     • total_points = 5

     • Stack: ['c', 'd', 'b', 'c', 'b']

   • Iteration 8:

     • Current character: 'a'

     • Forms another 'ba' with top of stack

     • Pop 'b' from stack, add 5 points

     • total_points = 10

     • Stack: ['c', 'd', 'b', 'c']

   • Iteration 9-10:

     • Push 'a' and 'b' onto stack

     • Stack: ['c', 'd', 'b', 'c', 'a', 'b']

   • Iteration 11:

     • Current character: 'a'

     • Forms 'ba' with top of stack

     • Pop 'b' from stack, add 5 points

     • total_points = 15

     • Stack: ['c', 'd', 'b', 'c', 'a']

   • Iteration 12:

     • Push 'b' onto stack

     • Final stack after the high-value sequence: ['c', 'd', 'b', 'c', 'a', 'b']

3. Second Loop (Processing low-value sequence 'ab'):

   • The remaining string 'cdbcab' is processed for 'ab' sequences

   • Iterations 1-5:

     • Characters 'c', 'd', 'b', 'c', 'a' are pushed onto the new stack

   • Iteration 6:

     • Current character: 'b'

     • Forms 'ab' with top of stack

     • Pop 'a' from stack, add 4 points

     • total_points = 19

     • Final stack: ['c', 'd', 'b', 'c']

4. Visual Aid:

   These images illustrate the stack's state during each iteration of the maximumGain1 function, showing how it processes the input string "cdbcbbaaabab":

   1. Dark blue cells represent characters newly pushed onto the stack.

   2. Light blue cells show characters already on the stack from previous iterations.

   3. Red cells indicate characters that form a high-value 'ba' or low-value 'ab' pair and will be removed.

   4. The total points accumulated are shown for each iteration.

   The visualization demonstrates how the function prioritizes removing 'ba' pairs first, then processes the remaining string for 'ab' pairs.

   

   

   Here's a summary of the stack changes and point accumulation for each substring type: High-Value Sequence 'ba':

   Iteration	Current Char	Stack After	Points Added	Total Points
   1-6	c,d,b,c,b,b	cdbcbb	0	0
   7	a	cdbcb	5	5
   8	a	cdbc	5	10
   9-10	a,b	cdbcab	0	10
   11	a	cdbca	5	15
   12	b	cdbcab	0	15

   Low-Value Sequence 'ab':

   Iteration	Current Char	Stack After	Points Added	Total Points
   1-5	c,d,b,c,a	cdbca	0	15
   6	b	cdbc	4	19

5. Result Calculation/Final Steps:

   • After processing both high-value and low-value sequences, the final total_points is 19.

   • This represents the maximum points that can be gained by removing 'ba' and 'ab' substrings from the input string.

The central concept of this solution is to use a single-pass, greedy approach with character counting to maximize point gain from substring removals. This method processes the string once, making immediate decisions about high-value pairs while deferring low-value pair decisions.

• Greedy Prioritization: The algorithm prioritizes removing the higher-value substring whenever possible, ensuring optimal point accumulation.

• Character Counting: Instead of using a stack, this approach keeps track of unpaired characters, allowing for efficient pair identification and removal.

• Deferred Processing: Low-value pair removals are deferred until necessary, ensuring that no high-value pair opportunities are missed.

1. Initialization and Prioritization:

   high_value_char, low_value_char = 'a', 'b' if x < y: x, y = y, x high_value_char, low_value_char = low_value_char, high_value_char total_points = 0 unpaired_high_count, unpaired_low_count = 0, 0

   The code determines which character pair has a higher value and sets up variables to track points and unpaired character counts. This setup is crucial for the greedy approach and efficient counting.

2. Single-Pass Processing:

   for char in s:

   The function processes the string in a single pass, making decisions for each character as it's encountered. This is a key efficiency improvement over the two-pass approach.

3. High-Value Character Handling:

   if char == high_value_char: unpaired_high_count += 1 elif char == low_value_char: if unpaired_high_count > 0: unpaired_high_count -= 1 total_points += x else: unpaired_low_count += 1

   When a high-value character is encountered, it's counted. For a low-value character, the algorithm immediately pairs it with a high-value character if available, adding points. Otherwise, it's counted as unpaired.

4. Resetting Counts on Non-Pair Characters:

   elif unpaired_high_count: if unpaired_low_count: total_points += min(unpaired_high_count, unpaired_low_count) * y unpaired_high_count, unpaired_low_count = 0, 0 else: unpaired_high_count = 0 elif unpaired_low_count: unpaired_low_count = 0

   When a character that's not part of any pair is encountered, the algorithm processes any accumulated low-value pairs and resets counts. This ensures that pairs are only formed within continuous sequences of relevant characters.

5. Final Processing:

   if unpaired_high_count and unpaired_low_count: total_points += min(unpaired_high_count, unpaired_low_count) * y return total_points

   After processing all characters, any remaining low-value pairs are accounted for, and the total points are returned.

• , where is the length of the input string. This is because the function processes each character in the string exactly once in a single pass. Each operation within the loop (counting, point calculation) is constant time, resulting in linear overall time complexity.

Space Complexity:

• , as the algorithm uses only a constant amount of extra space regardless of the input size. It maintains a fixed number of variables (counters and point totals) that do not grow with the input size.

Input:

Step-by-Step Walkthrough:

1. Initialization:

   • Since y > x, we swap the high and low value characters:

     • high_value_char = 'b', with points = 5

     • low_value_char = 'a', with points = 4

   • total_points, unpaired_high_count, and unpaired_low_count are all initialized to 0

2. Main Loop (Processing characters one by one):

   • Iterations 1-2:

     • Characters 'c' and 'd' are non-pair characters.

     • No action is taken, counts remain unchanged.

   • Iteration 3:

     • Current character: 'b' (high-value char)

     • Increment unpaired_high_count to 1

     • total_points remains 0

   • Iteration 4:

     • Current character: 'c' (non-pair char)

     • Reset unpaired_high_count to 0 as no matching low-value char was found

     • total_points remains 0

   • Iterations 5-6:

     • Current characters: 'b', 'b' (high-value chars)

     • Increment unpaired_high_count to 2

     • total_points remains 0

   • Iteration 7:

     • Current character: 'a' (low-value char)

     • Form a high-value pair 'ba'

     • Decrement unpaired_high_count to 1

     • Add 5 points, total_points = 5

   • Iteration 8:

     • Current character: 'a' (low-value char)

     • Form another high-value pair 'ba'

     • Decrement unpaired_high_count to 0

     • Add 5 points, total_points = 10

   • Iteration 9:

     • Current character: 'a' (low-value char)

     • No matching high-value char, increment unpaired_low_count to 1

     • total_points remains 10

   • Iteration 10:

     • Current character: 'b' (high-value char)

     • Increment unpaired_high_count to 1

     • total_points remains 10

   • Iteration 11:

     • Current character: 'a' (low-value char)

     • Form a high-value pair 'ba'

     • Decrement unpaired_high_count to 0

     • Add 5 points, total_points = 15

   • Iteration 12:

     • Current character: 'b' (high-value char)

     • Increment unpaired_high_count to 1

     • unpaired_low_count remains 1

     • total_points remains 15

3. Final Check for Remaining Pairs:

   • Both unpaired_high_count and unpaired_low_count are 1

   • Form a low-value pair 'ab'

   • Add 4 points, total_points = 19

4. Visual Aid:

   Here's a summary of the character processing and point accumulation:

   Iteration	Current Char	Unpaired High Count	Unpaired Low Count	Total Points
   1	c	0	0	0
   2	d	0	0	0
   3	b	1	0	0
   4	c	0	0	0
   5	b	1	0	0
   6	b	2	0	0
   7	a	1	0	5
   8	a	0	0	10
   9	a	0	1	10
   10	b	1	1	10
   11	a	0	1	15
   12	b	1	1	15
   Final	N/A	1	1	19

5. Result Calculation/Final Steps:

   • After processing all characters and checking for remaining pairs, the final total_points is 19.

   • This represents the maximum points that can be gained by removing 'ba' and 'ab' substrings from the input string.

The central concept of this solution is to leverage a sorting-based approach combined with a stack data structure to efficiently simulate robot collisions on a line. This approach allows for handling collisions in the correct order while minimizing unnecessary comparisons.

• Sorting by Position: The solution first sorts the robots by their positions, ensuring that collisions are processed in the correct spatial order.

• Stack for Right-moving Robots: A stack is used to keep track of right-moving robots, which allows for efficient collision detection with left-moving robots.

• Collision Simulation: The algorithm simulates collisions by iterating through the sorted robots and handling collisions between left-moving robots and any right-moving robots they encounter.

1. Sorting Robot Indices:

   sorted_robot_indices = sorted(range(len(positions)), key=lambda x: positions[x])

   This line creates a list of indices sorted based on the robots' positions. By sorting indices instead of the actual data, we maintain the original order of robots for the final output while still being able to process them in order of position.

2. Stack Initialization:

   right_moving_robot_stack = []

   This stack will store the indices of right-moving robots. It's used to efficiently handle collisions with left-moving robots.

3. Collision Simulation Loop:

   for current_robot_index in sorted_robot_indices: if directions[current_robot_index] == 'R': right_moving_robot_stack.append(current_robot_index) else: while (right_moving_robot_stack and healths[current_robot_index] > 0): colliding_robot_index = right_moving_robot_stack[-1] handle_collision(colliding_robot_index, current_robot_index)

   This loop iterates through the sorted robot indices. For right-moving robots, it simply adds them to the stack. For left-moving robots, it checks for collisions with right-moving robots on the stack and handles them using the handle_collision function. The main idea here is to process collisions in the correct order while avoiding unnecessary comparisons.

4. Collision Handling:

   def handle_collision(left_robot_index, right_robot_index): health_diff = healths[left_robot_index] - healths[right_robot_index] if health_diff < 0: healths[left_robot_index] = 0 healths[right_robot_index] -= 1 right_moving_robot_stack.pop() elif health_diff > 0: healths[right_robot_index] = 0 healths[left_robot_index] -= 1 else: healths[left_robot_index], healths[right_robot_index] = 0, 0 right_moving_robot_stack.pop()

   This function handles the collision between two robots. It compares their health, updates them accordingly, and removes robots from the simulation if their health reaches 0. Here we handle three cases:

   • The left robot has lower health: Remove the left robot (health set to 0) and decrease the left robot's health by 1 since it survives. Since the left robot is on the stack, it's also removed from the simulation.

   • The right robot has lower health: Remove the right robot (health set to 0) and decrease the left robot's health by 1. Since the left robot survives, it stays in the stack.

   • Both robots have the same health: Remove both robots (healths set to 0). Since the left robot is on the stack, it's also removed.

5. Result Calculation:

   surviving_healths = [health for health in healths if health > 0] return surviving_healths

   Finally, the function creates a list of surviving robot healths by filtering out robots with zero health, and returns this list.

Time Complexity:

• , where is the number of robots. This is because:

  1. Sorting the robot indices takes time.

  2. The main loop iterates through all robots once, which is .

  3. Each robot can be involved in at most one collision, so the total number of collision handlings is .

  4. The final filtering of surviving robots is .

The dominant factor is the initial sorting, hence the overall time complexity is .

Space Complexity:

• , where is the number of robots. This is because:

  1. We store the sorted indices of robots, which takes space.

  2. The stack of right-moving robots can, in the worst case, contain all robots, taking space.

  3. The final list of surviving healths is at most size .

All other variables use constant space, so the overall space complexity is .

Input:

Step-by-Step Walkthrough:

1. Initialization:

   • The function starts by printing the input parameters:

     • positions = [3, 5, 2, 6]

     • healths = [10, 12, 15, 12]

     • directions = "RLRL"

   • It then initializes key variables:

     • sorted_robot_indices = [2, 0, 1, 3]: This is the order in which robots will be processed, sorted by their positions.

     • right_moving_robot_stack = []: An empty stack to keep track of robots moving right.

2. Main Loop (Robot Collision Simulation):

   • Iteration 1 (Robot 3):

     • Position: 2, Health: 15, Direction: R

     • This robot is moving right, so it's added to the right_moving_robot_stack.

     • Updated right_moving_robot_stack: [2]

   • Iteration 2 (Robot 1):

     • Position: 3, Health: 10, Direction: R

     • This robot is also moving right, so it's added to the right_moving_robot_stack.

     • Updated right_moving_robot_stack: [2, 0]

   • Iteration 3 (Robot 2):

     • Position: 5, Health: 12, Direction: L

     • This robot is moving left, so we check for collisions with right-moving robots.

     • First collision with Robot 1 (Last right-moving robot):

       • Robot 1's health (10) is less than Robot 2's health (12).

       • Robot 1 is destroyed (health becomes 0).

       • Robot 2's health decreases by 1 to 11.

       • Robot 1 is removed from the right_moving_robot_stack.

     • Second collision with Robot 3 (the only remaining right-moving robot):

       • Robot 3's health (15) is greater than Robot 2's health (11).

       • Robot 2 is destroyed (health becomes 0).

       • Robot 3's health decreases by 1 to 14.

   • Iteration 4 (Robot 4):

     • Position: 6, Health: 12, Direction: L

     • This robot is moving left, so we check for collisions.

     • It collides with Robot 3 (the last remaining robot in the right_moving_robot_stack).

     • Collision handling:

       • Robot 3's health (14) is greater than Robot 4's health (12).

       • Robot 4 is destroyed (health becomes 0).

       • Robot 3's health decreases by 1 to 13.

3. Loop Termination:

   • The loop terminates after processing all robots in sorted_robot_indices.

   • Final state of right_moving_robot_stack: [2] (only Robot 3 remains)

4. Visual Aids: Here's a summary table of the iterations:

   Iteration	Robot	Position	Health	Direction	Right-moving Stack
   1	3	2	15	R	[2]
   2	1	3	10	R	[2, 0]
   3	2	5	0	L	[2]
   4	4	6	0	L	[2]

5. Result Calculation/Final Steps:

   • The function creates a list of surviving robot healths:

     • It iterates through the healths list and includes only non-zero values.

   • In this case, only Robot 3 survived with health of 13.

   • The final result is [13], which is then returned by the function.

The central concept of this solution is to leverage a stack-based approach to handle nested formulas within parentheses, combined with a defaultdict for efficient atom counting. This technique allows for parsing complex chemical formulas with nested structures in a single pass through the input string.

• Stack for Nested Formulas: The solution uses a stack of defaultdicts to keep track of atom counts at different nesting levels. This allows for handling nested formulas within parentheses efficiently.

• Single-Pass Parsing: The algorithm processes the formula string character by character, handling three main cases: opening parentheses, closing parentheses, and atoms. This approach allows for efficient parsing without the need for multiple passes.

• Dynamic Counting: By using a defaultdict, the solution can dynamically add new atoms and update their counts without explicitly checking for existence first.

1. Initialization:

   formula_length = len(formula) atoms_stack = [defaultdict(int)] index = 0

   The code initializes the length of the formula, creates a stack with an initial empty defaultdict for atom counts, and sets up an index to traverse the formula string.

2. Main Parsing Loop:

   while index < formula_length:

   This loop iterates through each character in the formula string, processing it based on its type.

3. Handling Opening Parenthesis:

   if formula[index] == '(': # Start of a new nested formula atoms_stack.append(defaultdict(int)) index += 1

   When an opening parenthesis is encountered, a new defaultdict is added to the stack to start counting atoms in a new nested level.

4. Handling Closing Parenthesis:

   elif formula[index] == ')': # End of a nested formula index += 1 start_index = index while index < formula_length and formula[index].isdigit(): index += 1 count_multiplier = int(formula[start_index:index] or 1) # Multiply and merge nested atom counts with parent formula nested_atom_counts = atoms_stack.pop() for atom, count in nested_atom_counts.items(): atoms_stack[-1][atom] += count * count_multiplier

   When a closing parenthesis is encountered, the code extracts any following number as a multiplier for the nested formula. If no number follows, it defaults to 1. The code then pops the nested atom counts from the stack, and adds these counts (multiplied by the multiplier) to the parent level's counts.

5. Handling Atoms:

   elif formula[index].isupper(): # Start of a new atom start_index = index index += 1 while index < formula_length and formula[index].islower(): index += 1 atom_name = formula[start_index:index] # Parse the count for this atom start_index = index while index < formula_length and formula[index].isdigit(): index += 1 count = int(formula[start_index:index] or 1) # Add atom count to current formula level atoms_stack[-1][atom_name] += count

   When an uppercase letter is encountered (indicating the start of an atom), the code extracts the full atom name and its count (If no digits follow, the count defaults to 1). Then, it adds this count to the current level's atom counts.

6. Result Calculation/Return:

   # Format the final result formatted_atom_counts = "".join(atom + (str(count) if count > 1 else "") for atom, count in sorted(atoms_stack[0].items())) return formatted_atom_counts

   The final result is calculated by sorting the atoms in the outermost defaultdict (representing the entire formula) and formatting each atom with its count if greater than 1.

Time Complexity:

• , where is the length of the formula string. The main loop iterates through each character once, which is . However, when processing closing parentheses, we may need to update counts for all atoms in the nested formula, which could be up to operations. In the worst case of deeply nested formulas, this could happen times, leading to a quadratic time complexity.

Space Complexity:

• , where is the length of the formula string. In the worst case of deeply nested formulas, the stack could grow to size . Each defaultdict in the stack stores at most 26 entries (one for each uppercase letter), so the space complexity remains linear with respect to the input size.

Input:

Step-by-Step Walkthrough:

1. Initialization:

   • Set formula_length = 15 (length of the input string)

   • Initialize atoms_stack = [defaultdict(int)] with one empty dictionary

2. Main Algorithm Process:

   • Iteration 1 (index 0):

     • Character: 'K' (uppercase)

     • Parse atom name: 'K'

     • Parse count: '32'

     • Update atoms_stack[-1]['K'] = 32: [{K: 32}]

   • Iteration 2 (index 3):

     • Character: '(' (opening parenthesis)

     • Append new defaultdict(int) to atoms_stack: [{K: 32}, {}]

   • Iteration 3 (index 4):

     • Character: 'O' (uppercase)

     • Parse atom name: 'O'

     • No explicit count, so use 1

     • Update atoms_stack[-1]['O'] = 1: [{K: 32}, {O: 1}]

   • Iteration 4 (index 5):

     • Character: 'N' (uppercase)

     • Parse atom name: 'N'

     • No explicit count, so use 1

     • Update atoms_stack[-1]['N'] = 1: [{K: 32}, {O: 1, N: 1}]

   • Iteration 5 (index 6):

     • Character: '(' (opening parenthesis)

     • Append new defaultdict(int) to atoms_stack: [{K: 32}, {O: 1, N: 1}, {}]

   • Iteration 6 (index 7):

     • Character: 'S' (uppercase)

     • Parse atom name: 'S'

     • No explicit count, so use 1

     • Update atoms_stack[-1]['S'] = 1: [{K: 32}, {O: 1, N: 1}, {S: 1}]

   • Iteration 7 (index 8):

     • Character: 'O' (uppercase)

     • Parse atom name: 'O'

     • Parse count: '3'

     • Update atoms_stack[-1]['O'] = 3: [{K: 32}, {O: 1, N: 1}, {S: 1, O: 3}]

   • Iteration 8 (index 10):

     • Character: ')' (closing parenthesis)

     • Parse count: '10'

     • Pop last dictionary from atoms_stack: {S: 1, O: 3}

     • Multiply counts: {'S': 1 * 10, 'O': 3 * 10}

     • Add to the previous level: atoms_stack[-1]['S'] += 10, atoms_stack[-1]['O'] += 30

     • Current state of atoms_stack: [{'K': 32}, {'O': 31, 'N': 1, 'S': 10}]

   • Iteration 9 (index 13):

     • Character: ')' (closing parenthesis)

     • Parse count: '2'

     • Pop last dictionary from atoms_stack: {'O': 31, 'N': 1, 'S': 10}

     • Multiply counts: {'O': 31 * 2, 'N': 1 * 2, 'S': 10 * 2}

     • Add to previous level: atoms_stack[-1]['O'] += 62, atoms_stack[-1]['N'] += 2, atoms_stack[-1]['S'] += 20

     • Current state of atoms_stack: [{'K': 32, 'O': 62, 'N': 2, 'S': 20}]

3. Loop Termination:

   • The loop terminates when index reaches formula_length (15)

   • Final state of atoms_stack[0]: {'K': 32, 'O': 62, 'N': 2, 'S': 20}

4. Visual Aid:

   Iteration	Current Char	Action	atoms_stack[-1] After Action
   1	'K'	Add K:32	{'K': 32}
   2	'('	New dict	{}
   3	'O'	Add O:1	{'O': 1}
   4	'N'	Add N:1	{'O': 1, 'N': 1}
   5	'('	New dict	{}
   6	'S'	Add S:1	{'S': 1}
   7	'O'	Add O:3	{'S': 1, 'O': 3}
   8	')'	Multiply by 10, merge	{'O': 31, 'N': 1, 'S': 10}
   9	')'	Multiply by 2, merge	{'K': 32, 'O': 62, 'N': 2, 'S': 20}

5. Result Calculation/Final Steps:

   • Sort the atoms alphabetically: K, N, O, S

   • For each atom, append its name and count (if > 1) to the result string

   • Final result: "K32N2O62S20"

The central concept of this solution is to leverage a two-pass parsing technique to efficiently count atoms in a complex chemical formula. This approach addresses the challenge of nested parentheses and varying multipliers by pre-calculating multipliers for each position in the formula.

• Two-Pass Strategy: The solution uses two passes through the formula string - a reverse pass to calculate multipliers, and a forward pass to count atoms.

• Multiplier Pre-calculation: By calculating multipliers in the first pass, the solution avoids repeated calculations and simplifies the atom counting process.

• Stack for Nested Structures: A stack is used to handle nested parentheses, allowing the solution to correctly apply multipliers at different levels of nesting.

• Efficient Atom Counting: The second pass uses the pre-calculated multipliers to count atoms accurately, even in complex nested structures.

1. Initialization:

   formula_length = len(formula) multiplier_stack = [1] multiplier_at_index = [1] * formula_length current_multiplier = 1 current_count_str = ""

   This section initializes key variables:

   • formula_length: Stores the length of the input formula for efficient access.

   • multiplier_stack: A stack to handle nested structures, initialized with 1 for the base level.

   • multiplier_at_index: A list to store pre-calculated multipliers for each character position.

   • current_multiplier: Keeps track of the current multiplier value as we parse the formula.

   • current_count_str: A string to accumulate digits when parsing numbers.

2. First Pass: Multiplier Calculation (Part 1):

   # First pass: Calculate multipliers index = formula_length - 1 while index >= 0: current_char = formula[index] if current_char.isdigit(): current_count_str += current_char elif current_char.isalpha(): current_count_str = ""

   This begins the reverse iteration through the formula:

   • We start from the end of the string (formula_length - 1) and move backwards.

   • For digits, we accumulate them in current_count_str (in reverse order).

   • For letters, we reset current_count_str, as we've finished parsing a number.

3. First Pass: Multiplier Calculation (Part 2):

   elif current_char == ')': # Start of a nested formula (reading right to left) new_multiplier = int(current_count_str[::-1] or 1) multiplier_stack.append(new_multiplier) current_multiplier *= new_multiplier current_count_str = "" elif current_char == '(': # End of a nested formula (reading right to left) current_multiplier //= multiplier_stack.pop()

   This section handles parentheses:

   • For a closing parenthesis ')', we're entering a new nested level:

     • Convert the accumulated digits to an integer (reversing the string first).

     • Push this new multiplier onto the stack and update current_multiplier.

   • For an opening parenthesis '(', we're exiting a nested level:

     • Divide current_multiplier by the top value from the stack.

4. First Pass: Multiplier Calculation (Part 3):

   # Store the current multiplier for each character multiplier_at_index[index] = current_multiplier index -= 1

   For each character, we store the current multiplier and move to the previous character.

5. Second Pass: Atom Counting (Part 1):

   # Second pass: Count atoms atom_map = defaultdict(int) index = 0 while index < formula_length: if formula[index].isupper(): # Start of a new atom start_index = index index += 1 if index < formula_length and formula[index].islower(): index += 1 atom_name = formula[start_index:index]

   This begins the forward pass to count atoms:

   • We use a defaultdict to store atom counts.

   • We identify atoms by their uppercase start and optional lowercase continuation.

6. Second Pass: Atom Counting (Part 2):

   # Get atom count start_index = index while index < formula_length and formula[index].isdigit(): index += 1 atom_count = int(formula[start_index:index] or 1) # Apply pre-calculated multiplier to atom count atom_map[atom_name] += atom_count * multiplier_at_index[index - 1] else: index += 1

   For each atom:

   • We parse its count (defaulting to 1 if no count is specified).

   • We multiply this count by the pre-calculated multiplier for its position.

   • We add this to the total count for the atom in our atom_map.

7. Result Formatting:

   # Format the final result formatted_atom_counts = "".join(atom + (str(count) if count > 1 else "") for atom, count in sorted(atom_map.items())) return formatted_atom_counts

   Finally, we format the result:

   • We sort the atoms alphabetically.

   • We create a string where each atom is followed by its count (if greater than 1).

   • We return this formatted string as the final result.

Time Complexity:

• , where is the length of the formula string, and is the number of unique atoms.

  • The two passes through the string (multiplier calculation and atom counting) each take time.

  • Sorting the unique atoms takes time.

  • In the worst case where is proportional to (e.g., each character is a unique atom), the time complexity becomes .

Space Complexity:

• , where is the length of the formula string.

  • The multiplier_at_index list requires space to store a multiplier for each character.

  • The atom_map dictionary can grow up to size in the worst case where each character is a unique atom.

  • The multiplier_stack can also grow up to size in cases of deeply nested parentheses.

Input:

Step-by-Step Walkthrough:

1. Initialization:

   • Set formula_length = 15 (length of the input string)

   • Initialize multiplier_stack = [1]

   • Initialize multiplier_at_index = [1] * 15 (list of 15 ones)

   • Set current_multiplier = 1

   • Set current_count_str = ""

2. First Pass: Calculate multipliers

   • Start from the end of the formula (index 14) and move left

   • Iteration 1 (index 14):

     • Character: '2'

     • Update current_count_str to "2"

     • Store current_multiplier (1) at multiplier_at_index[14]

   • Iteration 2 (index 13):

     • Character: ')'

     • Calculate new multiplier: 2

     • Update multiplier_stack to [1, 2]

     • Update current_multiplier to 1 * 2 = 2

     • Reset current_count_str to ""

     • Store current_multiplier (2) at multiplier_at_index[13]

   • Iterations 3-4 (index 12–11):

     • Characters: "01"

     • Update current_count_str to "10"

     • Store current_multiplier (2) at both indices

   • Iteration 5 (index 10):

     • Character: ')'

     • Calculate new multiplier: 10

     • Update multiplier_stack to [1, 2, 10]

     • Update current_multiplier to 2 * 10 = 20

     • Reset current_count_str to ""

     • Store current_multiplier (20) at multiplier_at_index[10]

   • Iterations 6-8 (index 9-7):

     • Characters: "3OS"

     • Process digits and reset for alphabets

     • Store current_multiplier (20) at all indices

   • Iteration 9 (index 6):

     • Character: '('

     • Update current_multiplier to 20 // 10 = 2

     • Update multiplier_stack to [1, 2]

     • Store current_multiplier (2) at multiplier_at_index[6]

   • Iterations 10-11 (index 5-4):

     • Characters: "NO"

     • Reset current_count_str for alphabets

     • Store current_multiplier (2) at both indices

   • Iteration 12 (index 3):

     • Character: '('

     • Update current_multiplier to 2 // 2 = 1

     • Update multiplier_stack to [1]

     • Store current_multiplier (1) at multiplier_at_index[3]

   • Iterations 13-15 (index 2–0):

     • Characters: "32K"

     • Process digits and reset for alphabet

     • Store current_multiplier (1) at all indices

   First Pass Summary:

   multiplier_at_index = [1, 1, 1, 1, 2, 2, 2, 20, 20, 20, 20, 2, 2, 2, 1]

3. Second Pass: Count atoms

   • Initialize atom_map = defaultdict(int)

   • Start from the beginning of the formula (index 0) and move right

   • Iteration 1 (index 0-2):

     • Atom: 'K'

     • Count: 32

     • Multiplier: 1

     • Update atom_map['K'] = 32 * 1 = 32

   • Iteration 2 (index 4):

     • Atom: 'O'

     • Count: 1 (implicit)

     • Multiplier: 2

     • Update atom_map['O'] = 1 * 2 = 2

     • Current state of atom_map: {'K': 32, 'O': 2}

   • Iteration 3 (index 5):

     • Atom: 'N'

     • Count: 1 (implicit)

     • Multiplier: 2

     • Update atom_map['N'] = 1 * 2 = 2

     • Current state of atom_map: {'K': 32, 'O': 2, 'N': 2}

   • Iteration 4 (index 7):

     • Atom: 'S'

     • Count: 1 (implicit)

     • Multiplier: 20

     • Update atom_map['S'] = 1 * 20 = 20

     • Current state of atom_map: {'K': 32, 'O': 2, 'N': 2, 'S': 20}

   • Iteration 5 (index 8-9):

     • Atom: 'O'

     • Count: 3

     • Multiplier: 20

     • Update atom_map['O'] += 3 * 20 = 60

     • Current state of atom_map: {'K': 32, 'O': 62, 'N': 2, 'S': 20}

   Second Pass Summary:

   atom_map = {'K': 32, 'O': 62, 'N': 2, 'S': 20}

4. Result Calculation/Final Steps:

   • Sort the atoms alphabetically: K, N, O, S

   • For each atom, append its name and count (if > 1) to the result string

   • Final result: "K32N2O62S20"

The central concept of this solution is to leverage regular expression (regex) tokenization combined with reverse processing to efficiently count atoms in a complex chemical formula. This approach addresses the challenge of nested parentheses and varying multipliers by tokenizing the formula and then processing these tokens in reverse order.

• Regex Tokenization: The solution uses a regular expression to break down the formula into meaningful tokens ( atoms, parentheses, and their associated counts).

• Reverse Processing: By processing the tokens from right to left, the solution simplifies the handling of nested structures and multipliers.

• Stack for Nested Structures: A stack is used to manage multipliers for nested parentheses, allowing the solution to correctly apply multipliers at different levels of nesting.

• Efficient Atom Counting: The reverse processing allows for immediate application of multipliers to atom counts, simplifying the counting process.

1. Initialization:

   multiplier_stack = [1] current_multiplier = 1 atom_map = defaultdict(int)

   This section initializes key variables:

   • multiplier_stack: A stack to handle nested structures, initialized with 1 for the base level.

   • current_multiplier: Keeps track of the current multiplier value as we process tokens.

   • atom_map: A defaultdict to store the count of each atom.

2. Regex Pattern Definition and Tokenization:

   # Regex pattern to match atoms, parentheses, and their counts pattern = r"([A-Z][a-z]*)(\d*)|(\()|(\))(\d*)" formula_components = re.findall(pattern, formula) # Process from right to left for correct nested parentheses handling formula_components.reverse()

   This section defines the regex pattern and tokenizes the formula:

   • The pattern matches atoms (uppercase followed by optional lowercase), their counts, and parentheses with their multipliers.

     • ([A-Z][a-z]*)(\d*): Matches atoms and their counts.

     • (\()|(\))(\d*): Matches opening and closing parentheses with their multipliers.

   • re.findall() is used to find all matches in the formula.

   • The resulting list is reversed for right-to-left processing.

3. Token Processing Loop:

   for atom, count, left_parenthesis, right_parenthesis, multiplicity\ in formula_components:

   This loop iterates through the tokenized and reversed formula components. Each iteration unpacks five possible parts: atom, its count, left parenthesis, right parenthesis, and multiplicity.

4. Atom Handling:

   if atom: # Handle atom and its count atom_count = int(count) if count else 1 atom_map[atom] += current_multiplier * atom_count

   When an atom is encountered:

   • Its count is determined (defaulting to 1 if not specified).

   • The atom's count is multiplied by the current multiplier and added to the atom_map.

5. Closing Parenthesis Handling:

   elif right_parenthesis: # Handle closing parenthesis and its multiplier multiplicity = int(multiplicity) if multiplicity else 1 multiplier_stack.append(multiplicity) current_multiplier *= multiplicity

   When a closing parenthesis is encountered (remember, we're processing in reverse):

   • Its multiplicity is determined (defaulting to 1 if not specified).

   • This multiplicity is pushed onto the stack and multiplied into the current multiplier.

6. Opening Parenthesis Handling:

   elif left_parenthesis: # Handle opening parenthesis (end of a group when processing in reverse) current_multiplier //= multiplier_stack.pop()

   When an opening parenthesis is encountered (end of a group in reverse processing):

   • The current multiplier is divided by the top value from the stack, effectively ending the nested multiplication.

7. Result Formatting:

   # Format the final result formatted_atom_counts = "".join(atom + (str(count) if count > 1 else "") for atom, count in sorted(atom_map.items())) return formatted_atom_counts

   Finally, the result is formatted:

   • Atoms are sorted alphabetically.

   • Each atom is followed by its count (if greater than 1).

   • The formatted string is returned as the final result.

Time Complexity:

• , where is the length of the formula string, and is the number of unique atoms.

  • The regex matching and reverse processing of tokens take time.

  • Sorting the unique atoms takes time.

  • In the worst case where is proportional to (e.g., each character is a unique atom), the time complexity becomes .

Space Complexity:

• , where is the length of the formula string.

  • The space is primarily used by the regex matches list (formula_components), which can contain up to elements.

  • The atom_map dictionary can grow up to size in the worst case where each character is a unique atom.

  • The multiplier_stack can also grow up to size in cases of deeply nested parentheses.

Input:

Step-by-Step Walkthrough:

1. Initialization:

   • Initialize multiplier_stack = [1]

   • Set current_multiplier = 1

   • Initialize atom_map = defaultdict(int)

   • Define regex pattern: r"([A-Z][a-z]*)(\d*)|(\()|(\))(\d*)"

2. Tokenization:

   • Apply regex to formula:

     formula_components = [ ('K', '32', '', '', ''), ('', '', '(', '', ''), ('O', '', '', '', ''), ('N', '', '', '', ''), ('', '', '(', '', ''), ('S', '', '', '', ''), ('O', '3', '', '', ''), ('', '', '', ')', '10'), ('', '', '', ')', '2') ]

   • Reverse formula_components

3. Process Components:

   • Start from the end of the formula and move left

   • Iteration 1:

     • Component: ('', '', '', ')', '2')

     • Right parenthesis found

     • Update multiplier_stack to [1, 2]

     • Update current_multiplier to 1 * 2 = 2

   • Iteration 2:

     • Component: ('', '', '', ')', '10')

     • Right parenthesis found

     • Update multiplier_stack to [1, 2, 10]

     • Update current_multiplier to 2 * 10 = 20

   • Iteration 3:

     • Component: ('O', '3', '', '', '')

     • Atom 'O' found with count 3

     • Update atom_map['O'] += 20 * 3 = 60

   • Iteration 4:

     • Component: ('S', '', '', '', '')

     • Atom 'S' found with implicit count 1

     • Update atom_map['S'] += 20 * 1 = 20

   • Iteration 5:

     • Component: ('', '', '(', '', '')

     • Left parenthesis found

     • Update current_multiplier to 20 // 10 = 2

     • Update multiplier_stack to [1, 2]

   • Iteration 6:

     • Component: ('N', '', '', '', '')

     • Atom 'N' found with implicit count 1

     • Update atom_map['N'] += 2 * 1 = 2

   • Iteration 7:

     • Component: ('O', '', '', '', '')

     • Atom 'O' found with implicit count 1

     • Update atom_map['O'] += 2 * 1 = 2

     • atom_map['O'] is now 62

   • Iteration 8:

     • Component: ('', '', '(', '', '')

     • Left parenthesis found

     • Update current_multiplier to 2 // 2 = 1

     • Update multiplier_stack to [1]

   • Iteration 9:

     • Component: ('K', '32', '', '', '')

     • Atom 'K' found with count 32

     • Update atom_map['K'] += 1 * 32 = 32

   Processing Summary:

   atom_map = {'O': 62, 'S': 20, 'N': 2, 'K': 32}

4. Result Calculation/Final Steps:

   • Sort the atoms alphabetically: K, N, O, S

   • For each atom, append its name and count (if > 1) to the result string

   • Final result: "K32N2O62S20"