June 2024 Weekly Exercises

The core idea of this solution is to optimize the search for common elements in multiple sorted arrays by focusing on the smallest array. It leverages binary search to efficiently check if elements from the smallest array exist in all other arrays. This approach reduces the overall search space and time complexity compared to a naive brute-force method.

• Binary Search: Utilizes the fact that the arrays are sorted to perform efficient logarithmic-time searches for elements.

• Smallest Array Focus: By iterating through the smallest array, we minimize the number of elements that need to be checked across all arrays, leading to a potential performance improvement.

• Common Subsequence Construction: An element is added to the longest common subsequence (LCS) only if it's found in all other arrays, ensuring the result is a valid subsequence shared by all.

1. Initialization:

   • Finds the smallest_array using the min function with a key based on array length.

   • Removes the smallest_array from the arrays list to avoid redundant searches.

   • Initializes an empty list common_subsequences to store the LCS.

2. Helper Function (binary_search):

   • This function takes an array array and a target number num.

   • It uses bisect_left to find the insertion point for num in array while maintaining sorted order.

   • If num is found at the insertion point, it returns True; otherwise, it returns False.

3. Main Loop (Finding Common Elements):

   • Iterates through each num in the smallest_array.

   • For each num:

     • A generator expression checks if num is present in all remaining arrays using binary_search.

     • The all function checks if all results from the generator are True.

     • If num is found in all arrays, it's added to common_subsequences.

4. Return Result:

   • Returns the common_subsequences list, representing the longest common subsequence found.

Time Complexity:

• , where:

  • is the number of elements in the smallest array.

  • is the total number of arrays.

  • is the average length of the arrays.

• The nested loops dominate the time complexity. The outer loop iterates through the smallest array (n iterations), and the inner loop implicitly iterates through all remaining arrays (m-1 iterations) within the generator expression. For each element in the smallest array, we perform a binary search in each of the other arrays, with an average time complexity of O(log(k)).

Space Complexity:

• , where is the length of the longest common subsequence.

• The main space usage is due to storing the elements of the LCS in the common_subsequences list. The space used by the binary_search function is constant.

Input: arrays = [[2, 3, 6, 8], [1, 2, 3, 5, 6, 7, 10], [2, 3, 4, 6, 9]]

Step-by-Step Walkthrough:

1. Initialization:

   • The function prints the initial arrays input: [[2, 3, 6, 8], [1, 2, 3, 5, 6, 7, 10], [2, 3, 4, 6, 9]]

   • An empty list common_subsequences is initialized to store the LCS.

2. Identifying Smallest Array:

   • The function identifies [2, 3, 6, 8] as the smallest array.

   • This array is removed from the arrays list, leaving [[1, 2, 3, 5, 6, 7, 10], [2, 3, 4, 6, 9]].

3. Main Loop (Finding Common Elements): The function iterates through the smallest array, [2, 3, 6, 8].

   • Iteration 1:

     • num = 2

     • binary_search is used to check for the presence of 2 in the remaining arrays:

       • binary_search([1, 2, 3, 5, 6, 7, 10], 2) returns True.

       • binary_search([2, 3, 4, 6, 9], 2) returns True.

     • Since 2 is found in both remaining arrays, it is added to common_subsequences: [2]

   • Iteration 2:

     • num = 3

     • binary_search([1, 2, 3, 5, 6, 7, 10], 3) returns True.

     • binary_search([2, 3, 4, 6, 9], 3) returns True.

     • '3' is added to common_subsequences: [2, 3]

   • Iteration 3:

     • num = 6

     • binary_search([1, 2, 3, 5, 6, 7, 10], 6) returns True.

     • binary_search([2, 3, 4, 6, 9], 6) returns True.

     • '6' is added to common_subsequences: [2, 3, 6]

   • Iteration 4:

     • num = 8

     • binary_search([1, 2, 3, 5, 6, 7, 10], 8) returns False.

     • Since '8' is not found in all arrays, the loop moves to the next iteration without updating common_subsequences.

4. Iteration Summary:

   Iteration	num	common_subsequences
   1	2	[2]
   2	3	[2, 3]
   3	6	[2, 3, 6]
   4	8	[2, 3, 6]

5. Result Calculation/Final Steps:

   • The function reaches the end of the loop, and the final common_subsequences is [2, 3, 6]. This is returned as the longest common subsequence among the input arrays.

The core idea of this solution is to use a two-pointer technique, similar to merging sorted arrays, to iteratively refine the longest common subsequence (LCS). It starts by assuming the first array is the initial LCS. Then, for each later array, it compares elements with the current LCS using two pointers, keeping only the elements that match in a new subsequence. This new subsequence then becomes the LCS for the next comparison.

• Two-Pointer Technique: This technique enables efficient traversal and comparison of sorted arrays, avoiding nested loops and reducing time complexity.

• Iterative Refinement: The LCS is continuously updated, ensuring that only the longest common elements persist across all arrays.

• Sorted Property: The algorithm leverages the sorted nature of the arrays, allowing for linear time complexity through pointer advancement based on element comparisons.

1. Initialization:

   • common_subsequences is initialized with the first array in the arrays list, representing the initial LCS.

2. Iterative Refinement Loop:

   • The for loop iterates through each array in arrays starting from the second array.

   • new_subsequence is created to store the potential LCS found in the current iteration.

   • Two pointers, array_index (for the current array) and common_subseq_index (for the current common_subsequences) are initialized to 0.

   • The lengths of the current array and the current common_subsequences are stored in array_length and common_subseq_length, respectively.

3. Two-Pointer Comparison:

   • The while loop continues as long as both pointers are within the bounds of their respective lists.

   • The following cases are handled inside the loop:

     • Match: If the elements at the current indices match, the element is appended to new_subsequence, and both pointers are incremented.

     • Element in array is Smaller: The array_index is incremented to find a potential match further in the array.

     • Element in common_subsequences is Smaller: The common_subseq_index is incremented to find a potential match further in the common subsequence.

4. Update and Continue:

   • After processing an array, common_subsequences is updated with new_subsequence, which now represents the longest common subsequence found so far.

   • The process repeats for the next array in arrays.

5. Return Result:

   • After all arrays are processed, the final common_subsequences is returned, representing the longest common subsequence between all the input arrays.

Time Complexity:

• , where is the length of the input string. This is because each character is processed once in the main loop, and while the inner while loop may seem to add additional iterations, each character is removed from the set at most once across all iterations.

Space Complexity:

• , because the set window_chars stores at most 26 characters (the size of the lowercase English alphabet). While this could be written as where is the size of the alphabet, since is fixed at 26, it's considered constant space.

Input: arrays = [[2, 3, 6, 8], [1, 2, 3, 5, 6, 7, 10], [2, 3, 4, 6, 9]]

Step-by-Step Walkthrough:

1. Initialization:

   • common_subsequences is initialized with the first array: [2, 3, 6, 8].

   • This represents the initial longest common subsequence (LCS), as it's the only one considered so far.

2. Main Loop (Iterative Refinement of Common Subsequence):

   • Iteration 1:

     • We compare common_subsequences (currently [2, 3, 6, 8]) with the second array: [1, 2, 3, 5, 6, 7, 10].

     • Initially, array_index = 0, common_subseq_index = 0

       • Comparing 2 (from common_subsequences) and 1 (from the second array): 1 is smaller, so we move to the next element in the second array (array_index = 1).

       • Comparing 2 and 2: Match found! We add 2 to new_subsequence and advance both pointers.

       • Comparing 3 and 3: Match found! We add 3 to new_subsequence and advance both pointers.

       • Comparing 6 and 5: 5 is smaller, so we move to the next element in the second array.

       • Comparing 6 and 6: Match found! We add 6 to new_subsequence and advance both pointers.

       • Comparing 8 and 7: 7 is smaller, so we move to the next element in the second array.

       • Comparing 8 and 10: 8 is smaller, but we've reached the end of common_subsequences. The loop terminates.

     • new_subsequence is now [2, 3, 6]. This becomes the new common_subsequences for the next iteration.

   • Iteration 2:

     • We compare the updated common_subsequences ([2, 3, 6]) with the third array: [2, 3, 4, 6, 9].

     • Initially, array_index = 0, common_subseq_index = 0

       • Comparing 2 and 2: Match found! We add 2 to new_subsequence and advance both pointers.

       • Comparing 3 and 3: Match found! We add 3 to new_subsequence and advance both pointers.

       • Comparing 6 and 4: 4 is smaller, so we move to the next element in the third array.

       • Comparing 6 and 6: Match found! We add 6 to new_subsequence and advance both pointers.

       • The loop terminates as we reach the end of 'common_subsequences.'

     • new_subsequence is now [2, 3, 6]. This is the final LCS.

3. Loop Termination: The loop terminates after iterating through all arrays in the input.

4. Iteration Summary:

   Iteration	Array Compared	Resulting Common Subsequence
   1	[1, 2, 3, 5, 6, 7, 10]	[2, 3, 6]
   2	[2, 3, 4, 6, 9]	[2, 3, 6]

5. Result Calculation/Final Steps:

   • After the loop, the final common_subsequences is [2, 3, 6]. This is returned as the longest common subsequence across all input arrays.

Time Complexity:

• , where:

  • is the total number of arrays.

  • is the average length of the arrays.

• In the worst case, the algorithm might iterate through all elements of each array once for comparison, resulting in linear time complexity with respect to the total number of elements.

Space Complexity:

• , where is the maximum length among the input arrays.

• The algorithm uses a new_subsequence list to store the potential LCS in each iteration. The maximum size of this list would be the maximum length among the input arrays. The common_subsequences list may also hold up to K elements at any given time.

The core idea of this solution is to count the occurrences of each unique character in the string and then calculate the number of substrings that can be formed by each character based on its frequency.

• Frequency Counting: The code first iterates through the string and counts how many times each character appears to use a defaultdict(int). This dictionary automatically initializes new keys (characters) with a default value of 0.

• Combinatorial Calculation: For each character, the code calculates the number of substrings it can form. If a character appears count times, it can be the start and end of count substrings of length 1, count - 1 substrings of length 2, and so on, down to one substring of length count. This forms an arithmetic series, and the sum of this series is count * (count + 1) // 2. This formula derives from n choose 2, which calculates the number of ways to choose 2 items from a set of n items; however, usually, n choose 2 is written as because we aren't taking into consideration subsets of size 1. That's why we add n to the result, making it .

1. Initialization:

   • result is initialized to 0. This variable will store the total number of substrings.

   • letter_counts is a defaultdict(int) to store the frequency of each letter.

2. Counting Letter Occurrences:

   • The code iterates through each letter in the string s.

   • For each letter, its count in letter_counts is incremented by 1.

3. Calculating Substring Counts:

   • The code iterates through the values (counts) of letter_counts.

   • For each count, it calculates the number of substrings possible for that letter using count * (count + 1) // 2 and adds it to the result.

4. Alternative Solution (Commented):

   • A more concise alternative solution is provided using the Counter class from the collections module which effectively does the same thing.

5. Result Calculation/Return:

   • The final result, representing the total number of valid substrings, is returned.

Input: s = "abacabd"

Step-by-Step Walkthrough:

1. Initialization:

   • result is set to 0. It will accumulate the total count of substrings.

   • letter_counts is a defaultdict(int), initially empty, designed to store the frequency of each letter encountered.

2. First Loop (Counting Letters):

   • Iteration 1:

     • The first letter 'a' is processed.

     • The count for 'a' in letter_counts becomes 1: {'a': 1}.

   • Iteration 2:

     • The letter 'b' is processed.

     • The count for 'b' in letter_counts becomes 1: {'a': 1, 'b': 1}.

   • Iteration 3:

     • Another 'a' is encountered.

     • The count for 'a' in letter_counts is incremented to 2: {'a': 2, 'b': 1}.

   • Iteration 4:

     • The letter 'c' is processed.

     • The count for 'c' in letter_counts becomes 1: {'a': 2, 'b': 1, 'c': 1}.

   • Iteration 5:

     • The letter 'a' is encountered again.

     • The count for 'a' in letter_counts is incremented to 3: {'a': 3, 'b': 1, 'c': 1}.

   • Iteration 6:

     • The letter 'b' is processed.

     • The count for 'b' in letter_counts becomes 2: {'a': 3, 'b': 2, 'c': 1}.

   • Iteration 7:

     • The letter 'd' is processed.

     • The count for 'd' in letter_counts becomes 1: {'a': 3, 'b': 2, 'c': 1, 'd': 1}.

3. Loop Termination (First Loop):

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

   • The final state of letter_counts is {'a': 3, 'b': 2, 'c': 1, 'd': 1}.

4. Second Loop (Calculating Substrings):

   • Iteration 1:

     • The first count encountered is 3 (the count of 'a').

     • The formula 3 * (3 + 1) // 2 = 6 calculates the number of substrings that can be formed with 'a' as the start and end letter.

     • This value (6) is added to the result.

   • Iteration 2:

     • The next count is 2 (the count of 'b').

     • The formula 2 * (2 + 1) // 2 = 3 calculates the substrings with 'b'.

     • This value (3) is added to the result, making it 9.

   • Iterations 3 and 4:

     • The remaining counts (1 for 'c' and 1 for 'd') are processed similarly, contributing 1 substring each to the result.

     • The formula 1 * (1 + 1) // 2 = 1 is used for these counts, contributing 2 to the result.

5. Result Calculation/Final Steps:

   • The second loop terminates after processing all counts in letter_counts.

   • The final value of result (11) represents the total number of substrings in "abacabd" that start and end with the same letter.

   • This result is returned by the function.

Time Complexity:

• , where is the length of the string s. This is because the code iterates through the string once to count the character frequencies and then iterates over the dictionary values, which has a maximum size of 26 (for lowercase English letters).

Space Complexity:

• . The space used by letter_counts is constant, as it stores a maximum of 26 entries for the lowercase English alphabet. Even if the input string is huge, the space usage remains the same.

The central concept of this solution is to leverage a two-pointer technique combined with a greedy approach to maximize the number of boxes that can be placed in the warehouse. The solution exploits the flexibility of placing boxes from either end of the warehouse to optimize box placement.

• Greedy Box Selection: The boxes are sorted in descending order, prioritizing larger boxes for placement first.

• Two-Pointer Technique: Two pointers (left and right) are used to track available spaces at both ends of the warehouse.

• Flexible Placement: The algorithm attempts to place each box at either end of the warehouse, whichever allows for placement.

• Early Termination: The process stops when all warehouse rooms have been considered (pointers meet).

1. Initialization: The function starts by initializing max_number to keep track of the number of boxes placed, and two pointers left_index and right_index to represent the current available positions from the left and right ends of the warehouse, respectively.

   max_number = 0 left_index, right_index = 0, len(warehouse) - 1

2. Sorting Boxes: The boxes are sorted in descending order to prioritize placing the largest boxes first.

   for box in sorted(boxes, reverse=True):

3. Placing Boxes: The function iterates through each sorted box and tries to place it either from the left or the right of the warehouse:

   • Left Placement: If the current box can fit in the leftmost available position (warehouse[left_index]), it is placed there, max_number is incremented, and left_index is moved to the next position.

     if box <= warehouse[left_index]: max_number += 1 left_index += 1

   • Right Placement: If the current box does not fit on the left but can fit in the rightmost available position (warehouse[right_index]), it is placed there, max_number is incremented, and right_index is moved to the previous position.

     elif box <= warehouse[right_index]: max_number += 1 right_index -= 1

4. Early Termination: If the two pointers meet (left_index == right_index), the loop terminates early as no more boxes can be placed.

   if left_index == right_index: return max_number

5. Result Calculation/Return: Finally, the function returns the total number of boxes placed.

   return max_number

Input: boxes = [5, 4, 3, 2, 1], warehouse = [4, 3, 1, 5, 2, 3]

Step-by-Step Walkthrough:

1. Initialization:

   • max_number = 0: Initialize the count of placed boxes.

   • left_index = 0: Set the left pointer to the first room.

   • right_index = 5: Set the right pointer to the last room (len(warehouse) - 1).

2. Main Loop (Placing Boxes):

   • Iteration 1 (Box height: 5):

     • Left room height (warehouse[0]) = 4, Right room height (warehouse[5]) = 3

     • The box doesn't fit on either side, so it's skipped.

     • max_number remains 0, indices unchanged.

   • Iteration 2 (Box height: 4):

     • Left room height (warehouse[0]) = 4, Right room height (warehouse[5]) = 3

     • The box fits on the left side (4 <= 4).

     • Place the box, increment max_number to 1, and left_index to 1.

   • Iteration 3 (Box height: 3):

     • Left room height (warehouse[1]) = 3, Right room height (warehouse[5]) = 3

     • The box fits on the left side (3 <= 3).

     • Place the box, increment max_number to 2, and left_index to 2.

   • Iteration 4 (Box height: 2):

     • Left room height (warehouse[2]) = 1, Right room height (warehouse[5]) = 3

     • The box doesn't fit on the left but fits on the right (2 <= 3).

     • Place the box, increment max_number to 3, and decrement right_index to 4.

   • Iteration 5 (Box height: 1):

     • Left room height (warehouse[2]) = 1, Right room height (warehouse[4]) = 2

     • The box fits on the left side (1 <= 1).

     • Place the box, increment max_number to 4, and left_index to 3.

3. Loop Termination:

   • The loop terminates after processing all boxes.

   • Final state: max_number = 4, left_index = 3, right_index = 4

4. Visual Aid:

   Room	Box Height	Left Room Height	Right Room Height	Action	Max Number
   1	5	warehouse[0] = 4	warehouse[5] = 3	Not placed	0
   2	4	warehouse[0] = 4	warehouse[5] = 3	Placed on the left, increase left_index	1
   3	3	warehouse[1] = 3	warehouse[5] = 3	Placed on the left, increase left_index	2
   4	2	warehouse[2] = 1	warehouse[5] = 3	Placed on the right, decrease right_index	3
   5	1	warehouse[2] = 1	warehouse[4] = 2	Placed on the left, increase left_index	4

5. Result Calculation/Final Steps:

   • The algorithm has placed four boxes in total.

   • The function returns max_number, which is 4.

The algorithm successfully places 4 out of 5 boxes in the warehouse, optimizing the placement by considering both ends of the warehouse and prioritizing larger boxes.

Time Complexity:

• O, where is the number of boxes. This is because:

  1. Sorting the boxes takes time.

  2. The later iteration through the sorted boxes takes time.

  3. The dominant factor is the sorting operation, hence overall.

Space Complexity:

• , where is the number of boxes. The space complexity is due to the built-in Tim Sort that in the worst case can use space.

In summary, this solution efficiently places the maximum number of boxes into the warehouse by sorting the boxes and using a two-pointer technique to use space from both ends, with a time complexity of and a space complexity of .

The central concept of this solution is to leverage a sliding window technique combined with a set data structure to efficiently count special substrings. This approach takes advantage of the problem's nature by counting non-special substrings and subtracting them from the total possible substrings.

• Sliding Window: The solution uses a variable-size sliding window to keep track of the current substring being considered.

• Set for Uniqueness: A set is used to maintain the unique characters in the current window, allowing for constant-time lookups.

• Subtraction Strategy: Instead of directly counting special substrings, the algorithm starts with the total number of all possible substrings and subtracts the count of non-special substrings.

1. Initialization:

   window_chars = set() left_index = 0 n = len(s) potential_special_substrings = n * (n + 1) // 2

   The code initializes an empty set window_chars to store unique characters, left_index to mark the start of the window, and calculates the total number of possible substrings . The total number of substrings works with that formula because for each character at position i, there are n - i substrings that end at that position; since we also count substrings of length 1, we add the total number o substrings of length 1 to the total to get the formula .

2. Iterating Through Characters:

   for current_char in s:

   The main loop iterates through each character in the input string.

3. Window Adjustment:

   while current_char in window_chars: window_chars.remove(s[left_index]) left_index += 1

   If the current character is already in the window, the window is shrunk from the left until the current character becomes unique in the window.

4. Counting Non-Special Substrings:

   potential_special_substrings -= left_index

   The number of non-special substrings ending at the current position is equal to left_index. This is subtracted from the total count.

5. Updating Window:

   window_chars.add(current_char)

   The current character is added to the set of window characters.

6. Result Calculation/Return:

   return potential_special_substrings

   After processing all characters, the remaining count in potential_special_substrings represents the number of special substrings.

Input:

Step-by-Step Walkthrough:

1. Initialization:

   • Set window_chars = set() to keep track of unique characters in the current window.

   • Set left_index = 0 to mark the start of the current window.

   • Set n = len(s) = 4 (length of the input string).

   • Calculate potential_special_substrings = n * (n + 1) // 2 = 4 * (4 + 1) // 2 = 10. This represents the total number of all possible substrings.

2. Main Loop (Iterate through each character in the string):

   • Iteration 1: Character 'a' at index 0

     • Current state: window_chars = set(), left_index = 0

     • Check if 'a' is in window_chars: It's not, so we don't need to shrink the window.

     • Calculate non-special substrings ending at this position: left_index = 0

     • Update potential_special_substrings = 10 - 0 = 10

     • Add 'a' to window_chars: window_chars = {'a'}

   • Iteration 2: Character 'b' at index 1

     • Current state: window_chars = {'a'}, left_index = 0

     • Check if 'b' is in window_chars: It's not, so we don't need to shrink the window.

     • Calculate non-special substrings ending at this position: left_index = 0

     • Update potential_special_substrings = 10 - 0 = 10

     • Add 'b' to window_chars: window_chars = {'a', 'b'}

   • Iteration 3: Character 'a' at index 2

     • Current state: window_chars = {'a', 'b'}, left_index = 0

     • Check if 'a' is in window_chars: It is, so we need to shrink the window.

       • Remove 's[left_index]' ('a') from window_chars

       • Increment left_index to 1

     • Calculate non-special substrings ending at this position: left_index = 1

     • Update potential_special_substrings = 10 - 1 = 9

     • Add 'a' to window_chars: window_chars = {'a', 'b'}

   • Iteration 4: Character 'b' at index 3

     • Current state: window_chars = {'a', 'b'}, left_index = 1

     • Check if 'b' is in window_chars: It is, so we need to shrink the window.

       • Remove 's[left_index]' ('b') from window_chars

       • Increment left_index to 2

     • Calculate non-special substrings ending at this position: left_index = 2

     • Update potential_special_substrings = 9 - 2 = 7

     • Add 'b' to window_chars: window_chars = {'a', 'b'}

3. Loop Termination:

   • The loop ends after processing all characters in the string.

   • Final state: window_chars = {'a', 'b'}, left_index = 2, potential_special_substrings = 7

4. Visual Aid:

   Here's a table summarizing each iteration:

   Position	Character	Left Index	Window Chars	Non-special Count	Potential Special
   1	a	0	['a']	0	10
   2	b	0	['a', 'b']	0	10
   3	a	1	['a', 'b']	1	9
   4	b	2	['a', 'b']	2	7

5. Result Calculation:

   • The algorithm starts with the total number of all possible substrings (10) and subtracts the count of non-special substrings at each step.

   • The final value of potential_special_substrings (7) represents the count of special substrings.

   • Therefore, the function returns 7 as the final count of special substrings in "abab".

The special substrings are: "a", "b", "a", "b", "ab", "ba", "ab".

The central concept of this solution is to leverage a sliding window technique combined with an array to track the last occurrence of each character. This approach efficiently counts special substrings by maintaining the most recent valid window for each character encountered.

• Sliding Window: The solution uses a variable-size sliding window to keep track of the current substring being considered.

• Last Occurrence Array: An array is used to store the index of the last occurrence of each character, allowing for constant-time lookups and updates.

• Counting Strategy: The algorithm counts special substrings directly by calculating the number of valid substrings ending at each character.

1. Initialization:

   special_substrings_count = 0 last_occurrence = [-1] * 26 start_index = 0

   The code initializes the count of special substrings, an array to store the last occurrence of each character (initialized to -1), and the start index of the current window.

2. Iterating Through Characters:

   for current_index, char in enumerate(s):

   The main loop iterates through each character in the input string, keeping track of both the character and its index.

3. Character Index Calculation:

   char_index = ord(char) - ord('a')

   Calculates the index for the current character in the last_occurrence array.

4. Window Adjustment:

   if last_occurrence[char_index] >= start_index: start_index = last_occurrence[char_index] + 1

   If the current character has appeared before within the current window, the window's start is moved to just after the last occurrence of this character.

5. Counting Special Substrings:

   special_substrings_count += current_index - start_index + 1

   Counts the number of special substrings ending at the current character. This is equal to the size of the current window.

6. Updating Last Occurrence:

   last_occurrence[char_index] = current_index

   Updates the last occurrence index for the current character.

7. Result Calculation/Return:

   return special_substrings_count

   After processing all characters, the total count of special substrings is returned.

Time Complexity:

• , where is the length of the input string. This is because the algorithm iterates through each character in the string exactly once, performing constant-time operations at each step.

Space Complexity:

• , because the last_occurrence array has a fixed size of 26 elements (one for each lowercase English letter), regardless of the input string's length. While this could be written as where is the size of the alphabet, since is fixed at 26, it's considered constant space.

Input:

Step-by-Step Walkthrough:

1. Initialization:

   • Set special_substrings_count = 0 to keep track of the number of special substrings.

   • Initialize last_occurrence = [-1] * 26, an array to store the last occurrence index of each lowercase letter.

   • Set start_index = 0 to mark the start of the current valid window.

   • Set n = len(s) = 4 (length of the input string).

2. Main Loop (Iterate through each character in the string):

   • Iteration 1: Character 'a' at index 0

     • Calculate char_index = ord('a') - ord('a') = 0

     • Check if 'a' occurred within the current window:

       • last_occurrence[0] = -1, which is less than start_index = 0, so no adjustment is needed.

     • Count special substrings:

       • New special substrings = current_index - start_index + 1 = 0 - 0 + 1 = 1

       • Update special_substrings_count = 0 + 1 = 1

     • Update last_occurrence[0] = 0

   • Iteration 2: Character 'b' at index 1

     • Calculate char_index = ord('b') - ord('a') = 1

     • Check if 'b' occurred within the current window:

       • last_occurrence[1] = -1, which is less than start_index = 0, so no adjustment is needed.

     • Count special substrings:

       • New special substrings = current_index - start_index + 1 = 1 - 0 + 1 = 2

       • Update special_substrings_count = 1 + 2 = 3

     • Update last_occurrence[1] = 1

   • Iteration 3: Character 'a' at index 2

     • Calculate char_index = ord('a') - ord('a') = 0

     • Check if 'a' occurred within the current window:

       • last_occurrence[0] = 0, which is not less than start_index = 0

       • Update start_index = last_occurrence[0] + 1 = 0 + 1 = 1

     • Count special substrings:

       • New special substrings = current_index - start_index + 1 = 2 - 1 + 1 = 2

       • Update special_substrings_count = 3 + 2 = 5

     • Update last_occurrence[0] = 2

   • Iteration 4: Character 'b' at index 3

     • Calculate char_index = ord('b') - ord('a') = 1

     • Check if 'b' occurred within the current window:

       • last_occurrence[1] = 1, which is not less than start_index = 1

       • Update start_index = last_occurrence[1] + 1 = 1 + 1 = 2

     • Count special substrings:

       • New special substrings = current_index - start_index + 1 = 3 - 2 + 1 = 2

       • Update special_substrings_count = 5 + 2 = 7

     • Update last_occurrence[1] = 3

3. Loop Termination:

   • The loop ends after processing all characters in the string.

   • Final state: special_substrings_count = 7, start_index = 2, last_occurrence = [2, 3, -1, -1, -1, ...]

4. Visual Aid:

   Here's a table summarizing each iteration:

   Position	Character	Char Index	Start Index	New Special	Total Special	Last Occurrence
   1	a	0	0	1	1	[0, -1, -1, -1, -1] + ["..."]
   2	b	1	0	2	3	[0, 1, -1, -1, -1] + ["..."]
   3	a	0	1	2	5	[2, 1, -1, -1, -1] + ["..."]
   4	b	1	2	2	7	[2, 3, -1, -1, -1] + ["..."]

5. Result Calculation:

   • The algorithm directly calculates the count of special substrings at each step.

   • For each character, it counts the number of special substrings ending at that character, which is the length of the current valid window.

   • The special_substrings_count is updated cumulatively, so the final value (7) represents the total count of special substrings in "abab".

The special substrings are: "a", "b", "a", "b", "ab", "ba", "ab".

The central concept of this solution is to leverage an iterative traversal of the Binary Search Tree (BST) to split it into two subtrees based on a target value. This approach efficiently maintains the BST structure while separating nodes into two categories: those with values less than or equal to the target, and those with values greater than the target.

• Iterative Traversal: The solution uses an iterative method with a stack to traverse the BST, avoiding the potential stack overflow issues of a recursive approach for very deep trees.

• In-place Reconstruction: Instead of creating new nodes, the algorithm reconstructs the two subtrees by modifying the existing tree structure, which is memory-efficient.

• Path Tracking: A stack is used to keep track of the traversal path, which is crucial for the reconstruction phase.

1. Initialization:

   smaller_equal_subtree, greater_subtree = None, None path_stack = []

   We initialize two variables to store the roots of our resulting subtrees and an empty stack to keep track of our traversal path.

2. Downward Traversal:

   current = root while current: path_stack.append(current) if current.val <= target: current = current.right else: current = current.left

   This loop traverses down the tree, always choosing the right child if the current node's value is less than or equal to the target, and the left child otherwise. Each visited node is added to the path_stack. This ensures we only traverse one path of the tree, which is key to the algorithm's efficiency.

3. Upward Reconstruction:

   while path_stack: node = path_stack.pop() if node.val <= target: node.right = smaller_equal_subtree smaller_equal_subtree = node else: node.left = greater_subtree greater_subtree = node

   We now pop nodes from the stack (which reverses our traversal order) and reconstruct our two subtrees. If a node's value is less than or equal to the target, we make it the new root of the smaller_equal_subtree and set its right child to the previous smaller_equal_subtree. If it's greater, we do the same for the greater_subtree, but with the left child. This preserves the BST property in both resulting trees.

4. Result Return:

   return [smaller_equal_subtree, greater_subtree]

   Finally, we return a list containing the roots of our two reconstructed subtrees.

Time Complexity:

• , where is the height of the tree. This is because we traverse down one path of the tree and then back up. In the worst case (a completely unbalanced tree), this becomes , where is the number of nodes.

Space Complexity:

• , where is the height of the tree. This is due to the stack used to store the path, which in the worst case (a completely unbalanced tree) stores all nodes, resulting in space complexity.

The space complexity is generally better than a recursive solution, especially for very deep trees, as it avoids the overhead of recursive function calls on the system stack.

Input:

A visual representation of the input BST:

Step-by-Step Walkthrough:

1. Initialization:

   • Initialize smaller_equal_subtree = None and greater_subtree = None.

   • Create an empty list path_stack = [] to store the traversal path.

   • Set current = root (the node with value 4).

2. Main Loop (Traversing down the tree):

   • Iteration 1:

     • current.val (4) > target (2)

     • Append current (4) to path_stack

     • Move to the left child: current = current.left (node with value 2)

     • path_stack is now [4]

   • Iteration 2:

     • current.val (2) <= target (2)

     • Append current (2) to path_stack

     • Move to the right child: current = current.right (node with value 3)

     • path_stack is now [4, 2]

   • Iteration 3:

     • current.val (3) > target (2)

     • Append current (3) to path_stack

     • Move to the left child: current = current.left (which is None)

     • path_stack is now [4, 2, 3]

3. Loop Termination:

   • The loop ends when current becomes None

   • At this point, path_stack contains [4, 2, 3]

4. Visual Aid: Traversal Summary

   Iteration	Current Node	Path Stack	Direction
   1	4	[4]	Left
   2	2	[4, 2]	Right
   3	3	[4, 2, 3]	Left

5. Result Calculation/Final Steps:

   • Now we reconstruct the subtrees by popping nodes from path_stack:

   • Pop 3:

     • 3 > 2, so set node.left = greater_subtree (None)

     • Update greater_subtree = node (3)

     • Visual representation of the greater subtree:

       

   • Pop 2:

     • 2 <= 2, so set node.right = smaller_equal_subtree (None)

     • Update smaller_equal_subtree = node (2)

     • Visual representation of the smaller or equal subtree:

       

   • Pop 4:

     • 4 > 2, so set node.left = greater_subtree (3)

     • Update greater_subtree = node (4)

     • Visual representation of the greater subtree:

       

   • The loop ends as path_stack is empty

   Final state:

   • smaller_equal_subtree: Root 2, with left child 1

   • greater_subtree: Root 4, with left child 3 and right subtree unchanged (6, 5, 7)

The function returns [smaller_equal_subtree, greater_subtree], which represents the two split subtrees based on target value 2.

The central concept of this solution is to leverage a recursive traversal of the Binary Search Tree (BST) to split it into two subtrees based on a target value. This approach efficiently maintains the BST structure while separating nodes into two categories: those with values less than or equal to the target, and those with values greater than the target.

• Recursive Traversal: The solution uses a recursive method to traverse the BST, which aligns closely with the tree's inherent recursive structure.

• In-place Reconstruction: Instead of creating new nodes, the algorithm reconstructs the two subtrees by modifying the existing tree structure, which is memory-efficient.

• Divide and Conquer: The problem is solved by breaking it down into smaller subproblems (subtrees) and combining their results.

1. Helper Function Definition:

   def split_tree(node: Optional[TreeNode]) -> List[Optional[TreeNode]]:

   A helper function is defined to perform the recursive splitting. It returns a list of two subtrees: one with nodes less than or equal to the target, and one with nodes greater than the target.

2. Base Case:

   if not node: return [None, None]

   If the current node is None, we've reached a leaf, so we return two None values representing empty subtrees.

3. Recursive Case for Nodes Less Than or Equal to Target:

   if node.val <= target: smaller_equal_subtree, greater_subtree = split_tree(node.right) node.right = smaller_equal_subtree return [node, greater_subtree]

   If the current node's value is less than or equal to the target, we recursively split its right subtree. We then update the current node's right child to be the smaller/equal part of the split right subtree. We return the current node as the root of the smaller/equal subtree along with the greater part of the split right subtree.

4. Recursive Case for Nodes Greater Than Target:

   else: smaller_equal_subtree, greater_subtree = split_tree(node.left) node.left = greater_subtree return [smaller_equal_subtree, node]

   If the current node's value is greater than the target, we recursively split its left subtree. We then update the current node's left child to be the greater part of the split left subtree. We return the smaller/equal part of the split left subtree along with the current node as the root of the greater subtree.

5. Main Function Call:

   return split_tree(root)

   The main function simply calls the helper function with the root of the tree and returns its result.

Time Complexity:

• , where is the height of the tree. This is because we traverse down one path of the tree, making a decision at each node whether to go left or right based on the target value. In the worst case (a completely unbalanced tree), this becomes , where is the number of nodes.

Space Complexity:

• , where is the height of the tree. This is due to the recursive call stack, which in the worst case (a completely unbalanced tree) can go as deep as the number of nodes, resulting in space complexity.

The space complexity could be a concern for very deep trees, as it might lead to stack overflow errors. However, for balanced trees, the space complexity remains logarithmic in the number of nodes.

Input:

A visual representation of the input BST:

Step-by-Step Walkthrough:

1. Initialization:

   • The function splitBST2 is called with root (node with value 4) and target = 2.

   • It immediately calls the helper function split_tree(root).

2. Main Recursive Process:

   • Recursive call at depth 0:

     • Current node: 4

     • 4 > 2, so we recurse on the left child

   • Recursive call at depth 1:

     • Current node: 2

     • 2 <= 2, so we recurse on the right child

   • Recursive call at depth 2:

     • Current node: 3

     • 3 > 2, so we recurse on the left child

   • Recursive call at depth 3:

     • Current node: None

     • Base case reached, return [None, None]

3. Unwinding the Recursion:

   • Back to depth 2 (node 3):

     • Received [None, None] from left child

     • Set node.left = None (greater subtree)

     • Return [None, 3] (smaller_equal_subtree, node)

     • Visual representation of the smaller or equal subtree:

       

   • Back to depth 1 (node 2):

     • Received [None, 3] from right child

     • Set node.right = None (smaller_equal_subtree)

     • Return [2, 3] (node, greater_subtree)

     • Visual representation of the greater subtree:

       

   • Back to depth 0 (node 4):

     • Received [2, 3] from left child

     • Set node.left = 3 (greater subtree)

     • Return [2, 4] (smaller_equal_subtree, node)

     • Visual representation of the updated greater subtree:

       

4. Visual Aid: Reconstruction Summary

   Depth	Processed Node	Subtree Added To	Smaller/Equal Subtree Root	Greater Subtree Root
   2	3	greater		3
   1	2	smaller_equal	2	3
   0	4	greater	2	4

5. Result Calculation/Final Steps:

   • The recursion completes, and split_tree(root) returns [2, 4].

   • This result is directly returned by splitBST2.

   Final state:

   • smaller_equal_subtree: Root 2, with left child 1 and right child None

   • greater_subtree: Root 4, with left child 3 and right subtree unchanged (6, 5, 7)

The function returns [smaller_equal_subtree, greater_subtree], which represents the two split subtrees based on target value 2. The recursive approach efficiently splits the tree in a single pass, leveraging the BST property to make decisions at each node.

The central concept of this solution is to leverage an iterative traversal of the Binary Search Tree (BST) to split it into two subtrees based on a target value. This approach reconstructs the two subtrees in-place using pointer manipulation, without using additional data structures like stacks or recursion.

• Iterative Traversal: The solution uses an iterative method to traverse the BST, avoiding potential stack overflow issues associated with recursive approaches for very deep trees.

• In-Place Reconstruction: The algorithm reconstructs the two subtrees by modifying the existing tree structure through pointer manipulation, which is memory-efficient.

• Dummy Nodes: The use of dummy nodes simplifies the process of building the new subtrees, eliminating the need for special cases when adding the first node to each subtree.

1. Initialization:

   dummy_smaller_equal = TreeNode() dummy_greater = TreeNode() current_smaller_equal = dummy_smaller_equal current_greater = dummy_greater current_node = root

   We initialize dummy nodes for both subtrees and set up pointers to the current positions in each subtree. current_node is initialized to the root of the original tree.

2. Main Traversal Loop:

   while current_node is not None:

   This loop continues until we've processed all nodes in the original tree.

3. Processing Nodes Less Than or Equal to Target:

   if current_node.val <= target: current_smaller_equal.right = current_node current_smaller_equal = current_node next_node = current_node.right current_smaller_equal.right = None # Cut off right subtree

   If the current node's value is less than or equal to the target, we add it to the smaller/equal subtree. We update pointers, save the right child as the next node to process, and cut off the right subtree to maintain the BST property.

4. Processing Nodes Greater Than Target:

   else: current_greater.left = current_node current_greater = current_node next_node = current_node.left current_greater.left = None # Cut off left subtree

   If the current node's value is greater than the target, we add it to the greater subtree. We update pointers, save the left child as the next node to process, and cut off the left subtree to maintain the BST property.

5. Moving to the Next Node:

   current_node = next_node

   We move to the next node to process, which was saved earlier as either the left or right child of the current node.

6. Result Preparation and Return:

   smaller_equal_subtree = dummy_smaller_equal.right greater_subtree = dummy_greater.left return [smaller_equal_subtree, greater_subtree]

   We extract the actual roots of our new subtrees (skipping the dummy nodes) and return them as a list.

Time Complexity:

• , where is the height of the tree. This is because the algorithm traverses a single path from the root to a leaf, making a decision at each node whether to go left or right based on the target value.

• In the best and average cases (balanced BST), this results in time complexity, where is the number of nodes in the tree.

• In the worst case (completely unbalanced BST), this could become .

Space Complexity:

• , as it uses only a constant amount of extra space regardless of input size. The algorithm uses a fixed number of pointers and doesn't rely on any additional data structures that grow with the input size.

This approach is particularly helpful for large trees or in memory-constrained environments, as it avoids the risk of stack overflow and uses minimal extra memory.

Input:

A visual representation of the input BST:

Step-by-Step Walkthrough:

1. Initialization:

   • Create dummy_smaller_equal and dummy_greater as dummy TreeNodes.

   • Set current_smaller_equal = dummy_smaller_equal

   • Set current_greater = dummy_greater

   • Set current_node = root (node with value 4)

2. Main Loop:

   • Iteration 1:

     • current_node.val (4) > target (2)

     • Add to greater subtree:

       • Set current_greater.left = current_node (4)

       • Update current_greater = current_node (4)

     • Set next_node = current_node.left (2)

     • Cut off left subtree: current_greater.left = None

     • Update current_node = next_node (2)

     • Visual representation of the greater subtree with dummy node:

       

   • Iteration 2:

     • current_node.val (2) <= target (2)

     • Add to smaller_equal subtree:

       • Set current_smaller_equal.right = current_node (2)

       • Update current_smaller_equal = current_node (2)

     • Set next_node = current_node.right (3)

     • Cut off right subtree: current_smaller_equal.right = None

     • Update current_node = next_node (3)

     • Visual representation of the smaller or equal subtree with dummy node:

       

   • Iteration 3:

     • current_node.val (3) > target (2)

     • Add to greater subtree:

       • Set current_greater.left = current_node (3)

       • Update current_greater = current_node (3)

     • Set next_node = current_node.left (None)

     • Cut off left subtree: current_greater.left = None

     • Update current_node = next_node (None)

     • Visual representation of the greater subtree with dummy node:

       

3. Loop Termination:

   • The loop ends when current_node becomes None

4. Visual Aid: Iteration Summary

   Iteration	Current Smaller/Equal	Current Greater	Next Node
   1		4	2
   2	2	4	3
   3	2	3

5. Result Calculation/Final Steps:

   • Set smaller_equal_subtree = dummy_smaller_equal.right (2)

   • Set greater_subtree = dummy_greater.left (4)

   Final state:

   • smaller_equal_subtree: Root 2, with left child 1 and right child None

   • greater_subtree: Root 4, with left child 3 and right subtree unchanged (6, 5, 7)

The final visual representation of the two subtrees:

• Smaller or Equal Subtree:

  

• Greater Subtree:

  

The function returns [smaller_equal_subtree, greater_subtree], which represents the two split subtrees based on target value 2. This iterative approach efficiently splits the tree in a single pass without using recursion or additional stack space, making it memory-efficient and suitable for very deep trees.