June 2024, Week 4: June 17th - June 23rd

The core idea of this solution is to find if the given integer c can be decomposed into the sum of two perfect squares using a two-pointer approach. Here's the breakdown of the concept:

• Two-Pointer Approach:

  • The function initializes two pointers: start_index at 0 and end_index at the square root of c. These pointers represent potential values of 'a' and 'b'.

  • In each iteration, it calculates the squares_sum of the squares of these pointers.

  • If squares_sum equals c, it means we've found a pair of integers (a, b) whose squares add up to c.

  • If squares_sum is less than c, we increment start_index to increase the sum (since a^2 is the smaller term).

  • If squares_sum is greater than c, we decrement end_index to decrease the sum (since b^2 is the larger term).

• Mathematical Basis:

  • The algorithm is based on the fact that if a number c can be expressed as the sum of two squares, the numbers a and b must each be less than or equal to the square root of c. This ensures we only check relevant values.

1. Initialization:

   • start_index is set to 0, representing the smallest possible square.

   • end_index is set to the integer part of the square root of c, representing the largest possible square within the range.

2. Main Loop (Two-Pointer Search):

   • The while loop runs as long as start_index is less than or equal to end_index.

   • In each iteration:

     • squares_sum is calculated as start_index * start_index + end_index * end_index.

     • Decision Point (Conditional Statements):

       • If squares_sum == c, the function returns True (pair found).

       • If squares_sum < c, we increase start_index by 1.

       • If squares_sum > c, we decrease end_index by 1.

3. Result Calculation/Return:

   • If the loop completes without finding a match, the function returns False (no pair exists).

Time Complexity:

• where is the input integer. In the worst case, the while loop will run up to the square root of times.

Space Complexity:

• (Constant): The algorithm uses only a fixed number of variables (start_index, end_index, squares_sum), and this number doesn't grow with the input size.

Input: c = 98

Step-by-Step Walkthrough:

1. Initialization:

   • start_index is set to 0.

   • end_index is set to the integer part of the square root of 98, which is 9.

2. Main Loop (Two-Pointer Search):

   • While (Iteration 1):

     • squares_sum =

     • Since 81 < 98, we increment start_index to 1.

   • While (Iteration 2):

     • squares_sum =

     • Since 82 < 98, we increment start_index to 2.

   • While (Iteration 3):

     • squares_sum =

     • Since 85 < 98, we increment start_index to 3.

   • While (Iteration 4):

     • squares_sum =

     • Since 90 < 98, we increment start_index to 4.

   • While (Iteration 5):

     • squares_sum =

     • Since 97 < 98, we increment start_index to 5.

   • While (Iteration 6):

     • squares_sum =

     • Since 106 > 98, we decrement end_index to 8.

   • While (Iteration 7):

     • squares_sum =

     • Since 89 < 98, we increment start_index to 6.

   • While (Iteration 8):

     • squares_sum =

     • Since 100 > 98, we decrement end_index to 7.

   • While (Iteration 9):

     • squares_sum =

     • Since 85 < 98, we increment start_index to 7.

   • While (Iteration 10):

     • squares_sum =

     • Since 98 == 98, we have found a valid pair (7, 7), and the function returns True.

3. Loop Termination: The loop terminates when squares_sum == c, indicating a valid pair of integers has been found.

4. Iteration Summary:

   start_index	end_index	squares_sum	Comparison	Action
   0	9	81	81 < 98	start_index += 1
   1	9	82	82 < 98	start_index += 1
   2	9	85	85 < 98	start_index += 1
   3	9	90	90 < 98	start_index += 1
   4	9	97	97 < 98	start_index += 1
   5	9	106	106 > 98	end_index -= 1
   5	8	89	89 < 98	start_index += 1
   6	8	100	100 > 98	end_index -= 1
   6	7	85	85 < 98	start_index += 1
   7	7	98	98 == 98	Return True

5. Result Calculation/Final Steps:

   • The function directly returns True as soon as it finds the valid pair (7, 7). It does not need to calculate or return any additional values.

   • The pair (7, 7) satisfies the condition , confirming that 98 can be expressed as the sum of two squares.

The core idea of this solution is to leverage Fermat's Theorem on sums of two squares to determine if a given number can be expressed as such. Fermat's theorem states:

In simpler terms, a number can be expressed as the sum of two squares unless it has a prime factor of the form (such as 3, 7, 11, etc.) that appears an odd number of times in its prime factorization.

This is because a prime of the form cannot be expressed as the sum of two squares (as opposed to primes of the form which can always be expressed as the sum of two squares). When a prime of the form appears with an even exponent in the prime factorization of , its contribution to the overall sum of squares can be paired off and effectively neutralized. However, if such a prime appears with an odd exponent, it cannot be paired off, and its presence fundamentally prevents from being expressed as the sum of two squares. This is because the square of any number is congruent to 0 or 1 (mod 4), but never 3.

Another important concept is the Brahmagupta-Fibonacci identity, which states that the product of two sums of two squares is itself a sum of two squares. This property allows us to combine the squares of two numbers to form a new sum of squares:

This identity is crucial in the context of Fermat's theorem and the decomposition of numbers into sums of squares. When a prime factor of the form appears with an even exponent, it can be paired off to form a new sum of squares: , and this can be combined with other sums of squares to form a larger sum of squares. However, if appears with an odd exponent, it cannot be paired off. The number cannot be expressed as a sum of squares due to the presence of the lone term (which is congruent to 3 mod 4).

• Prime Factorization: The solution systematically checks for prime factors of up to its square root.

• Counting Divisibility: For each prime factor, it counts how many times it divides evenly (tracked by divisors_count).

• Checking Fermat's Condition: If a prime factor of the form has an odd divisors_count, the number cannot be a sum of squares (returns False).

• Base Case: After iterating through smaller primes, the remaining value of is either 1 or a prime. It checks if this remaining prime is of the form . If it is, it returns False; otherwise, it returns True.

1. Initialization:

   • index is initialized to 2, the smallest prime number.

2. Main Loop (Checking Prime Factors):

   • The while loop iterates as long as the square of index (potential prime factor) is less than or equal to c.

   • Inner Loop (Counting Divisors):

     • If c is divisible by index, a nested while loop repeatedly divides c by index to count its occurrences.

   • Decision Points (Fermat's Condition):

     • After the inner loop, it checks if divisors_count is odd (indicating an odd power) and if index is of the form 4k + 3. If both are true, the function returns False.

3. Base Case (After Main Loop):

   • If the main loop completes without returning, c is either 1 or a prime.

   • It checks if the remaining c is of the form 4k + 3. If so, it returns False; otherwise, True.

Time Complexity:

• . The main loop runs up to times. For each iteration, the inner loop might run up to times in the worst case (for a prime that divides c many times).

Space Complexity:

• (Constant): The algorithm uses a fixed number of variables (index, divisors_count, c), regardless of the input size.

Input: c = 98

Step-by-Step Walkthrough:

1. Initialization:

   • index is initialized to 2, the smallest prime number.

2. Main Loop (Checking Prime Factors):

   • While (Iteration 1):

     • index = 2

     • c is divisible by index (98 % 2 == 0).

     • Inner loop counts the divisors: divisors_count = 1, c = 49 (after division)

     • divisors_count is 1 and index (2) is not of the form 4k+3, so the loop continues.

     • index is incremented to 3.

   • While (Iteration 2):

     • index = 3

     • c is not divisible by index (49 % 3 != 0), so the loop continues.

     • index is incremented to 4.

   • While (Iteration 3):

     • index = 4

     • c is not divisible by index (49 % 4 != 0), so the loop continues.

     • index is incremented to 5.

   • While (Iteration 4):

     • index = 5

     • c is not divisible by index (49 % 5 != 0), so the loop continues.

     • index is incremented to 6.

   • While (Iteration 5):

     • index = 6

     • c is not divisible by index (49 % 6 != 0), so the loop continues.

     • index is incremented to 7.

   • While (Iteration 6):

     • index = 7

     • c is divisible by index (49 % 7 == 0).

     • Inner loop counts the divisors:

       • divisors_count = 1, c = 7

       • divisors_count = 2, c = 1

     • index (7) is of the form 4k+3, but since divisors_count is even (2), the loop continues.

     • index is incremented to 8.

3. Loop Termination: The loop terminates after iteration 6 because index * index (64) is not less than or equal to the current value of c (1).

4. Iteration Summary/Visual Aids:

   c	index	divisors_count	index % 4 == 3	Result
   108	2	1	False	Continue
   49	3	0	False	Continue
   49	4	0	False	Continue
   49	5	0	False	Continue
   49	6	0	False	Continue
   49	7	2	True	Continue (Even exponent)

5. Result Calculation/Final Steps:

   • After the loop, c is 1. Since 1 is not of the form 4k + 3, the function returns True, indicating that 98 can be expressed as the sum of two squares .

6. Visualizing the Pairing of Squares:

   • Since the factorization of 98 is , the prime factor 7 (of the form ) appears with an even exponent (2), allowing it to be paired off and expressed as a sum of squares.

   • This can be visualized using the Brahmagupta-Fibonacci identity to combine the squares of 7 into a new sum of squares:

     • We know that , and we can express as . With the factorization, we get the product of these two sums of squares:

       • .

     • Using the identity, we can combine these to get .

The core idea of this solution is to leverage a precomputed list of maximum profits achievable for each difficulty level up to the maximum ability of the workers. This allows a quick lookup of the best possible profit for any worker's ability, ensuring efficient assignment of workers to jobs.

• Precomputation of Maximum Profits: The solution first builds a list to store the maximum-profit achievable for each difficulty level up to the maximum worker ability.

• Accumulation of Profits: It ensures that for any difficulty level, the profit reflects the highest possible profit up to that level. This way, a worker can always find the best job they can perform.

• Efficient Assignment: By using the precomputed list, the solution quickly sums up the maximum possible profit for each worker based on their ability.

1. Initialization:

   • max_ability = max(worker): Finds the maximum ability among all workers. This is the highest difficulty level we need to consider.

   • max_profit_per_diff = [0] * (max_ability + 1): Creates a list to store the maximum-profit achievable for each difficulty level from 0 to max_ability. It's initialized with zeros.

2. Build Maximum Profit per Difficulty:

   • for diff, prof in zip(difficulty, profit): Iterates through each job's difficulty (diff) and profit (prof).

   • if diff <= max_ability: Checks if the current job's difficulty is within the range of worker abilities we need to consider.

   • max_profit_per_diff[diff] = max(max_profit_per_diff[diff], prof): Updates the maximum profit for the current job's difficulty level if the current job offers a higher profit.

3. Accumulate Maximum Profit:

   • for index in range(1, max_ability + 1): Iterates through each difficulty level from 1 to max_ability.

   • max_profit_per_diff[index] = max(max_profit_per_diff[index], max_profit_per_diff[index - 1]): Updates the maximum profit for the current difficulty level to be the maximum of either its current value or the maximum profit of the previous difficulty level. This ensures that even if a job has a lower difficulty, it can still be chosen if it offers a higher profit.

4. Result Calculation/Return:

   • sum(max_profit_per_diff[ability] for ability in worker): Iterates through each worker's ability and sums up their maximum achievable profits based on the precomputed max_profit_per_diff list.

Time Complexity:

• , where:

  • is the length of difficulty and profit (number of jobs)

  • is the length of worker (number of workers)

  • is the maximum value in worker

This is because the algorithm iterates over difficulty and profit once , over the range of possible abilities , and over worker once .

Space Complexity:

This is because the algorithm uses a list max_profit_per_diff to store maximum profits per difficulty level up to the maximum ability. The size of this list is directly proportional to max_ability.

Input:

Step-by-Step Walkthrough:

1. Initialization:

   • The function starts by finding the maximum ability of the workers: max_ability = max(worker), which is 12.

   • An array max_profit_per_diff of size max_ability + 1 is initialized to store the maximum profit for each difficulty level up to max_ability: max_profit_per_diff = [0] * (max_ability + 1).

2. Main Loop (Building Max Profit Array):

   • Iteration 1:

     • Current Job: (difficulty = 5, profit = 10)

     • Since 5 <= 12, update max_profit_per_diff[5]: 0 -> 10.

   • Iteration 2:

     • Current Job: (difficulty = 12, profit = 30)

     • Since 12 <= 12, update max_profit_per_diff[12]: 0 -> 30.

   • Iteration 3:

     • Current Job: (difficulty = 2, profit = 20)

     • Since 2 <= 12, update max_profit_per_diff[2]: 0 -> 20.

   • Iteration 4:

     • Current Job: (difficulty = 6, profit = 25)

     • Since 6 <= 12, update max_profit_per_diff[6]: 0 -> 25.

   • Iteration 5:

     • Current Job: (difficulty = 15, profit = 50)

     • Since 15 > 12, skip this job.

   • Iteration 6:

     • Current Job: (difficulty = 7, profit = 35)

     • Since 7 <= 12, update max_profit_per_diff[7]: 0 -> 35.

   • Iteration 7:

     • Current Job: (difficulty = 9, profit = 40)

     • Since 9 <= 12, update max_profit_per_diff[9]: 0 -> 40.

   Iteration Summary:

   max_profit_per_diff: [0, 0, 20, 0, 0, 10, 25, 35, 0, 40, 0, 0, 30]

3. Second Loop (Propagating Maximum Profits):

   • Iteration 1/12:

     • Here we update the maximum profit for each difficulty level based on the previous level's maximum profit.

     • Update max_profit_per_diff[1]: 0 -> 0.

   • Iteration 2/12:

     • Since the previous maximum profit is 0, the maximum profit for 2 remains 20.

     • Update max_profit_per_diff[2]: 20 -> 20.

   • Iteration 3/12:

     • Since the previous maximum profit is 20 and this level's profit is 0, the maximum profit for 3 gets updated to 20.

     • Update max_profit_per_diff[3]: 0 -> 20.

   • Iteration 4/12 - 12/12:

     • The loop continues, updating the maximum profit for each difficulty level based on the previous level's maximum profit.

   • The final updated max_profit_per_diff array is:

     max_profit_per_diff: [0, 0, 20, 20, 20, 20, 25, 35, 35, 40, 40, 40, 40]

4. Result Calculation/Final Steps:

   • Worker 1:

     • Ability: 10

     • Maximum Profit: 40

     • Current Total Profit: 0 + 40 = 40

   • Worker 2:

     • Ability: 5

     • Maximum Profit: 20

     • Current Total Profit: 40 + 20 = 60

   • Worker 3:

     • Ability: 7

     • Maximum Profit: 35

     • Current Total Profit: 60 + 35 = 95

   • Worker 4:

     • Ability: 12

     • Maximum Profit: 40

     • Current Total Profit: 95 + 40 = 135

   • Worker 5:

     • Ability: 8

     • Maximum Profit: 35

     • Current Total Profit: 135 + 35 = 170

   • The function returns the total maximum profit of 170.

The central concept of this solution is to leverage sorting and binary search to efficiently match workers to the most profitable jobs they can perform. The solution transforms the job list to reflect the highest profit available up to each difficulty level, enabling quick lookups.

• Sorting Jobs by Difficulty: The jobs are sorted by difficulty to facilitate binary search and cumulative-profit calculations.

• Cumulative-Profit Calculation: A transformation is applied to the sorted jobs to ensure each job entry reflects the highest profit up to that difficulty level.

• Binary Search for Worker Assignment: For each worker, a binary search is used to quickly find the highest-paying job they can perform.

1. Initialization:

   • jobs = sorted(zip(difficulty, profit)): Creates a list of tuples where each tuple contains the difficulty and profit of a job. It then sorts this list in ascending order based on the difficulty.

2. Job Transformation:

   • max_profit_so_far = 0: Initializes a variable to keep track of the maximum profit seen so far.

   • for index, job in enumerate(jobs): Iterates over the sorted jobs.

     • max_profit_so_far = max(max_profit_so_far, job[1]): Updates max_profit_so_far if the current job has a higher profit.

     • jobs[index] = (job[0], max_profit_so_far): Replaces the profit in the current job tuple with max_profit_so_far. This transformation ensures that for each job, the profit stored in the tuple represents the maximum profit achievable up to that job's difficulty level.

3. Profit Calculation:

   • total_profit = 0: Initializes the total profit earned by all workers.

   • for ability in worker: Iterates over the abilities of each worker.

     • index = bisect_right(jobs, (ability, float('inf'))): Performs a binary search to find the rightmost index in the jobs list where the tuple (ability, infinity) would be inserted while maintaining the sorted order. This essentially finds the index of the first job with a difficulty greater than the worker's ability.

     • if index > 0: Checks if the worker is qualified for at least one job.

       • total_profit += jobs[index - 1][1]: If the worker is qualified, add the profit of the highest-paying job they can do (located at index - 1) to the total profit.

4. Result Calculation/Return:

   • Returns the calculated total_profit.

Time Complexity:

• , where:

  • is the number of jobs

  • is the number of workers

This is because the algorithm:

• Sorts the jobs:

• Transforms the job list:

• Performs binary search for each worker:

Space Complexity:

This is because the algorithm uses additional space for the jobs list which has the same length as the input lists difficulty and profit.

Input:

Step-by-Step Walkthrough:

1. Initialization:

   • The jobs list is created by pairing corresponding difficulty and profit values and then sorting them in ascending order based on difficulty: [(2, 20), (5, 10), (6, 25), (7, 35), (9, 40), (12, 30), (15, 50)].

   • A variable max_profit_so_far is initialized to 0. It will track the highest profit encountered as the jobs are processed.

2. Main Loop (Transforming Jobs with Max Profit):

   • The loop iterates through each job in the sorted jobs list.

   • Iteration 1:

     • Job: (2, 20)

     • max_profit_so_far is updated to 20 (max of 0 and 20).

     • The job in jobs is replaced with (2, 20).

   • Iteration 2:

     • Job: (5, 10)

     • max_profit_so_far remains 20.

     • The job in jobs is replaced with (5, 20).

   • Iterations 3-7: Similar logic is applied to the remaining jobs. max_profit_so_far gets updated as needed, ensuring that the profit associated with each job represents the maximum profit achievable up to that difficulty level.

3. Iteration Summary (Transformed Jobs):

   • The provided table illustrates how the jobs list is transformed during the main loop. The "Max Profit So Far" column shows how this value is tracked and used to update the profit component of each job.

   Iteration	Original Job	Max Profit So Far	Updated Job
   1	(2, 20)	20	(2, 20)
   2	(5, 10)	20	(5, 20)
   3	(6, 25)	25	(6, 25)
   4	(7, 35)	35	(7, 35)
   5	(9, 40)	40	(9, 40)
   6	(12, 30)	40	(12, 40)
   7	(15, 50)	50	(15, 50)

4. Calculating Total Profit (Worker Loop):

   • The loop iterates through the worker list.

   • Iteration 1 (Worker with ability 10):

     • Binary search (bisect_right) is used to find the rightmost job in jobs with a difficulty less than or equal to the worker's ability (10). This leads to index 5.

     • Since the index is greater than 0, the profit of the job at index 4 is added to total_profit (40).

   • Iteration 2 (Worker with ability 5):

     • Binary search finds index 2.

     • The profit of the job at index 1 (20) is added to total_profit. The total profit is now 60.

   • Iteration 3 (Worker with ability 7):

     • Binary search finds index 4.

     • The profit of the job at index 3 (35) is added to total_profit. The total profit is now 95.

   • **Iteration 4

     • Binary search finds index 6.

     • The profit of the job at index 5 (40) is added to total_profit. The total profit is now 135.

   • **Iteration 5

     • Binary search finds index 4.

     • The profit of the job at index 3 (35) is added to total_profit. The total profit is now 170.

5. Final Result:

   • The function returns the calculated total_profit, which is 170 in this example. This represents the maximum achievable profit by assigning the workers to the available jobs.

The core idea of this solution is to use sorting and a two-pointer technique to efficiently assign the most profitable jobs to workers based on their abilities. The solution processes jobs and workers in a sorted order, ensuring that each worker is matched with the best possible job they can perform.

• Sorting Jobs and Workers: Both the jobs and workers are sorted to facilitate efficient traversal and profit calculation.

• Two-Pointer Technique: The solution uses a two-pointer technique to iterate through the sorted job list and keep track of the maximum profit available up to the current worker's ability. This technique ensures that the solution only processes each job and worker once, contributing to its efficiency.

• Accumulating Profits: By maintaining a running maximum of job profits, the solution ensures that each worker is assigned the best possible job they can perform, maximizing total profit.

1. Initialization:

   • jobs = sorted(zip(difficulty, profit)): Creates a list of tuples where each tuple contains the difficulty and profit of a job. It then sorts this list in ascending order based on the difficulty.

   • total_profit = 0: Initializes the total profit earned by all workers.

   • max_profit_so_far = 0: Initializes a variable to keep track of the maximum profit seen so far for any job within the current worker's ability.

   • index = 0: Initializes an index to keep track of the current job being considered.

2. Iterative Profit Calculation:

   • for ability in sorted(worker): Iterates over the abilities of each worker in ascending order.

     • while index < len(jobs) and jobs[index][0] <= ability: This loop continues as long as there are jobs left to consider, and the current job's difficulty is less than or equal to the worker's ability.

       • max_profit_so_far = max(max_profit_so_far, jobs[index][1]): Updates max_profit_so_far if the current job has a higher profit than any previously seen job within the worker's ability range.

       • index += 1: Moves on to the next job.

     • total_profit += max_profit_so_far: Adds the maximum profit achievable by the current worker to the total_profit.

3. Result Calculation/Return:

   • Returns the calculated total_profit.

Time Complexity:

• , where:

  • is the number of jobs

  • is the number of workers

This is because the algorithm:

• Sorts the jobs:

• Sorts the workers:

• Iterates over jobs and workers: While the nested loops might seem like , note that the index for jobs only increases and never goes back. So, in total, we iterate over each job and each worker only once, resulting in .

Since the sorting operations dominate, the overall time complexity is .

Space Complexity:

This is because the algorithm uses additional space for the jobs list which has the same length as the input lists difficulty and profit ( elements). The space complexity is dominated by the storage of the sorted job list, hence .

Input:

Step-by-Step Walkthrough:

1. Initialization:

   • The code combines difficulty and profit into pairs and sorts them by difficulty: jobs = [(2, 20), (5, 10), (6, 25), (7, 35), (9, 40), (12, 30), (15, 50)].

   • The worker array is also sorted: [5, 7, 8, 10, 12].

   • total_profit is initialized to 0.

   • max_profit_so_far (to track the highest profit seen so far) and index (to track the current position in jobs) are both set to 0.

2. Main Loop (Calculate Total Profit):

   • The loop iterates over the sorted worker array: worker = [5, 7, 8, 10, 12].

   • Iteration 1 (Worker with ability 5):

     • Current max_profit_so_far: 0

     • The while loop finds the jobs that the worker can do (difficulty <= 5).

       • Job (2, 20) is within the worker's ability. Update max_profit_so_far to 20.

       • Job (5, 10) is within the worker's ability. Keep max_profit_so_far at 20.

       • Job (6, 25) is not within the worker's ability (difficulty > 5). Exit the loop.

     • The worker's profit (20) is added to total_profit, making it 20.

   • Iteration 2 (Worker with ability 7):

     • Current max_profit_so_far: 20

     • The while loop finds the jobs that the worker can do (up to difficulty 7).

       • Job (6, 25) is within the worker's ability. Update max_profit_so_far to 25.

       • Job (7, 35) is within the worker's ability. Update max_profit_so_far to 35.

       • Job (9, 40) is not within the worker's ability (difficulty > 7). Exit the loop

     • The worker's profit (35) is added to total_profit, making it 55 (20 + 35).

   • Iteration 3 (Worker with ability 8):

     • Current max_profit_so_far: 35

     • The while loop finds the jobs that the worker can do (up to difficulty 8).

       • Since the worker's ability is 8 and the next job's difficulty is 9, the loop doesn't execute.

     • The worker's profit (35) is added to total_profit, making it 90 (55 + 35).

   • Iteration 4 (Worker with ability 10):

     • Current max_profit_so_far: 35

     • The while loop finds the jobs that the worker can do (up to difficulty 10).

       • Job (9, 40) is within the worker's ability. Update max_profit_so_far to 40.

       • Job (12, 30) is not within the worker's ability (difficulty > 10). Exit the loop.

     • The worker's profit (40) is added to total_profit, making it 130 (90 + 40).

   • Iteration 5 (Worker with ability 12):

     • Current max_profit_so_far: 40

     • The while loop finds the jobs that the worker can do (up to difficulty 12).

       • Job (12, 30) is within the worker's ability. Keep max_profit_so_far at 40.

       • Job (15, 50) is not within the worker's ability (difficulty > 12). Exit the loop.

     • The worker's profit (40) is added to total_profit, making it 170 (130 + 40).

3. Iteration Summary:

   • The table demonstrates how max_profit_so_far and total_profit change with each iteration of the loop.

   Iteration	Worker Ability	Max Profit So Far	Total Profit
   1	5	20	20
   2	7	35	55
   3	8	35	90
   4	10	40	130
   5	12	40	170

4. Result Calculation/Final Steps:

   • After the loop completes, the final value of total_profit (170) is returned, representing the maximum-profit achievable.

The solution leverages binary search to efficiently find the minimum number of days required to make m bouquets. It exploits the monotonic property of the problem: if it's possible to make m bouquets after waiting x days, it's also possible after waiting x + 1 days.

• Binary Search on Days: The search space is the range of possible days (from the minimum to the maximum bloom day). The solution repeatedly checks the midpoint of this range.

• Bouquet Formation Check: For each candidate day, the code simulates bouquet formation using adjacent flowers that have bloomed by that day.

• Search Space Reduction: If enough bouquets can be made, the search space is narrowed to earlier days. Otherwise, it's narrowed to later days.

1. Base Case Check: The code first checks if it's even possible to make m bouquets with k adjacent flowers each, given the total number of flowers. If not, it returns -1.

2. can_make_bouquets(day) Function:

   • This function simulates the garden on a given day.

   • It iterates through the bloom days of each flower.

   • If a flower has bloomed by the given day, it's added to a potential bouquet (flowers counter incremented).

   • Once k adjacent flowers are found, a bouquet is formed (bouquet_count incremented).

   • If m or more bouquets are formed, the function immediately returns True.

   • If a flower hasn't bloomed, the current potential bouquet is reset (flowers set to 0).

   • Finally, the function returns whether m or more bouquets were formed.

3. Binary Search Initialization:

   • left_index starts at the earliest bloom day (minimum of bloom_day).

   • right_index starts at the latest bloom day (maximum of bloom_day).

4. Binary Search Loop:

   • The loop continues as long as there is a range of days to explore (left_index < right_index).

   • mid_index is calculated as the middle of the current range.

   • The can_make_bouquets function checks if m bouquets can be formed by mid_index day.

   • If yes, the answer might be found earlier (or at mid_index), so right_index is set to mid_index.

   • If no, the answer must be later, so left_index is set to mid_index + 1.

5. Result:

   • When the loop ends, left_index holds the minimum day when m bouquets can be formed. This value is returned as the final result.

Time Complexity:

• , where n is the length of bloom_day and D is the difference between the maximum and minimum bloom days.

  • The binary search takes time to find the minimum number of days.

  • In each iteration, the can_make_bouquets function takes time to simulate the garden.

Space Complexity:

• , the solution uses a constant amount of extra space, regardless of the input size. This is because only a few variables are used to track the indices and counters.

Input: bloom_day = [3, 2, 4, 9, 10, 4, 3, 4], m = 3, k = 2

Step-by-Step Walkthrough:

1. Initialization:

   • The function first checks if it's possible to make three bouquets of size 2, given 8 flowers. Since 3 * 2 = 6 and 6 <= 8, it's possible, and the function proceeds.

   • The binary search range is initialized:

     • left_index = 2 (the minimum value in bloom_day)

     • right_index = 10 (the maximum value in bloom_day)

2. Main Loop (Binary Search):

   • Iteration 1:

     • mid_index = (2 + 10) // 2 = 6

     • can_make_bouquets(6) is called:

       • The function simulates forming bouquets with flowers that bloom by day 6.

       • It finds only two bouquets can be made. (See iteration summary table below)

         Flower	Bloom Day	Bloom Day <= 6	Action	Flowers	Bouquets (2 size)
         1	3	Yes	Increment Flowers	1	0
         2	2	Yes	Make Bouquet and Reset Flowers	2	1
         3	4	Yes	Increment Flowers	1	1
         4	9	No	Reset Flowers	0	1
         5	10	No	Reset Flowers	0	1
         6	4	Yes	Increment Flowers	1	1
         7	3	Yes	Make Bouquet and Reset Flowers	2	2
         8	4	Yes	Increment Flowers	1	2

       • Since this is less than m = 3, it returns False.

     • The binary search range is updated:

       • left_index = 7 (since we need to look at later bloom days)

   • Iteration 2:

     • mid_index = (7 + 10) // 2 = 8

     • can_make_bouquets(8) is called:

       • The function simulates forming bouquets with flowers that bloom by day 8.

       • It again finds only two bouquets can be made. (See the iteration summary table below)

         Flower	Bloom Day	Bloom Day <= 8	Action	Flowers	Bouquets (2 size)
         1	3	Yes	Increment Flowers	1	0
         2	2	Yes	Make Bouquet and Reset Flowers	2	1
         3	4	Yes	Increment Flowers	1	1
         4	9	No	Reset Flowers	0	1
         5	10	No	Reset Flowers	0	1
         6	4	Yes	Increment Flowers	1	1
         7	3	Yes	Make Bouquet and Reset Flowers	2	2
         8	4	Yes	Increment Flowers	1	2

       • Since this is less than m = 3, it returns False.

     • The binary search range is updated:

       • left_index = 9

   • Iteration 3:

     • mid_index = (9 + 10) // 2 = 9

     • can_make_bouquets(9) is called:

       • The function simulates forming bouquets with flowers that bloom by day 9.

       • This time, it successfully makes three bouquets. (See the iteration summary table below)

         Flower	Bloom Day	Bloom Day <= 9	Action	Flowers	Bouquets (2 size)
         1	3	Yes	Increment Flowers	1	0
         2	2	Yes	Make Bouquet and Reset Flowers	2	1
         3	4	Yes	Increment Flowers	1	1
         4	9	Yes	Make Bouquet and Reset Flowers	2	2
         5	10	No	Reset Flowers	0	2
         6	4	Yes	Increment Flowers	1	2
         7	3	Yes	Make Bouquet and Reset Flowers	2	3

       • It returns True.

     • The binary search range is updated:

       • right_index = 9

3. Loop Termination:

   • The loop terminates because left_index and right_index both become 9.

4. Binary Search Iteration Summary:

   Iteration	Left Index	Right Index	Mid Index	Can Make Bouquets?	Action
   1	2	10	6	No	left = mid + 1 (7)
   2	7	10	8	No	left = mid + 1 (9)
   3	9	10	9	Yes	right = mid (9)

5. Result Calculation/Final Steps:

   • The final value of left_index is 9, indicating that you need to wait 9 days to make three bouquets.

   • The function returns 9 as the result.

This solution builds upon the core idea of binary search like the previous one, but with an optimization focused on the search space. Instead of searching within the full range of bloom days (which could be large), it leverages the fact that only the unique bloom days matter.

• Binary Search on Unique Bloom Days: The search space is reduced to the sorted set of unique bloom days. This potentially drastically reduces the number of iterations in the binary search.

• Bouquet Formation Check: The can_make_bouquets function remains largely the same, but it now implicitly handles the gaps between bloom days by counting how many bouquets can be formed from the consecutive bloomed flowers.

• Optimized Search: The binary search operates on the indices of the sorted unique bloom days, making the comparisons more efficient.

1. Base Case Check: Identical to the previous solution, this checks if making m bouquets is even possible.

2. can_make_bouquets(day) Function:

   • The function's logic is slightly modified from the previous version.

   • Instead of resetting flowers to 0 whenever a bloom day is exceeded, it now adds the current flowers count ( divided by k) to bouquet_count. This efficiently accounts for all bouquets formed up to the current point.

   • The rest of the logic remains the same: check if enough bouquets are made and return the result.

3. Unique Bloom Days Preparation:

   • unique_bloom_days is created by:

     • Converting the bloom_day list to a set to get unique days.

     • Sorting the set to have a defined order for binary search.

4. Binary Search Initialization and Loop:

   • left_index and right_index are initialized for the indices of unique_bloom_days.

   • The loop and its logic remain almost the same as before, but now it operates on the indices of unique_bloom_days.

5. Result:

   • After the loop, unique_bloom_days[left_index] gives the minimum bloom day required.

Time Complexity:

• , where n is the length of bloom_day. This is an improvement over the previous in cases where there are many repeated bloom days.

  • Creating the set takes time, and sorting it takes , where u is the number of unique bloom days.

  • The binary search now takes iterations.

  • Each iteration of can_make_bouquets still takes time.

  • In total, we have .

Since , the overall time complexity simplifies to .

Space Complexity:

• in the worst case, to store the set of unique bloom days. This is potentially much better than the previous if there are many repeated values. If u is small compared to n, it can be considered .

Input: bloom_day = [3, 2, 4, 9, 10, 4, 3, 4], m = 3, k = 2

Step-by-Step Walkthrough:

1. Initialization:

   • The function first checks if it's possible to make three bouquets of size 2, given 8 flowers. Since and , it's possible, and the function proceeds.

   • The unique bloom days are extracted and sorted: unique_bloom_days = [2, 3, 4, 9, 10]

   • The binary search range is initialized to the indices of unique_bloom_days:

     • left_index = 0 (the first index)

     • right_index = 4 (the last index)

2. Main Loop (Binary Search):

   • Iteration 1:

     • mid_index = (0 + 4) // 2 = 2

     • can_make_bouquets(unique_bloom_days[2]), which is can_make_bouquets(4), is called:

       • The function simulates forming bouquets with flowers that bloom by day 4.

       • It finds that only two bouquets can be made. (See iteration summary table below)

         Flower	Bloom Day	Bloom Day <= 4	Action	Flowers	Bouquets (2 size)
         1	3	Yes	Increment Flowers	1	0
         2	2	Yes	Increment Flowers	2	0
         3	4	Yes	Increment Flowers	3	0
         4	9	No	Reset Flowers/Calculate Bouquets	0	1
         5	10	No	Reset Flowers/Calculate Bouquets	0	1
         6	4	Yes	Increment Flowers	1	1
         7	3	Yes	Increment Flowers	2	1
         8	4	Yes	Final Bouquet Calculation	3	2

       • Since this is less than m = 3, it returns False.

     • The binary search range is updated:

       • left_index = 3 (since we need to look at later bloom days)

   • Iteration 2:

     • mid_index = (3 + 4) // 2 = 3

     • can_make_bouquets(unique_bloom_days[3]), which is can_make_bouquets(9), is called:

       • The function simulates forming bouquets with flowers that bloom by day 9.

       • This time, it successfully makes three bouquets. (See iteration summary table below)

       Flower	Bloom Day	Bloom Day <= 9	Action	Flowers	Bouquets (2 size)
       1	3	Yes	Increment Flowers	1	0
       2	2	Yes	Increment Flowers	2	0
       3	4	Yes	Increment Flowers	3	0
       4	9	Yes	Increment Flowers	4	0
       5	10	No	Reset Flowers/Calculate Bouquets	0	2
       6	4	Yes	Increment Flowers	1	2
       7	3	Yes	Increment Flowers	2	2
       8	4	Yes	Final Bouquet Calculation	3	3

       • It returns True.

     • The binary search range is updated:

       • right_index = 3

3. Loop Termination:

   • The loop terminates because left_index and right_index both become 3.

4. Binary Search Iteration Summary:

   Iteration	Left Index	Right Index	Mid Index	Can Make Bouquets?	Action
   1	0	4	2	No	left = mid + 1 (3)
   2	3	4	3	Yes	right = mid (3)

5. Result Calculation/Final Steps:

   • The final value of left_index is 3.

   • The function returns unique_bloom_days[3], which is 9, indicating that you need to wait 9 days to make three bouquets.

The central concept of this solution is to leverage binary search to efficiently find the maximum possible minimum distance between any two balls. This approach works because the problem has a monotonic property: if we can place m balls with a minimum distance d, we can also place them with any distance smaller than d.

• Binary Search on Answer: Instead of checking all possible distances, we use binary search to narrow down the range of potential answers quickly.

• Greedy Ball Placement: For each potential minimum distance, we use a greedy approach to place balls as far apart as possible.

• Monotonicity: The ability to place balls at a given minimum distance implies we can place them at any smaller distance, allowing for binary search.

1. Helper Function (can_place_balls):

   • Initializes remaining_balls to m - 1, representing the number of balls left to place after the first.

   • Sets next_valid_position to the first position plus the min_distance.

   • Iterates through the remaining positions:

     • If the current position (pos) is greater than or equal to next_valid_position, a ball is placed, remaining_balls is decremented, and next_valid_position is updated.

     • If all balls are placed (remaining_balls becomes 0), it returns True.

   • Returns False if not all balls could be placed, True otherwise.

2. Initialization:

   • Sorts the position array in ascending order.

   • Initializes the search space with start_index = 1 (minimum possible force) and end_index = (position[-1] - position[0]) // (m - 1) (maximum possible average distance rounded down).

3. Main Loop (Binary Search):

   • While start_index < end_index:

     • Calculates the mid_index with a slight bias towards the right to ensure progress.

     • Calls can_place_balls with mid_index as the minimum distance.

     • If can_place_balls returns True:

       • Updates start_index to mid_index, as a larger minimum distance might be possible.

     • Otherwise:

       • Updates end_index to mid_index - 1, as the minimum distance must be smaller.

4. Result Calculation/Return:

   • After the loop ends, start_index will have converged to the maximum possible minimum distance, which is returned.

Time Complexity:

• , where is the number of positions and range is .

• This comes from two main operations:

  1. Sorting the positions:

  2. Binary search with ball placement check: iterations, each doing an check.

• Combined, this gives

Space Complexity:

• , where is the number of positions.

• This is due to the space required for sorting the position array, as Python's sort() method uses TimSort, which can require additional space in the worst case.

• The rest of the algorithm uses only a constant amount of extra space.

Input: position = [64, 16, 128, 8, 2, 32, 1, 4], m = 4

Step-by-Step Walkthrough:

1. Initialization:

   • The position list is sorted: [1, 2, 4, 8, 16, 32, 64, 128]

   • start_index is set to 1 (minimum possible distance)

   • end_index is set to (128 - 1) // (4 - 1) = 42 (maximum possible average distance)

2. Binary Search Loop:

   • While 1 < 42 (Iteration 1):

     • mid_index = 1 + (1 + 42) // 2 = 22

     • can_place_balls(22) returns True

       • Balls are placed at positions 1, 32, 64, 128

     • start_index is updated to 22

   • While 22 < 42 (Iteration 2):

     • mid_index = 1 + (22 + 42) // 2 = 33

     • can_place_balls(33) returns False

       • Only 3 balls can be placed (at 1, 64, 128)

     • end_index is updated to 32

   • Iteration 3:

     • mid_index = 1 + (22 + 32) // 2 = 28

     • can_place_balls(28) returns True

       • Balls are placed at positions 1, 32, 64, 128

     • start_index is updated to 28

   • Iteration 4:

     • mid_index = 1 + (28 + 32) // 2 = 31

     • can_place_balls(31) returns True

       • Balls are placed at positions 1, 32, 64, 128

     • start_index is updated to 31

   • Iteration 5:

     • mid_index = 1 + (31 + 32) // 2 = 32

     • can_place_balls(32) returns False

       • Only 3 balls can be placed (at 1, 64, 128)

     • end_index is updated to 31

3. Loop Termination:

   • The loop terminates when start_index (31) is no longer less than end_index (31)

4. Visual Aid: Iteration Summary Table

   Iteration	Start Index	End Index	Mid Index	Can Place
   1	1	42	22	True
   2	22	42	33	False
   3	22	32	28	True
   4	28	32	31	True
   5	31	32	32	False

5. Result Calculation/Final Steps:

   • The algorithm returns start_index, which is 31

   • This means the maximum possible minimum distance between any two balls is 31

The final result indicates that we can place four balls in the baskets such that the minimum distance between any two balls is 31, and this is the maximum possible minimum distance we can achieve. We can place them at positions 1, 32, 64, and 128. The distance between each position is 31, 32, and 64, respectively; hence, the minimum distance is 31.

The central concept of this solution is to leverage a sliding window technique to maximize customer satisfaction given the constraint of a limited period where the bookstore owner can suppress their grumpiness. The solution exploits the pattern in the data by separating normally satisfied customers from potentially satisfied customers during the owner's non-grumpy period.

• Two-part Satisfaction Calculation: The solution divides customer satisfaction into two parts: those naturally satisfied when the owner isn't grumpy, and additional customers that could be satisfied during the "non-grumpy" window.

• Sliding Window for Optimization: A sliding window of size minutes is used to find the optimal period for the owner to use their "non-grumpy" technique, maximizing additional customer satisfaction.

• Single-pass Approach: The solution cleverly combines the calculation of normally satisfied customers and the sliding window technique into a single pass through the data, optimizing for efficiency.

1. Initialization:

   satisfied_customers = 0 potential_customers = 0

   These variables track normally satisfied customers and potential additional satisfied customers in the current window.

2. Initial Window Calculation:

   for customer, grump in zip(customers[:minutes], grumpy[:minutes]): if grump: potential_customers += customer else: satisfied_customers += customer

   This loop calculates initial satisfaction for the first minutes window, separating normally satisfied from potential additional customers.

3. Maximum Potential Tracking:

   max_potential_customers = potential_customers

   Initializes the maximum potential additional satisfied customers with the first window's value.

4. Sliding Window Implementation:

   for minute in range(minutes, len(customers)):

   This loop slides the window through the remaining data, one customer at a time.

5. Window Update:

   if grumpy[minute]: potential_customers += customers[minute] else: satisfied_customers += customers[minute]

   Updates satisfaction counts for the new customer entering the window.

6. Window Tail Removal:

   if grumpy[minute - minutes]: potential_customers -= customers[minute - minutes]

   Removes the contribution of the customer leaving the window if they were part of the potential satisfaction count.

7. Maximum Update:

   max_potential_customers = max(max_potential_customers, potential_customers)

   Updates the maximum potential additional satisfied customers if the current window is better.

8. Result Calculation/Return:

   return satisfied_customers + max_potential_customers

   Combines normally satisfied customers with the maximum additional satisfied customers from the optimal "non-grumpy" window.

Time Complexity:

• ), where is the number of minutes (length of the customers list). This is because the algorithm makes a single pass through the data, performing constant-time operations at each step.

Space Complexity:

• , as the algorithm uses a fixed amount of additional space (a few variables) regardless of the input size. It doesn't create any data structures that scale with the input.

Input:

Step-by-Step Walkthrough:

1. Initialization:

   • satisfied_customers = 0

   • potential_customers = 0

2. Main Loop (Initial Window Calculation):

   Since minutes = 3, the initial window covers the first three minutes:

   • Iteration 1 (minute 0):

     • customer = 1, grump = 0

     • Owner is not grumpy, so add to satisfied_customers

     • satisfied_customers = 1, potential_customers = 0

   • Iteration 2 (minute 1):

     • customer = 0, grump = 1

     • Owner is grumpy, but no customers to add to potential_customers

     • satisfied_customers = 1, potential_customers = 0

   • Iteration 3 (minute 2):

     • customer = 1, grump = 0

     • Owner is not grumpy, so add to satisfied_customers

     • satisfied_customers = 2, potential_customers = 0

   • Initial Window Summary:

     [1, 0, 1, 2, 1, 1, 7, 5] Customers [0, 1, 0, 1, 0, 1, 0, 1] Grumpy [S P S] Window S: Satisfied = 2, P: Potential = 0, Max Potential = 0

3. Sliding Window Loop:

   • Iteration 4 (minute 3):

     • customer = 2, grump = 1

     • Owner is grumpy, add customers to potential_customers

       • potential_customers = 0 + 2 = 2

     • Current window: [1, 3]

     • Update max_potential_customers = 2

       [1, 0, 1, 2, 1, 1, 7, 5] Customers [0, 1, 0, 1, 0, 1, 0, 1] Grumpy [P S P] Window S: Satisfied = 2, P: Potential = 2, Max Potential = 2

   • Iteration 5 (minute 4):

     • customer = 1, grump = 0

     • Owner is not grumpy, add customer to satisfied_customers

       • satisfied_customers = 2 + 1 = 3

     • Current window: [2, 4]

     • No change to potential_customers

       [1, 0, 1, 2, 1, 1, 7, 5] Customers [0, 1, 0, 1, 0, 1, 0, 1] Grumpy [S P S] Window S: Satisfied = 3, P: Potential = 2, Max Potential = 2

   • Iteration 6 (minute 5):

     • customer = 1, grump = 1

     • Owner is grumpy, add customers to potential_customers

       • potential_customers = 2 + 1 = 3

     • Current window: [3, 5]

     • Update max_potential_customers = 3

       [1, 0, 1, 2, 1, 1, 7, 5] Customers [0, 1, 0, 1, 0, 1, 0, 1] Grumpy [P S P] Window S: Satisfied = 3, P: Potential = 3, Max Potential = 3

   • Iteration 7 (minute 6):

     • customer = 7, grump = 0

     • Owner is not grumpy, add to satisfied_customers

     • satisfied_customers = 10

     • Current window: [4, 6]

       • Remove 2 from potential_customers (from minute 3)

       • potential_customers = 3 - 2 = 1

       • The maximum potential customers remain at 3

       [1, 0, 1, 2, 1, 1, 7, 5] Customers [0, 1, 0, 1, 0, 1, 0, 1] Grumpy [S P S] Window S: Satisfied = 10, P: Potential = 1, Max Potential = 3

   • Iteration 8 (minute 7):

     • customer = 5, grump = 1

     • Owner is grumpy, add customers to potential_customers

     • potential_customers = 5 + 1 = 6

     • Current window: [5, 7]

     • Update max_potential_customers = 6

       [1, 0, 1, 2, 1, 1, 7, 5] Customers [0, 1, 0, 1, 0, 1, 0, 1] Grumpy [P S P] Window S: Satisfied = 10, P: Potential = 6, Max Potential = 6

4. Visual Aid:

   Minute	Window	Customers	Grumpy	Satisfied Customers	Potential Customers	Max Potential
   0	-	1	0	1	0	0
   1	-	0	1	1	0	0
   2	[0, 2]	1	0	2	0	0
   3	[1, 3]	2	1	2	2	2
   4	[2, 4]	1	0	3	2	2
   5	[3, 5]	1	1	3	3	3
   6	[4, 6]	7	0	10	1	3
   7	[5, 7]	5	1	10	6	6

5. Result Calculation:

   • Final satisfied_customers = 10

   • Final max_potential_customers = 6

   • Total satisfied customers = satisfied_customers + max_potential_customers = 10 + 6 = 16

The algorithm returns 16 as the maximum number of customers that can be satisfied throughout the day.

The central concept of this solution is to use a sliding window technique combined with a counting approach to efficiently find nice subarrays. The solution exploits the fact that once a nice subarray is found, any extension of it that doesn't include new odd numbers is also nice.

• Sliding Window: The solution maintains a window that expands when odd numbers are encountered and contracts when the odd count exceeds k.

• Incremental Counting: Instead of recounting for each subarray, it incrementally updates the count of nice subarrays.

1. Initialization:

   total_nice_subarrays = 0 current_nice_subarrays = 0 odd_count = 0 start_index = 0

   The function initializes several variables:

   • total_nice_subarrays to store the final count of nice subarrays.

   • current_nice_subarrays to track the number of valid subarrays ending at the current position.

   • odd_count to count the number of odd numbers in the current window.

   • start_index to mark the beginning of the current window.

2. Iterate Through Array:

   for num in nums:

   This loop processes each number in the input array.

3. Handle Odd Numbers:

   if num % 2 == 1: odd_count += 1 current_nice_subarrays = 0

   When an odd number is encountered, the odd count is incremented and the current nice subarray count is reset.

4. Window Adjustment and Counting:

   while odd_count == k: current_nice_subarrays += 1 odd_count -= nums[start_index] % 2 start_index += 1

   When k odd numbers are found, this loop counts nice subarrays by moving the start of the window. It increments the count for each valid start position and stops when an odd number is removed from the window.

5. Accumulate Nice Subarray Count:

   total_nice_subarrays += current_nice_subarrays

   The count of nice subarrays ending at the current position is added to the total.

6. Return Result:

   return total_nice_subarrays

   The function returns the total count of nice subarrays found.

Time Complexity:

• , where is the length of the input array nums. This is because each element in the array is processed at most twice: once in the outer for loop and potentially once in the inner while loop.

Space Complexity:

• , as the algorithm uses only a constant amount of extra space regardless of the input size. It maintains a few variables (total_nice_subarrays, current_nice_subarrays, odd_count, start_index) that do not grow with the input size.

Input:

Step-by-Step Walkthrough:

1. Initialization:

   • total_nice_subarrays = 0: Keeps track of all nice subarrays found

   • current_nice_subarrays = 0: Counts nice subarrays ending at the current position

   • odd_count = 0: Tracks the number of odd integers in the current window

   • start_index = 0: Marks the start of the current window

2. Main Loop (Iterating through nums):

   • Iteration 1 (num = 2):

     • 2 is even, so no changes to odd_count or current_nice_subarrays

     • No nice subarrays found yet

   • Iteration 2 (num = 1):

     • 1 is odd, so odd_count increases to 1

     • current_nice_subarrays resets to 0 (new odd number encountered)

     • Still no nice subarrays (need two odd numbers)

   • Iteration 3 (num = 2):

     • 2 is even, no changes to counters

     • No nice subarrays yet

   • Iteration 4 (num = 1):

     • 1 is odd, odd_count increases to 2

     • current_nice_subarrays resets to 0

     • odd_count == k, so we enter the while loop:

       1. current_nice_subarrays increases to 1

       2. Check if nums[0] (2) is odd - it's not, so odd_count remains 2

       3. start_index increases to 1

       4. Still odd_count == k, so loop continues

       5. current_nice_subarrays increases to 2

       6. nums[1] (1) is odd, so odd_count decreases to 1

       7. start_index increases to 2

     • total_nice_subarrays becomes 2

   • Iteration 5 (num = 2):

     • 2 is even, no changes to odd_count

     • total_nice_subarrays increases to 4 (2 + 2)

   • Iteration 6 (num = 1):

     • 1 is odd, odd_count increases to 2

     • current_nice_subarrays resets to 0

     • odd_count == k, enter while loop:

       1. current_nice_subarrays increases to 1

       2. nums[2] (2) is not odd, odd_count stays 2

       3. start_index increases to 3

       4. Still odd_count == k, continue loop

       5. current_nice_subarrays increases to 2

       6. nums[3] (1) is odd, odd_count decreases to 1

       7. start_index increases to 4

     • total_nice_subarrays becomes 6 (4 + 2)

   • Iteration 7 (num = 2):

     • 2 is even, no changes to odd_count

     • total_nice_subarrays increases to 8 (6 + 2)

   • Iteration 8 (num = 2):

     • 2 is even, no changes to odd_count

     • total_nice_subarrays increases to 10 (8 + 2)

3. Loop Termination:

   • The loop ends after processing all elements in nums

   • Final state: total_nice_subarrays = 10, odd_count = 1, start_index = 4

4. Result Calculation:

   • The final result is the value of total_nice_subarrays, which is 10

   • This represents the total number of subarrays containing exactly k = 2 odd numbers

The central concept of this solution is to leverage a two-pass approach to efficiently count nice subarrays. It exploits the structure of the array by focusing on the distribution of odd numbers and the even numbers between them.

• Even Number Counting: The solution counts the number of even integers between each pair of odd integers, including the ends of the array.

• Combinatorial Calculation: It uses these counts to calculate the total number of nice subarrays without explicitly listing them.

1. Initialization:

   even_counts = [] current_even_count, odd_count = 1, 0

   even_counts will store the count of even numbers between odd numbers, and current_even_count is initialized to 1 to account for the start of the array.

2. Count Even Numbers and Identify Odd Numbers:

   for num in nums: if num % 2 == 0: current_even_count += 1 else: even_counts.append(current_even_count) current_even_count = 1 even_counts.append(current_even_count)

   This loop counts even numbers between odd numbers. When an odd number is encountered, the current count of even numbers is appended to even_counts, and the count is reset. The final even_counts.append(current_even_count) accounts for even numbers at the end of the array.

3. Calculate Nice Subarrays:

   total_nice_subarrays = 0 for left_even_count, right_even_count in zip(even_counts, even_counts[k:]): total_nice_subarrays += left_even_count * right_even_count

   This loop calculates the number of nice subarrays. It pairs the count of even numbers before a sequence of k odd numbers with the count after, multiplying them to get the number of nice subarrays for that sequence.

4. Return Result:

   return total_nice_subarrays

   The function returns the total count of nice subarrays found.

Time Complexity:

• , where n is the length of the input array nums. This is because the algorithm makes two passes through the data: one to count even numbers between odd numbers, and another to calculate the number of nice subarrays.

Space Complexity:

• , as the algorithm uses only a constant amount of extra space regardless of the input size. It maintains a few variables (even_counts, current_even_count, odd_count) that do not grow with the input size.

Input:

Step-by-Step Walkthrough:

1. Initialization:

   • even_counts = []: List to store counts of consecutive even numbers

   • current_even_count = 1: Initialize count of the current sequence of even numbers

2. First Pass: Counting Even Numbers

   • Iteration 1 (num = 2):

     • 2 is even, so current_even_count increases to 2

     • even_counts remains empty: []

   • Iteration 2 (num = 1):

     • 1 is odd, so we append current_even_count (2) to even_counts

     • Reset current_even_count to 1

     • even_counts becomes [2]

   • Iteration 3 (num = 2):

     • 2 is even, so current_even_count increases to 2

     • even_counts remains [2]

   • Iteration 4 (num = 1):

     • 1 is odd, so we append current_even_count (2) to even_counts

     • Reset current_even_count to 1

     • even_counts becomes [2, 2]

   • Iteration 5 (num = 2):

     • 2 is even, so current_even_count increases to 2

     • even_counts remains [2, 2]

   • Iteration 6 (num = 1):

     • 1 is odd, so we append current_even_count (2) to even_counts

     • Reset current_even_count to 1

     • even_counts becomes [2, 2, 2]

   • Iteration 7 (num = 2):

     • 2 is even, so current_even_count increases to 2

     • even_counts remains [2, 2, 2]

   • Iteration 8 (num = 2):

     • 2 is even, so current_even_count increases to 3

     • even_counts remains [2, 2, 2]

3. Loop Termination:

   • After processing all numbers, append final current_even_count (3) to even_counts

   • Final state of even_counts: [2, 2, 2, 3]

4. Second Pass: Calculating Nice Subarrays

   • Iteration 1:

     • left_even_count = 2 (from even_counts[0])

     • right_even_count = 2 (from even_counts[2])

     • Calculate nice subarrays: 2 * 2 = 4

     • total_nice_subarrays becomes 4

   • Iteration 2:

     • left_even_count = 2 (from even_counts[1])

     • right_even_count = 3 (from even_counts[3])

     • Calculate nice subarrays: 2 * 3 = 6

     • total_nice_subarrays increases to 10 (4 + 6)

5. Visual Aid:

   Iteration Summary Table:

   Subarray	Left Even Count	Right Even Count	Nice Subarrays	Total Nice Subarrays
   1	2	2	4	4
   2	2	3	6	10

6. Result Calculation/Final Steps:

   • The final result is the value of total_nice_subarrays, which is 10

   • This represents the total number of subarrays containing exactly two odd numbers

The central concept of this solution is to leverage a sliding window approach combined with a sorted list to efficiently find the longest subarray that satisfies the given condition. This technique allows us to maintain a dynamic view of the current subarray while keeping track of its minimum and maximum elements.

• Sliding Window: The solution uses two pointers (left_index and the current iteration index) to define a window that expands and contracts as we iterate through the array.

• Sorted List: A sorted list is maintained to represent the elements in the current window, allowing for quick access to the minimum and maximum elements.

• Dynamic Window Adjustment: As new elements are added, the window expands. If the condition is violated, the window contracts from the left side.

1. Initialization:

   left_index = 0 sorted_nums = []

   The left_index variable keeps track of the left boundary of our sliding window. sorted_nums is an empty list that will store the elements of the current window in sorted order.

2. Iterating Through the Array:

   for num in nums:

   This loop processes each element of the input array nums sequentially.

3. Inserting New Element:

   bisect.insort(sorted_nums, num)

   The current element num is inserted into sorted_nums at its correct position, maintaining the sorted order. bisect.insort is used for efficient insertion in a sorted list.

4. Checking and Adjusting Window:

   if sorted_nums[-1] - sorted_nums[0] > limit: sorted_nums.pop(bisect.bisect(sorted_nums, nums[left_index]) - 1) left_index += 1

   This block checks if the difference between the maximum (last element) and minimum (first element) of the sorted list exceeds the limit. If so, it removes the leftmost element of the window from sorted_nums and moves the left boundary of the window.

5. Result Calculation/Return:

   return len(nums) - left_index

   The function returns the length of the longest valid subarray, which is the total length of the array minus the left index of the sliding window.

Time Complexity:

• , where is the length of the input array nums. This is because for each element in the array , we perform an insertion into a sorted list using bisect.insort ( in the worst case). However, in practice, it can be much faster if the subarrays tend to be short.

Space Complexity:

• , where is the length of the input array nums. This is because in the worst case, the sorted_nums list might contain all elements of the input array.

Input:

Step-by-Step Walkthrough:

1. Initialization:

   • Set left_index = 0

   • Initialize empty sorted_nums list

2. Main Loop (Iterate through nums):

   • Iteration 1 (nums[0] = 10):

     • Insert 10 into sorted_nums: sorted_nums = [10]

     • No window adjustment needed (10 - 10 = 0 <= 5)

     • Current subarray length: 1

   • Iteration 2 (nums[1] = 1):

     • Insert 1 into sorted_nums: sorted_nums = [1, 10]

     • Window adjustment: 10 - 1 = 9 > 5

       • Remove 10 from sorted_nums: sorted_nums = [1]

       • Update left_index to 1

     • Current subarray length: 1

   • Iteration 3 (nums[2] = 2):

     • Insert 2 into sorted_nums: sorted_nums = [1, 2]

     • No window adjustment needed (2 - 1 = 1 <= 5)

     • Current subarray length: 2

   • Iteration 4 (nums[3] = 4):

     • Insert 4 into sorted_nums: sorted_nums = [1, 2, 4]

     • No window adjustment needed (4 - 1 = 3 <= 5)

     • Current subarray length: 3

   • Iteration 5 (nums[4] = 7):

     • Insert 7 into sorted_nums: sorted_nums = [1, 2, 4, 7]

     • Window adjustment: 7 - 1 = 6 > 5

       • Remove 1 from sorted_nums: sorted_nums = [2, 4, 7]

       • Update left_index to 2

     • Current subarray length: 3

   • Iteration 6 (nums[5] = 2):

     • Insert 2 into sorted_nums: sorted_nums = [2, 2, 4, 7]

     • No window adjustment needed (7 - 2 = 5 <= 5)

     • Current subarray length: 4

3. Loop Termination:

   • The loop ends after processing all elements in nums

   • Final state: left_index = 2

4. Visual Aid: Iteration Summary Table

   Iteration	Current Number	Left Index	Sorted Nums	Current Length
   1	10	0	[10]	1
   2	1	1	[1]	1
   3	2	1	[1, 2]	2
   4	4	1	[1, 2, 4]	3
   5	7	2	[2, 4, 7]	3
   6	2	2	[2, 2, 4, 7]	4

5. Result Calculation/Final Steps:

   • The algorithm calculates the result as len(nums) - left_index

   • Result = 6 - 2 = 4

The longest subarray where the absolute difference between any two elements is at most 5 has a length of 4, which corresponds to the subarray [2, 4, 7, 2].

The central concept of this solution is to leverage a sliding window approach combined with two heaps (min-heap and max-heap) to efficiently find the longest subarray that satisfies the given condition. This technique allows us to maintain a dynamic view of the current subarray while keeping track of its minimum and maximum elements in a more efficient manner.

• Sliding Window: The solution uses two pointers (left_index and right_index) to define a window that expands and contracts as we iterate through the array.

• Dual Heap Structure: Two heaps (min-heap and max-heap) are used to efficiently track the minimum and maximum elements in the current window.

• Dynamic Window Adjustment: As new elements are added, the window expands. If the condition is violated, the window contracts from the left side until the condition is satisfied again.

1. Initialization:

   max_heap = [] min_heap = [] left_index = 0 max_length = 0

   Two empty heaps are initialized: max_heap for maximum elements and min_heap for minimum elements. left_index tracks the left boundary of the sliding window, and max_length stores the length of the longest valid subarray found so far.

2. Iterating Through the Array:

   for right_index, num in enumerate(nums):

   This loop processes each element of the input array nums sequentially, using right_index as the current position.

3. Adding Elements to Heaps:

   heapq.heappush(max_heap, (-num, right_index)) heapq.heappush(min_heap, (num, right_index))

   The current element num is added to both heaps. Note that for max_heap, we use the negative of num to create a max-heap using Python's min-heap implementation.

4. Adjusting Window:

   while -max_heap[0][0] - min_heap[0][0] > limit: left_index = min(max_heap[0][1], min_heap[0][1]) + 1 while max_heap[0][1] < left_index: heapq.heappop(max_heap) while min_heap[0][1] < left_index: heapq.heappop(min_heap)

   This block checks if the difference between the maximum and minimum elements exceeds the limit. If so, it moves the left boundary of the window and removes elements from the heaps that are now outside the window.

5. Updating Maximum Length:

   max_length = max(max_length, right_index - left_index + 1)

   After each iteration, update the max_length if the current window is longer than the previous maximum.

6. Result Calculation/Return:

   return max_length

   The function returns the length of the longest valid subarray.

Time Complexity:

• , where is the length of the input array nums. This is because for each element in the array , we perform heap operations (push and pop) which take time. In the worst case, we might need to perform these operations for each element.

Space Complexity:

• , where is the length of the input array nums. This is because in the worst case, all elements might be stored in the heaps simultaneously.

Input:

Step-by-Step Walkthrough:

1. Initialization:

   • Initialize empty max_heap and min_heap

   • Set left_index = 0 and max_length = 0

2. Main Loop (Iterate through nums):

   • Iteration 1 (nums[0] = 10):

     • Push (10, 0) to max_heap as (-10, 0) and to min_heap as (10, 0)

       • The heaps become:

         • max_heap = [(-10, 0)]

         • min_heap = [(10, 0)]

     • No window adjustment needed (10 - 10 = 0 <= 5)

     • Update max_length to 1

   • Iteration 2 (nums[1] = 1):

     • Push (1, 1) to both heaps

       • The heaps become:

         • max_heap = [(-10, 0), (-1, 1)]

         • min_heap = [(1, 1), (10, 0)]

     • Window adjustment: 10 - 1 = 9 > 5

       • Update left_index to 1

       • Remove (-10, 0) from max_heap: max_heap = [(-1, 1)]

     • max_length remains 1

   • Iteration 3 (nums[2] = 2):

     • Push (2, 2) to both heaps

       • The heaps become:

         • max_heap = [(-1, 1), (-2, 2)]

         • min_heap = [(1, 1), (2, 2), (10, 0)]

     • No window adjustment needed (2 - 1 = 1 <= 5)

     • Update max_length to 2

   • Iteration 4 (nums[3] = 4):

     • Push (4, 3) to both heaps

       • The heaps become:

         • max_heap = [(-1, 1), (-2, 2), (-4, 3)]

         • min_heap = [(1, 1), (2, 2), (4, 3), (10, 0)]

     • No window adjustment needed (4 - 1 = 3 <= 5)

     • Update max_length to 3

   • Iteration 5 (nums[4] = 7):

     • Push (7, 4) to both heaps

       • The heaps become:

         • max_heap = [(-1, 1), (-2, 2), (-4, 3), (-7, 4)]

         • min_heap = [(1, 1), (2, 2), (4, 3), (7, 4), (10, 0)]

     • Window adjustment: 7 - 1 = 6 > 5

       • Update left_index to 2

       • Remove (1, 1) from min_heap: min_heap = [(2, 2), (4, 3), (7, 4), (10, 0)]

     • max_length remains 3

   • Iteration 6 (nums[5] = 2):

     • Push (2, 5) to both heaps

       • The heaps become:

         • max_heap = [(-2, 2), (-4, 3), (-7, 4), (-2, 5)]

         • min_heap = [(2, 2), (4, 3), (7, 4), (10, 0)]

     • No window adjustment needed (7 - 2 = 5 <= 5)

     • Update max_length to 4

3. Loop Termination:

   • The loop ends after processing all elements in nums

   • Final state: left_index = 2, max_length = 4

4. Visual Aid: Iteration Summary Table

   Iteration	Current Number	Left Index	Current Length	Max Length	Max Heap (top 3)	Min Heap (top 3)
   1	10	0	1	1	[(10, 0)]	[(10, 0)]
   2	1	1	1	1	[(1, 1)]	[(1, 1), (10, 0)]
   3	2	1	2	2	[(2, 2), (1, 1)]	[(1, 1), (10, 0), (2, 2)]
   4	4	1	3	3	[(4, 3), (1, 1), (2, 2)]	[(1, 1), (4, 3), (2, 2)]
   5	7	2	3	3	[(7, 4), (4, 3), (2, 2)]	[(2, 2), (4, 3), (7, 4)]
   6	2	2	4	4	[(7, 4), (4, 3), (2, 2)]	[(2, 2), (2, 5), (7, 4)]

5. Result Calculation/Final Steps:

   • The algorithm maintains the max_length variable throughout the iterations

   • After processing all elements, max_length holds the length of the longest valid subarray

   • The final result is max_length = 4, corresponding to the subarray [2, 4, 7, 2]

The central concept of this solution is to leverage a sliding window approach combined with two monotonic queues (implemented using deques) to efficiently find the longest subarray that satisfies the given condition. This technique allows us to maintain a dynamic view of the current subarray while keeping track of its minimum and maximum elements in a highly efficient manner.

• Sliding Window: The solution uses two pointers (left_index and right_index) to define a window that expands and contracts as we iterate through the array.

• Monotonic Queues: Two deques are used: a monotonically decreasing queue for maximum values and a monotonically increasing queue for minimum values.

• Dynamic Window Adjustment: As new elements are added, the window expands. If the condition is violated, the window contracts from the left side.

1. Initialization:

   max_queue = deque() min_queue = deque() left_index = 0

   Two empty deques are initialized: max_queue for maximum elements and min_queue for minimum elements. left_index tracks the left boundary of the sliding window.

2. Iterating Through the Array:

   for right_index, num in enumerate(nums):

   This loop processes each element of the input array nums sequentially, using right_index as the current position.

3. Maintaining Monotonic Queues:

   while max_queue and num > max_queue[-1]: max_queue.pop() max_queue.append(num) while min_queue and num < min_queue[-1]: min_queue.pop() min_queue.append(num)

   These blocks maintain the monotonic property of the queues. For max_queue, we remove elements smaller than the current number from the right side before adding the current number. For min_queue, we remove elements larger than the current number.

4. Adjusting Window:

   if max_queue[0] - min_queue[0] > limit: if max_queue[0] == nums[left_index]: max_queue.popleft() if min_queue[0] == nums[left_index]: min_queue.popleft() left_index += 1

   This block checks if the difference between the maximum and minimum elements exceeds the limit. If so, it moves the left boundary of the window and updates the queues if necessary.

5. Result Calculation/Return:

   return len(nums) - left_index

   The function returns the length of the longest valid subarray, which is the total length of the array minus the final position of the left index.

Time Complexity:

• , where is the length of the input array nums. This is because each element is processed once and all operations on the deques (append, pop, pop-left) are . Although there are nested loops, each element is pushed and popped at most once from each deque, ensuring linear time complexity.

Space Complexity:

• , where is the length of the input array nums. This is because in the worst case, the deques might need to store all elements of the input array simultaneously.

Input:

Step-by-Step Walkthrough:

1. Initialization:

   • Initialize empty max_queue and min_queue as deques

   • Set left_index = 0

2. Main Loop (Iterate through nums):

   • Iteration 1 (nums[0] = 10):

     • Update max_queue: Append 10, max_queue = [10]

     • Update min_queue: Append 10, min_queue = [10]

     • No window adjustment needed (10 - 10 = 0 <= 5)

     • Current window length: 1

   • Iteration 2 (nums[1] = 1):

     • Update max_queue: Append 1, max_queue = [10, 1]

     • Update min_queue: Remove 10, append 1, min_queue = [1]

     • Window adjustment: 10 - 1 = 9 > 5

       • Remove 10 from max_queue: max_queue = [1]

       • Update left_index to 1

     • Current window length: 1

   • Iteration 3 (nums[2] = 2):

     • Update max_queue: Remove 1, append 2, max_queue = [2]

     • Update min_queue: Append 2, min_queue = [1, 2]

     • No window adjustment needed (2 - 1 = 1 <= 5)

     • Current window length: 2

   • Iteration 4 (nums[3] = 4):

     • Update max_queue: Remove 2, append 4, max_queue = [4]

     • Update min_queue: Append 4, min_queue = [1, 2, 4]

     • No window adjustment needed (4 - 1 = 3 <= 5)

     • Current window length: 3

   • Iteration 5 (nums[4] = 7):

     • Update max_queue: Remove 4, append 7, max_queue = [7]

     • Update min_queue: Append 7, min_queue = [1, 2, 4, 7]

     • Window adjustment: 7 - 1 = 6 > 5

       • Remove 1 from min_queue: min_queue = [2, 4, 7]

       • Update left_index to 2

     • Current window length: 3

   • Iteration 6 (nums[5] = 2):

     • Update max_queue: Append 2, max_queue = [7, 2]

     • Update min_queue: Remove 7 and 4, append 2, min_queue = [2, 2]

     • No window adjustment needed (7 - 2 = 5 <= 5)

     • Current window length: 4

3. Loop Termination:

   • The loop ends after processing all elements in nums

   • Final state: left_index = 2

4. Visual Aid: Iteration Summary Table

   Iteration	Current Number	Left Index	Current Length	Max Queue	Min Queue
   1	10	0	1	[10]	[10]
   2	1	1	1	[1]	[1]
   3	2	1	2	[2]	[1, 2]
   4	4	1	3	[4]	[1, 2, 4]
   5	7	2	3	[7]	[2, 4, 7]
   6	2	2	4	[7, 2]	[2, 2]

5. Result Calculation/Final Steps:

   • The algorithm calculates the result as len(nums) - left_index

   • Result = 6 - 2 = 4