July 2024, Week 1: July 1st - July 7th

The central concept of this solution is to leverage a single-pass iteration with a counter to efficiently detect three consecutive odd numbers in the array. This approach exploits the sequential nature of the problem and the binary property of numbers (odd or even).

• Counter Reset Mechanism: The solution uses a counter that increments for odd numbers and resets for even numbers, allowing it to track consecutive odd numbers efficiently.

• Early Termination: The algorithm returns True as soon as it detects three consecutive odd numbers, avoiding unnecessary iterations through the rest of the array.

1. Input Validation:

   if len(arr) < 3: return False

   This check ensures that the array has at least three elements. If not, it's impossible to have three consecutive odd numbers, so we return False immediately. This optimization prevents unnecessary processing for small arrays.

2. Counter Initialization:

   consecutive_odds = 0

   We initialize a counter consecutive_odds to keep track of the number of consecutive odd integers encountered. This single variable is key to the algorithm's space efficiency.

3. Array Iteration and Odd Number Detection:

   for num in arr: if num % 2: consecutive_odds += 1 else: consecutive_odds = 0

   We iterate through each number in the array. The modulo operation num % 2 checks if a number is odd (remainder 1 when divided by 2). If it's odd, we increment the counter. If it's even, we reset the counter to 0. This reset is crucial as it ensures we're only counting consecutive odd numbers.

4. Consecutive Odd Check and Early Return:

   if consecutive_odds == 3: return True

   After each update to consecutive_odds, we check if it has reached 3. If so, we've found three consecutive odd numbers, and we can immediately return True. This early return optimizes the function by avoiding unnecessary iterations once the condition is met.

5. Final Return:

   return False

   If we've iterated through the entire array without finding three consecutive odd numbers, we return False.

Time Complexity:

• , where is the length of the input array arr. This is because in the worst case, we need to iterate through all elements of the array once. Each operation inside the loop (modulo, increment, comparison) takes constant time, so the overall time complexity is linear with respect to the array length.

Space Complexity:

• , or constant space. This is because we only use a single integer variable consecutive_odds regardless of the input size. The space used does not grow with the size of the input array, making this solution very memory-efficient.

Input:

Step-by-Step Walkthrough:

1. Initialization:

   • The function threeConsecutiveOdds1 is called with the input array [1, 2, 34, 3, 4, 5, 7, 23, 12].

   • The variable consecutive_odds is initialized to 0. This variable will keep track of the count of consecutive odd numbers encountered.

2. Input Validation:

   • The function checks if the length of the input array is less than 3.

   • In this case, the array length is 9, which is valid (greater than or equal to 3), so the function proceeds.

3. Main Loop:

   • Iteration 1:

     • Current number: 1

     • 1 is odd (1 % 2 = 1), so consecutive_odds is incremented to 1.

     • 1 < 3, so the function continues to the next iteration.

   • Iteration 2:

     • Current number: 2

     • 2 is even (2 % 2 = 0), so consecutive_odds is reset to 0.

   • Iteration 3:

     • Current number: 34

     • 34 is even (34 % 2 = 0), so consecutive_odds remains 0.

   • Iteration 4:

     • Current number: 3

     • 3 is odd (3 % 2 = 1), so consecutive_odds is incremented to 1.

   • Iteration 5:

     • Current number: 4

     • 4 is even (4 % 2 = 0), so consecutive_odds is reset to 0.

   • Iteration 6:

     • Current number: 5

     • 5 is odd (5 % 2 = 1), so consecutive_odds is incremented to 1.

   • Iteration 7:

     • Current number: 7

     • 7 is odd (7 % 2 = 1), so consecutive_odds is incremented to 2.

   • Iteration 8:

     • Current number: 23

     • 23 is odd (23 % 2 = 1), so consecutive_odds is incremented to 3.

     • At this point, consecutive_odds equals 3, so the function immediately returns True.

4. Visual Aid:

   • The iteration summary table shows the progression of the consecutive_odds count for each element in the array.

   Element	Number	Odd/Even	Consecutive Odds
   1	1	Odd	1
   2	2	Even	0
   3	34	Even	0
   4	3	Odd	1
   5	4	Even	0
   6	5	Odd	1
   7	7	Odd	2
   8	23	Odd	3

5. Result Calculation/Final Steps:

   • The function returns True as soon as it encounters three consecutive odd numbers (5, 7, 23).

   • The last element (12) is not processed because the function has already found the desired pattern and returned.

The central concept of this solution is to leverage sorted arrays and a two-pointer technique to efficiently find the intersection of two integer arrays while maintaining element frequency. This approach exploits the ordered nature of sorted arrays to compare elements linearly and identify common elements.

• Sorting: By sorting both input arrays, we create a structure where identical elements are grouped together, making comparisons easier.

• Two-Pointer Technique: Using two pointers, one for each sorted array, we can traverse both arrays simultaneously, comparing elements and identifying matches.

• Frequency Preservation: The algorithm naturally preserves the frequency of elements in the intersection by including a match in the result each time it occurs in both arrays.

1. Array Swapping for Optimization:

   if len(nums1) > len(nums2): nums1, nums2 = nums2, nums1

   This step ensures that nums1 is always the shorter (or equal length) array. This optimization can reduce space complexity in cases where one array is significantly shorter than the other.

2. Sorting the Arrays:

   nums1.sort() nums2.sort()

   Sorting both arrays is crucial for the two-pointer approach to work. It allows us to compare elements sequentially and efficiently identify matches.

3. Initialization:

   intersection = [] index1, index2 = 0, 0

   We initialize an empty list intersection to store the result, and two pointers index1 and index2 to traverse nums1 and nums2 respectively.

4. Two-Pointer Traversal:

   while index1 < len(nums1) and index2 < len(nums2): if nums1[index1] < nums2[index2]: index1 += 1 elif nums1[index1] > nums2[index2]: index2 += 1 else: intersection.append(nums1[index1]) index1 += 1 index2 += 1

   This is the core of the algorithm. We compare elements from both arrays:

   • If the element in nums1 is smaller, we move the index1 pointer.

   • If the element in nums2 is smaller, we move the index2 pointer.

   • If they are equal, we've found an intersection element. We add it to the result and move both pointers. This process naturally handles frequency, as we only add an element when we find a match in both arrays.

5. Result Return:

   return intersection

   After the traversal, intersection contains all common elements with their correct frequencies, so we return it.

Time Complexity:

• , where and are the lengths of nums1 and nums2 respectively. This is because:

  1. Sorting both arrays takes and time.

  2. The two-pointer traversal is linear, .

  3. The sorting step dominates the time complexity.

Space Complexity:

• , where and are the lengths of nums1 and nums2. This is because:

  1. The result array intersection will at most contain as many elements as the shorter input array.

  2. The space used for sorting (which could be or for Python's built-in sort) is not counted as extra space in this analysis, as it's considered part of the input modification.

  3. Apart from the result array, we only use a constant amount of extra space for pointers and temporary variables.

Input:

Step-by-Step Walkthrough:

1. Initialization:

   • The function first checks if nums1 is longer than nums2. In this case, len(nums1) = 3 and len(nums2) = 5, so no swap occurs.

   • Both arrays are then sorted:

     • nums1 becomes [4, 5, 9]

     • nums2 becomes [4, 4, 8, 9, 9]

   • Initialize intersection = [], index1 = 0, and index2 = 0

2. Main Loop (Two-pointer comparison):

   • Iteration 1:

     • Compare nums1[0] = 4 and nums2[0] = 4

     • They are equal, so add 4 to intersection

     • Increment both index1 and index2

     • Current state: intersection = [4], index1 = 1, index2 = 1

   • Iteration 2:

     • Compare nums1[1] = 5 and nums2[1] = 4

     • 5 > 4, so increment only index2

     • Current state: intersection = [4], index1 = 1, index2 = 2

   • Iteration 3:

     • Compare nums1[1] = 5 and nums2[2] = 8

     • 5 < 8, so increment only index1

     • Current state: intersection = [4], index1 = 2, index2 = 2

   • Iteration 4:

     • Compare nums1[2] = 9 and nums2[2] = 8

     • 9 > 8, so increment only index2

     • Current state: intersection = [4], index1 = 2, index2 = 3

   • Iteration 5:

     • Compare nums1[2] = 9 and nums2[3] = 9

     • They are equal, so add 9 to intersection

     • Increment both index1 and index2

     • Current state: intersection = [4, 9], index1 = 3, index2 = 4

3. Loop Termination:

   • After iteration 5, index1 = 3, which equals len(nums1), so the loop terminates

   • The final state of variables:

     • intersection = [4, 9]

     • index1 = 3

     • index2 = 4

4. Visual Aid:

   Iteration Summary Table:

   Iteration	index1	index2	nums1 value	nums2 value	Action	Current Intersection
   1	0	0	4	4	Add to intersection, increment both	[4]
   2	1	1	5	4	Increment index2	[4]
   3	1	2	5	8	Increment index1	[4]
   4	2	2	9	8	Increment index2	[4]
   5	2	3	9	9	Add to intersection, increment both	[4, 9]

5. Result Calculation/Final Steps:

   • The algorithm has already built the intersection array during the loop iterations

   • No further calculations are needed

   • The function returns the final intersection array: [4, 9]

The central concept of this solution is to leverage a hash table (specifically, Python's Counter class) to efficiently find the intersection of two integer arrays while maintaining element frequency. This approach exploits the average-case lookup time of hash tables to quickly identify common elements.

• Hash Table Usage: By creating a frequency map of the shorter array, we can quickly check for the presence of elements and their remaining counts.

• Single Pass through Longer Array: We only need to iterate through the longer array once, checking each element against the frequency map.

• Frequency Preservation: The algorithm naturally preserves the frequency of elements in the intersection by decrementing the count in the frequency map each time a match is found.

1. Array Swapping for Optimization:

   if len(nums1) > len(nums2): nums1, nums2 = nums2, nums1

   This step ensures that nums1 is always the shorter (or equal length) array. This optimization reduces space complexity by creating the frequency map for the smaller array.

2. Creating the Frequency Map:

   count_nums1 = Counter(nums1)

   We use Python's Counter class to create a frequency map of nums1. This hash-based data structure allows for average-case lookups and updates.

3. Initialization of Result Array:

   intersection = []

   We initialize an empty list intersection to store the common elements.

4. Iterating through the Longer Array:

   for num in nums2: if count_nums1[num] > 0: intersection.append(num) count_nums1[num] -= 1

   This is the core of the algorithm. For each number in nums2:

   • We check if it exists in count_nums1 and has a count greater than 0.

   • If it does, we add it to the intersection list and decrement its count in count_nums1. This process naturally handles frequency, as we only add an element when it's still available in the frequency map.

5. Result Return:

   return intersection

   After the iteration, intersection contains all common elements with their correct frequencies, so we return it.

Time Complexity:

• , where and are the lengths of nums1 and nums2 respectively. This is because:

  1. Creating the Counter for nums1 takes time.

  2. Iterating through nums2 takes time.

  3. Each lookup and update in the Counter is on average.

Space Complexity:

• , where and are the lengths of nums1 and nums2. This is because:

  1. The Counter will contain at most as many elements as the shorter input array (which we ensure is nums1).

  2. The result array intersection will also contain at most as many elements as the shorter input array.

  3. By swapping arrays if necessary, we guarantee that we're using the minimum possible space for the Counter.

Input:

Step-by-Step Walkthrough:

1. Initialization:

   • The function first checks if nums1 is longer than nums2. In this case, len(nums1) = 3 and len(nums2) = 5, so no swap occurs.

   • Create a Counter object count_nums1 from nums1:

     count_nums1 = Counter({4: 1, 9: 1, 5: 1})

   • Initialize intersection = []

2. Main Loop (Iterating through nums2):

   • Iteration 1:

     • Processing num = 9

     • count_nums1[9] = 1, which is > 0

     • Add 9 to intersection

     • Decrement count_nums1[9] to 0

     • Current state: intersection = [9]

   • Iteration 2:

     • Processing num = 4

     • count_nums1[4] = 1, which is > 0

     • Add 4 to intersection

     • Decrement count_nums1[4] to 0

     • Current state: intersection = [9, 4]

   • Iteration 3:

     • Processing num = 9

     • count_nums1[9] = 0, which is not > 0

     • No action taken

     • Current state: intersection = [9, 4]

   • Iteration 4:

     • Processing num = 8

     • count_nums1[8] = 0 (default for a missing key), which is not > 0

     • No action taken

     • Current state: intersection = [9, 4]

   • Iteration 5:

     • Processing num = 4

     • count_nums1[4] = 0, which is not > 0

     • No action taken

     • Current state: intersection = [9, 4]

3. Loop Termination:

   • The loop terminates after processing all elements in nums2

   • The final state of variables:

     • intersection = [9, 4]

     • count_nums1 = Counter({4: 0, 9: 0, 5: 1})

4. Visual Aid:

   Iteration Summary Table:

   Iteration	Number	Count after action	Action	Current Intersection
   1	9	0	Add 9 to intersection, decrement count	[9]
   2	4	0	Add 4 to intersection, decrement count	[9, 4]
   3	9	0	No action	[9, 4]
   4	8	0	No action	[9, 4]
   5	4	0	No action	[9, 4]

5. Result Calculation/Final Steps:

   • The algorithm has already built the intersection array during the loop iterations

   • No further calculations are needed

   • The function returns the final intersection array: [9, 4]

The central concept of this solution is to leverage heap operations to efficiently identify the elements that have the most significant impact on the range of the array. By focusing on the four smallest and four largest elements, we can explore all possible combinations of three moves that minimize the difference between the maximum and minimum values.

• Edge Case Handling: The solution first addresses the scenario where the array has 4 or fewer elements, recognizing that in this case, all elements can be made equal.

• Extrema Extraction: Using heap operations, the algorithm efficiently extracts the four smallest and four largest elements from the array.

• Optimal Move Combination: By considering all possible ways to apply three moves to these extreme elements, the solution determines the minimal achievable difference.

1. Edge Case Handling:

   if len(nums) <= 4: return 0

   This check addresses the scenario where we have four or fewer elements. In such cases, we can always make all elements equal using at most three moves, resulting in a minimum difference of 0.

2. Extrema Extraction:

   smallest_four = heapq.nsmallest(4, nums) largest_four = heapq.nlargest(4, nums)

   Here, we use heap operations to efficiently extract the four smallest and four largest elements from the input array. These elements are crucial because they represent the boundaries of our possible solutions after applying up to three moves.

3. Optimal Move Calculation:

   return min( largest_four[0] - smallest_four[3], # Change 3 smallest largest_four[1] - smallest_four[2], # Change 2 smallest, 1 largest largest_four[2] - smallest_four[1], # Change 1 smallest, 2 largest largest_four[3] - smallest_four[0] # Change 3 largest )

   This step calculates and compares four scenarios, each representing a different way to apply the three allowed moves:

   • Change the 3 smallest elements: We compare the largest element with the 4th smallest.

   • Change 2 smallest and 1 largest: We compare the 2nd largest with the 3rd smallest.

   • Change 1 smallest and 2 largest: We compare the 3rd largest with the 2nd smallest.

   • Change the 3 largest elements: We compare the 4th largest with the smallest element.

   By taking the minimum of these four scenarios, we find the smallest possible difference achievable with at most three moves.

Time Complexity:

• , where is the number of elements in the input array nums. This is because both heapq.nsmallest() and heapq.nlargest() operations have a time complexity of , where is 4 in our case. Since is a constant, this simplifies to . The final min() operation is performed on a constant number of elements, so it doesn't affect the overall time complexity.

Space Complexity:

• , because we only store a constant number of elements (8 in total: 4 smallest and 4 largest) regardless of the input size. The space used does not grow with the input size, making it constant space complexity.

Input:

Step-by-Step Walkthrough:

1. Initialization:

   • The function starts by initializing n as the length of the input list nums.

   • n = len(nums) = 8

2. Edge Case Check:

   • The function checks if n <= 4. In this case, n = 8, so this condition is false, and we proceed with the main algorithm.

3. Extracting Smallest and Largest Elements:

   • The function uses heapq.nsmallest(4, nums) to find the four smallest elements: [0, 4, 5, 6]

   • It then uses heapq.nlargest(4, nums) to find the four largest elements: [21, 10, 8, 7]

   • These operations efficiently extract the extremes without fully sorting the list.

4. Calculating Possible Differences:

   • The function calculates four different scenarios, each representing a different strategy for applying the three allowed moves:

   • Iteration 1 (Change 3 smallest):

     • Largest: 21 (largest element)

     • Smallest: 6 (4th smallest element)

     • Difference: 21 - 6 = 15

   • Iteration 2 (Change 2 smallest, 1 largest):

     • Largest: 10 (2nd largest element)

     • Smallest: 5 (3rd smallest element)

     • Difference: 10 - 5 = 5

   • Iteration 3 (Change 1 smallest, 2 largest):

     • Largest: 8 (3rd largest element)

     • Smallest: 4 (2nd smallest element)

     • Difference: 8 - 4 = 4

   • Iteration 4 (Change 3 largest):

     • Largest: 7 (4th largest element)

     • Smallest: 0 (smallest element)

     • Difference: 7 - 0 = 7

5. Visual Aid - Difference Summary:

   Strategy	Largest	Smallest	Difference
   Change 3 smallest	21	6	15
   Change 2 smallest, 1 largest	10	5	5
   Change 1 smallest, 2 largest	8	4	4
   Change 3 largest	7	0	7

6. Finding Minimum Difference:

   • The function identifies the minimum difference from the calculated scenarios: 4

   • The corresponding strategy is "Change 1 smallest, 2 largest."

7. Result Calculation/Final Steps:

   • The minimum difference 4 is returned as the final result.

   • This result represents the smallest possible difference between the maximum and minimum elements in the array after applying at most three moves.

   • In this case, the optimal strategy would be to change the smallest element and the two largest elements, resulting in a final array where the difference between the largest and smallest elements is 4.

The central concept of this solution is to use two pointers to traverse the linked list while simultaneously modifying it in-place. This approach allows us to merge nodes between zeros efficiently without creating a new list.

• Two-Pointer Technique: The solution uses two pointers: result_node for building the modified list and traversal_node for iterating through the original list.

• In-Place Modification: Instead of creating a new list, the solution modifies the existing list, which saves space and simplifies the process.

• Running Sum: A current_sum variable is used to accumulate the sum of values between zeros.

1. Initialization:

   result_node = head traversal_node = head.next_node current_sum = 0

   We initialize result_node to the head (which is a zero node), traversal_node to the next node to start processing, and current_sum to zero. This setup allows us to start summing from the first non-zero node.

2. Traversal and Summing:

   while traversal_node: if traversal_node.val == 0: result_node = result_node.next_node result_node.val = current_sum current_sum = 0 else: current_sum += traversal_node.val traversal_node = traversal_node.next_node

   This is the core of the algorithm. We traverse the list with traversal_node, summing values until we hit a zero. When we encounter a zero:

   • We move result_node to the next node (which will store our sum)

   • We set its value to the accumulated sum

   • We reset current_sum for the next group If the current node is not zero, we simply add its value to current_sum.

3. Finalizing the List:

   result_node.next_node = None return head.next_node

   After the traversal, we set the next_node of the last result node to None, effectively trimming any remaining nodes. We return head.next_node because the original head was a zero node that we don't want in our final list.

Time Complexity:

• , where is the number of nodes in the original linked list. This is because we traverse the list exactly once, performing constant-time operations at each node.

Space Complexity:

• , as we are modifying the list in-place and only using a constant amount of extra space for our pointers and the current_sum variable, regardless of the input size.

Input:

A visual representation of the input linked list:

Step-by-Step Walkthrough:

1. Initialization:

   • result_node is set to the head of the list (value 0)

   • traversal_node is set to the next node (value 3)

   • current_sum is initialized to 0

2. Main Loop (Merging nodes between zeros):

   • Iteration 1:

     • Current state: result_node.val = 0, traversal_node.val = 3, current_sum = 0

     • The traversal node is not 0, so we add its value to current_sum

     • current_sum becomes 3

     • Move traversal_node to the next node

   • Iteration 2:

     • Current state: result_node.val = 0, traversal_node.val = 1, current_sum = 3

     • The traversal node is not 0, so we add its value to current_sum

     • current_sum becomes 4

     • Move traversal_node to the next node

   • Iteration 3:

     • Current state: result_node.val = 0, traversal_node.val = 0, current_sum = 4

     • The traversal node is 0, triggering the merging process:

       1. Move result_node to the next node

       2. Set result_node.val to the current sum (4)

       3. Reset current_sum to 0

     • The list is updated:

       

     • Move traversal_node to the next node

   • Iteration 4:

     • Current state: result_node.val = 4, traversal_node.val = 4, current_sum = 0

     • The traversal node is not 0, so we add its value to current_sum

     • current_sum becomes 4

     • Move traversal_node to the next node

   • Iteration 5:

     • Current state: result_node.val = 4, traversal_node.val = 5, current_sum = 4

     • The traversal node is not 0, so we add its value to current_sum

     • current_sum becomes 9

     • Move traversal_node to the next node

   • Iteration 6:

     • Current state: result_node.val = 4, traversal_node.val = 2, current_sum = 9

     • The traversal node is not 0, so we add its value to current_sum

     • current_sum becomes 11

     • Move traversal_node to the next node

   • Iteration 7:

     • Current state: result_node.val = 4, traversal_node.val = 0, current_sum = 11

     • The traversal node is 0, triggering the merging process:

       1. Move result_node to the next node

       2. Set result_node.val to the current sum (11)

       3. Reset current_sum to 0

     • The list is updated:

       

     • Move traversal_node to the next node

3. Loop Termination:

   • The loop terminates when traversal_node becomes None, indicating we've reached the end of the list

4. Final List Adjustment:

   • Set result_node.next_node to None, effectively removing any remaining nodes

5. Visual Aid: Iteration Summary:

   Iteration	Result Node Val	Traversal Node Val	Current Sum	Action
   1	0	1	3	Added 3 to sum
   2	0	0	4	Added 1 to sum
   3	4	4	0	Zero found, sum applied
   4	4	5	4	Added 4 to sum
   5	4	2	9	Added 5 to sum
   6	4	0	11	Added 2 to sum
   7	11	None	0	Zero found, sum applied

6. Result Calculation/Final Steps:

   • The function returns head.next_node, which is the start of the merged list

   • The output list is [4, 11]

A visual representation of the output linked list:

The central concept of this solution is to leverage a single-pass traversal with a sliding window of three nodes to identify critical points in a linked list. This approach allows for efficient detection of local maxima and minima while simultaneously calculating the minimum and maximum distances between these critical points.

• Sliding Window: The algorithm uses a three-node window (previous, current, and next) that slides through the linked list, allowing for local comparisons to identify critical points.

• Critical Point Detection: By comparing the values of three consecutive nodes, the algorithm can determine if the current node is a local maximum or minimum.

• Distance Tracking: The solution keeps track of the first and last critical points, as well as the minimum distance between any two adjacent critical points.

1. Initial Check and Initialization:

   if not head.next_node.next_node: return [-1, -1] prev_node = head current_node = head.next_node first_critical_index, last_critical_index = None, None min_distance = float('inf') current_index = 1

   The function first checks if the list has at least three nodes, as required for a critical point. It then initializes variables to keep track of the previous and current nodes, the indices of the first and last critical points, and the minimum distance between critical points.

2. Main Loop - Traversing the List:

   while current_node.next_node: current_index += 1 # Check if we have a critical point if ( # Local maxima prev_node.val < current_node.val > current_node.next_node.val # Local minima or prev_node.val > current_node.val < current_node.next_node.val ):

   This loop traverses the list, incrementing the current_index for each node. It checks if the current node is a critical point by comparing it with its previous and next nodes. A critical point is identified if the current node is a local maxima or minima.

3. Critical Point Processing:

   if not first_critical_index: first_critical_index = current_index else: min_distance = min(min_distance, current_index - last_critical_index) last_critical_index = current_index

   When a critical point is found, the code updates the first_critical_index if it's the first one encountered. For later critical points, it updates the min_distance by comparing it with the distance to the last critical point. The last_critical_index is always updated to the current index.

4. Window Sliding:

   prev_node = current_node current_node = current_node.next_node

   At the end of each iteration, the window slides forward by updating the prev_node and current_node.

5. Result Calculation and Return:

   if min_distance != float('inf'): max_distance = last_critical_index - first_critical_index return [min_distance, max_distance] return [-1, -1]

   After the traversal, if at least two critical points were found (min_distance was updated), the function calculates the max_distance as the difference between the last and first critical point indices. It returns these values as a list. If fewer than two critical points were found, it returns [-1, -1] as specified in the problem statement.

Time Complexity:

• , where is the number of nodes in the linked list. This is because the algorithm performs a single pass through the entire linked list, processing each node exactly once.

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 (prev_node, current_node, first_critical_index, last_critical_index, min_distance, current_index) throughout the execution, and does not use any data structures that grow with the input size.

Input:

A visual representation of the input linked list:

Step-by-Step Walkthrough:

1. Initialization:

   • We start by checking if the linked list has at least three nodes. In this case, it does, so we proceed.

   • Initialize variables:

     • prev_node = head (value: 5)

     • current_node = head.next_node (value: 3)

     • first_critical_index = None

     • last_critical_index = None

     • min_distance = float('inf')

     • current_index = 1

2. Main Loop:

   • Iteration 1:

     • current_index is incremented to 2

     • Current state: prev_node = 5, current_node = 3, next_node = 1

     • Check if we have a critical point:

       • Local maxima: 5 < 3 > 1? No

       • Local minima: 5 > 3 < 1? No

     • Not a critical point, so we move to the next iteration

     • Update: prev_node = 3, current_node = 1

   • Iteration 2:

     • current_index is incremented to 3

     • Current state: prev_node = 3, current_node = 1, next_node = 2

     • Check if we have a critical point:

       • Local maxima: 3 < 1 > 2? No

       • Local minima: 3 > 1 < 2? Yes

     • Critical point found (local minima)

     • Set first_critical_index = 3

     • Set last_critical_index = 3

     • Update: prev_node = 1, current_node = 2

   • Iteration 3:

     • current_index is incremented to 4

     • Current state: prev_node = 1, current_node = 2, next_node = 5

     • Check if we have a critical point:

       • Local maxima: 1 < 2 > 5? No

       • Local minima: 1 > 2 < 5? No

     • Not a critical point, so we move to the next iteration

     • Update: prev_node = 2, current_node = 5

   • Iteration 4:

     • current_index is incremented to 5

     • Current state: prev_node = 2, current_node = 5, next_node = 1

     • Check if we have a critical point:

       • Local maxima: 2 < 5 > 1? Yes

     • Critical point found (local maxima)

     • Update min_distance since this is not the first critical point:

       • min_distance = min(inf, 5 - 3) = 2

     • Update last_critical_index = 5

     • Update: prev_node = 5, current_node = 1

   • Iteration 5:

     • current_index is incremented to 6

     • Current state: prev_node = 5, current_node = 1, next_node = 2

     • Check if we have a critical point:

       • Local maxima: 5 > 1 < 2? No

       • Local minima: 5 < 1 < 2? Yes

     • Critical point found (local minima)

     • Update min_distance:

       • min_distance = min(2, 6 - 5) = 1

     • Update last_critical_index = 6

     • Update: prev_node = 1, current_node = 2

3. Loop Termination:

   • The loop terminates as current_node.next_node is None

   • Final state of variables:

     • first_critical_index = 3

     • last_critical_index = 6

     • min_distance = 1

4. Visual Aids:

   Iteration Summary:

   Node Index	Prev	Current	Next	Is Critical	First Critical	Last Critical	Min Distance
   2	5	3	1	False			inf
   3	3	1	2	True	3	3	inf
   4	1	2	5	False	3	3	inf
   5	2	5	1	True	3	5	2
   6	5	1	2	True	3	6	1

   Here is a visual representation of the critical points:

   

5. Result Calculation/Final Steps:

   • We found at least two critical points, so we proceed to calculate the result

   • Calculate max_distance:

     • max_distance = last_critical_index - first_critical_index = 6 - 3 = 3

   • The final result is [min_distance, max_distance] = [1, 3]

   This result means:

   • The minimum distance between any two distinct critical points is 1 (between indices 5 and 6)

   • The maximum distance between any two distinct critical points is 3 (between indices 3 and 6)

The central concept of this solution is to leverage modular arithmetic to efficiently calculate the pillow's position without simulating the entire process. This approach exploits the cyclic nature of the pillow's movement.

• Cycle Length Calculation: The solution recognizes that the pillow's movement forms a complete cycle every 2(n-1) seconds, where n is the number of people.

• Position within Cycle: By using the modulo operation, the solution determines the position within a single cycle, avoiding the need to iterate through all time steps.

• Direction Determination: The solution uses a simple comparison to determine whether the pillow is moving forward or backward within the cycle.

1. Cycle Position Calculation:

   cycle_position = time % ((n - 1) * 2)

   This line calculates the position within a single back-and-forth cycle. (n - 1) * 2 represents the total time for one complete cycle (forward and backward). The modulo operation % gives us the remainder, which is the position within the current cycle.

2. Direction and Position Determination:

   if cycle_position < n: return cycle_position + 1 else: return 2 * n - (cycle_position + 1)

   This conditional block determines the direction of the pillow's movement and calculates its position:

   • If cycle_position < n, the pillow is moving forward. The position is simply cycle_position + 1 (adding 1 because people are labeled from 1 to n, not 0 to n-1).

   • Otherwise, the pillow is moving backward. The calculation 2 * n - (cycle_position + 1) determines the position by considering the total path length (2n) and subtracting the current progress (cycle_position + 1).

3. Result Return: The function implicitly returns the calculated position, which represents the index of the person holding the pillow after the given time.

Time Complexity:

• , where is the number of people and time is the elapsed time. This is because the solution performs a fixed number of arithmetic operations regardless of the input values. The modulo operation and conditional check are both constant time operations.

Space Complexity:

• , as the solution uses only a constant amount of additional memory (the cycle_position variable) regardless of the input size. No data structures that grow with input size are used.

Input:

Step-by-Step Walkthrough:

1. Initialization:

   • The function passThePillow1 is called with n = 5 (number of people) and time = 21 (seconds elapsed).

   • The cycle length is calculated: (n - 1) * 2 = (5 - 1) * 2 = 8. This represents the time it takes for the pillow to make a complete back-and-forth cycle.

2. Main Calculation:

   • The cycle position is calculated using the modulo operation: cycle_position = time % ((n - 1) * 2) = 21 % 8 = 5

   • This means that after 21 seconds, we are 5 steps into the current cycle.

3. Decision Point:

   • The function checks if cycle_position (5) < n (5).

   • Since 5 is not less than 5, the condition is false.

   • This indicates that the pillow is moving backward at this point in time.

4. Result Calculation:

   • For backward movement, the formula 2 * n - (cycle_position + 1) is used: 2 * 5 - (5 + 1) = 10 - 6 = 4

   • This calculation determines the position of the pillow counting from the start of the line when it's moving backward.

5. Visual Aid: Here's a summary of the calculation:

   Step	Description	Value
   Input n	Number of people	5
   Input time	Elapsed time	21
   Cycle length	(n - 1) * 2	8
   Cycle position	time % cycle_length	5
   Movement direction	Based on cycle_position < n	Backward
   Result	2 * n - (cycle_position + 1)	4

6. Final Result:

   • The function returns 4, which means after 21 seconds, the pillow is with the fourth person in the line.

The central concept of this solution is to leverage a greedy simulation approach to maximize the number of water bottles drunk. The solution iteratively drinks all available full bottles and exchanges empty bottles for new full ones until no more exchanges are possible.

• Greedy Strategy: The algorithm always drinks all available full bottles and exchanges as many empty bottles as possible in each iteration.

• Simulation: The process of drinking and exchanging is simulated step by step until the termination condition is met.

1. Initialization:

   total_drinks = num_bottles empty_bottles = num_bottles

   We initialize total_drinks and empty_bottles with num_bottles. This represents drinking all initially available full bottles, which becomes our starting point for total drinks and empty bottles.

2. Exchange Loop:

   while empty_bottles >= num_exchange:

   This loop continues as long as we have enough empty bottles to make an exchange. It's the core of our simulation.

3. Bottle Exchange Calculation:

   exchanged_bottles, unused_bottles = divmod(empty_bottles, num_exchange)

   Here, we use divmod() to efficiently calculate two values:

   • exchanged_bottles: The number of full bottles we get by exchanging empty ones.

   • unused_bottles: The number of empty bottles left that couldn't be exchanged.

4. Updating Total Drinks:

   total_drinks += exchanged_bottles

   We increment total_drinks by exchanged_bottles, as these new full bottles will be immediately consumed.

5. Updating Empty Bottles:

   empty_bottles = exchanged_bottles + unused_bottles

   We update empty_bottles for the next iteration. It includes the bottles we just drank (exchanged_bottles) and the leftover empty bottles from the previous round (unused_bottles).

6. Result Return:

   return total_drinks

   Once we can no longer exchange bottles, we return the total number of bottles drunk.

Time Complexity:

• , where is num_bottles and is num_exchange. This is because in each iteration, the number of empty bottles is effectively divided by num_exchange. The number of iterations is logarithmic in the initial number of bottles.

Space Complexity:

• , as the algorithm uses only a constant amount of extra space regardless of the input size. It maintains only a few variables (total_drinks, empty_bottles, exchanged_bottles, unused_bottles) throughout its execution.

Input:

Step-by-Step Walkthrough:

1. Initialization:

   • The function starts with num_bottles = 15 and num_exchange = 4.

   • We initialize total_drinks = 15 (all initial bottles are drunk).

   • We set empty_bottles = 15 (all initial bottles become empty after drinking).

2. Main Loop (Exchanging empty bottles for full ones):

   • Iteration 1:

     • We check if empty_bottles (15) >= num_exchange (4), which is true.

     • We calculate divmod(15, 4), which returns (3, 3):

       • exchanged_bottles = 3 (we can exchange 3 sets of 4 empty bottles)

       • unused_bottles = 3 (3 empty bottles remain after exchanges)

     • We update total_drinks:

       • total_drinks = 15 + 3 = 18 (add newly acquired full bottles)

     • We update empty_bottles:

       • empty_bottles = 3 + 3 = 6 (new empties plus unused empties)

   • Iteration 2:

     • We check if empty_bottles (6) >= num_exchange (4), which is true.

     • We calculate divmod(6, 4), which returns (1, 2):

       • exchanged_bottles = 1 (we can exchange 1 set of 4 empty bottles)

       • unused_bottles = 2 (2 empty bottles remain after exchanges)

     • We update total_drinks:

       • total_drinks = 18 + 1 = 19 (add newly acquired full bottle)

     • We update empty_bottles:

       • empty_bottles = 1 + 2 = 3 (new empty plus unused empties)

3. Loop Termination:

   • After the second iteration, empty_bottles (3) < num_exchange (4).

   • This condition fails the while loop check, so the loop terminates.

   • At this point, we have drunk all possible bottles and can't exchange anymore.

4. Visual Aid:

   Iteration	Starting Empty Bottles	Exchanged Bottles	Unused Bottles	Total Drinks	Ending Empty Bottles
   1	15	3	3	18	6
   2	6	1	2	19	3

5. Result Calculation/Final Steps:

   • The final result is the value of total_drinks, which is 19.

   • This represents:

     • 15 initial bottles

     • 3 bottles from the first exchange

     • 1 bottle from the second exchange

   • The function returns this value (19) as the maximum number of water bottles that can be drunk.

The central concept of this solution is to leverage a mathematical formula to directly calculate the maximum number of water bottles that can be drunk. Instead of simulating the process, this approach uses the relationship between the initial number of bottles, the exchange rate, and the maximum possible exchanges to compute the result in a single step.

• Mathematical Insight: The solution recognizes that the total number of bottles drunk is the sum of the initial bottles and the additional bottles gained through exchanges.

• Direct Calculation: By deriving a formula that captures the exchange process, we can avoid iterative simulation and compute the result instantly.

1. Formula Application:

   return num_bottles + (num_bottles - 1) // (num_exchange - 1)

   This single line of code encapsulates the entire solution. Let's break it down:

   • num_bottles: This represents the initial number of full bottles, which are all drunk.

   • (num_bottles - 1) // (num_exchange - 1): This calculates the additional bottles obtained through exchanges.

2. Understanding the Formula:

   • num_bottles - 1: This represents the number of empty bottles after drinking the initial bottles, minus one. We subtract one because in the last exchange, we might not have enough empties for a full exchange.

   • num_exchange - 1: This represents the net cost of each exchange in terms of empty bottles. Each exchange gives us 1 full bottle but costs num_exchange empty ones, so the net cost is num_exchange - 1.

   • The integer division // gives us the maximum number of complete exchanges possible.

3. Result Calculation: The function returns the sum of the initial bottles and the additional bottles from exchanges, giving us the total number of bottles that can be drunk.

Time Complexity:

• , because the function performs a fixed number of arithmetic operations regardless of the input values. It doesn't use any loops or recursive calls, making it constant time.

Space Complexity:

• , as the function uses only a constant amount of extra space. It doesn't create any data structures or variables that grow with the input size. All calculations are done in-place with the input parameters.

Input:

Step-by-Step Walkthrough:

1. Initialization:

   • The function starts with num_bottles = 15 and num_exchange = 4.

2. Main Calculation Steps:

   • Step 1: Calculate initial empty bottles

     • We calculate initial_empty = num_bottles - 1

     • This gives us initial_empty = 15 - 1 = 14

     • We subtract 1 because in the last exchange, we might not have enough empties for a full exchange

   • Step 2: Calculate net cost of exchange

     • We calculate net_exchange_cost = num_exchange - 1

     • This gives us net_exchange_cost = 4 - 1 = 3

     • We subtract 1 because each exchange gives us 1 full bottle but costs num_exchange empty ones, so the net cost is num_exchange - 1

   • Step 3: Calculate additional bottles from exchanges

     • We calculate additional_bottles = initial_empty // net_exchange_cost

     • This gives us additional_bottles = 14 // 3 = 4

     • The integer division // gives us the maximum number of complete exchanges possible

   • Step 4: Calculate total bottles drunk

     • We calculate total_bottles = num_bottles + additional_bottles

     • This gives us total_bottles = 15 + 4 = 19

     • This sum represents the initial bottles plus all bottles obtained through exchanges

3. Visual Aid:

   Item	Value
   Initial Bottles	15
   Exchange Rate	4
   Initial Empty Bottles	14
   Net Exchange Cost	3
   Additional Bottles	4
   Total Bottles Drunk	19

4. Result Calculation/Final Steps:

   • The final result is the value of total_bottles, which is 19.

   • This represents:

     • 15 initial bottles

     • 4 additional bottles from exchanges

   • The function returns this value (19) as the maximum number of water bottles that can be drunk.

5. Mathematical Insight:

   • This method uses a direct formula to calculate the result, avoiding the need for iteration.

   • The formula num_bottles + (num_bottles - 1) // (num_exchange - 1) captures the entire exchange process in one calculation.

   • It's equivalent to our step-by-step calculation but more concise and efficient.