This function computes the Longest Common Subsequence (LCS) of two strings using dynamic programming. The LCS is the longest sequence of characters that appear in the same relative order, but not necessarily contiguously...

The `longestCommonSubsequence` function employs a dynamic programming approach to find the LCS. It constructs a 2D table `dp` where `dp[i][j]` stores the length of the LCS for the first `i` characters of `text1` and the first `j` characters of `text2`. The recurrence relation is: if `text1[i-1] == text2[j-1]`, then `dp[i][j] = dp[i-1][j-1] + 1`; otherwise, `dp[i][j] = max(dp[i-1][j], dp[i][j-1])`. The base cases are `dp[0][j] = 0` and `dp[i][0] = 0`. After filling the table, the LCS string is reconstructed by backtracking from `dp[m][n]`. The time complexity is O(m*n) and the space complexity is also O(m*n), where m and n are the lengths of the input strings. Edge cases like empty strings are implicitly handled by the table initialization.

function longestCommonSubsequence(text1, text2): m = length of text1 n = length of text2 dp table of size (m+1)x(n+1), initialized to 0 for i from 1 to m: for j from 1 to n: if text1[i-1] == text2[j-1]: dp[i][j] = dp[i-1...

This function checks if a given binary tree adheres to the properties of a Binary Search Tree (BST). A BST is a binary tree where for each node, all values in its left subtree are less than the node's value, and all valu...

The `isValidBST` function determines if a binary tree is a valid BST by employing a recursive helper function `validate`. The `validate` function takes a node and its valid range (`low`, `high`) as arguments. The base case is when the node is null, which is considered a valid BST (returns `true`). For a non-null node, its value must be strictly between `low` and `high`. If it violates this, the tree is invalid. Otherwise, the function recursively calls itself for the left child, updating the `high` bound to the current node's value, and for the right child, updating the `low` bound to the current node's value. This ensures that all nodes in the left subtree are less than the current node and all nodes in the right subtree are greater. The time complexity is O(N), where N is the number of nodes, as each node is visited once. The space complexity is O(H) in the average case (balanced tree) and O(N) in the worst case (skewed tree) due to the recursion stack, where H is the height of the tree.

class TreeNode: value left_child right_child function isValidBST(root): return validate(root, -infinity, +infinity) function validate(node, low, high): if node is null: return true if node.value <= low or node.value >= h...

This function finds a peak element in an array. A peak element is an element that is strictly greater than its neighbors. The array may contain multiple peaks; in that case, returning the index to any one of the peaks is...

The `findPeakElement` function utilizes a binary search approach to efficiently locate a peak element in an array. A peak element is defined as an element greater than its adjacent elements. The algorithm iteratively narrows down the search space by comparing the middle element with its right neighbor. If the middle element is greater, the peak must lie in the left half (including the middle element); otherwise, it must be in the right half. This process continues until the search space is reduced to a single element, which is guaranteed to be a peak. The time complexity is O(log n) due to the binary search, and the space complexity is O(1) as it uses a constant amount of extra space. Edge cases such as empty or single-element arrays are handled explicitly.

function findPeakElement(nums): if nums is empty: return -1 if nums has 1 element: return 0 left = 0 right = length of nums - 1 while left < right: mid = floor((left + right) / 2) if nums[mid] > nums[mid + 1]: right = mi...

This implementation simulates a queue data structure using two stacks. A queue follows a First-In, First-Out (FIFO) principle, while stacks follow a Last-In, First-Out (LIFO) principle. By strategically using two stacks,...

The `MyQueue` class uses two stacks: `stack1` for enqueue operations and `stack2` for dequeue operations. When an element is pushed (`push`), it's simply added to `stack1`. When an element needs to be dequeued (`pop` or `peek`), if `stack2` is empty, all elements from `stack1` are transferred to `stack2`. This reversal ensures that the element that was pushed first into `stack1` (and is at the bottom of `stack1`) becomes the top element of `stack2`. The `pop` operation then removes and returns the top element from `stack2`. If `stack2` is still empty after the transfer, it means the queue is empty. The `peek` operation is similar to `pop` but only returns the top element without removing it. The `empty` method checks if both stacks are empty. The amortized time complexity for `push` is O(1). For `pop` and `peek`, the worst-case complexity is O(N) when a transfer is needed, but the amortized complexity is O(1) because each element is pushed and popped from each stack at most once. The space complexity is O(N) to store the elements.

class MyQueue: stack1 (for enqueue) stack2 (for dequeue) constructor(): initialize stack1 and stack2 as empty stacks push(x): push x onto stack1 pop(): if stack2 is empty: while stack1 is not empty: move top element from...

This function calculates the total number of vowels present in a given string. It iterates through each character of the string, converts it to lowercase, and checks if it is one of the five standard English vowels (a, e...

The `countVowels` function iterates through the input string once. For each character, it performs a constant-time lookup in a `Set` to determine if it's a vowel. The string is converted to lowercase upfront to handle both uppercase and lowercase vowels. The time complexity is O(N), where N is the length of the string, because each character is processed once. The space complexity is O(1) because the `Set` of vowels has a fixed size, and we only use a few variables for counting and iteration. Edge cases like empty or null strings are handled by returning 0.

function countVowels(input_string): if input_string is null or empty: return 0 vowels = set containing 'a', 'e', 'i', 'o', 'u' count = 0 lowercase_string = convert input_string to lowercase for each character in lowercas...

This function finds the Kth smallest element in a matrix where both rows and columns are sorted in ascending order. It leverages a min-heap (priority queue) to efficiently explore the matrix. By starting with the smalles...

The `kthSmallest` function uses a min-heap to efficiently find the Kth smallest element. It initializes the heap with the first element from each row. Then, it repeatedly extracts the minimum element from the heap `k` times. After extracting an element `(value, row, col)`, if there is a next element in the same row (`matrix[row][col + 1]`), it is added to the heap. This process ensures that at each step, the smallest available element across all considered rows is at the top of the heap. The time complexity is O(min(N, K) + K log min(N, M)), where N is the number of rows and M is the number of columns. This is because we initially push N elements (or K if K < N) and then perform K pop/push operations, each taking O(log min(N, M)) time (where min(N,M) is the heap size). The space complexity is O(min(N, M)) for storing elements in the heap. Edge cases like empty matrices or invalid `k` are handled.

function kthSmallest(matrix, k): rows = number of rows in matrix cols = number of columns in matrix if matrix is empty or has empty rows: return -1 if k is invalid (k < 1 or k > rows * cols): return -1 // Use a min-prior...

This function implements a modified binary search algorithm to find the index of the first occurrence of a specific target value within a sorted array. If the target value appears multiple times, it specifically returns...

The `findFirstOccurrence` function adapts the standard binary search. When the `target` is found at `mid`, instead of terminating, it records `mid` as a potential `result` and continues searching in the left half (`high = mid - 1`) to find an even earlier occurrence. If `arr[mid]` is less than `target`, the search continues in the right half (`low = mid + 1`). If `arr[mid]` is greater than `target`, the search continues in the left half (`high = mid - 1`). The loop terminates when `low` exceeds `high`. The `result` variable holds the index of the first occurrence found, or -1 if the target was never encountered. The time complexity is O(log N) because it's a binary search. The space complexity is O(1) as it uses a constant amount of extra space.

function findFirstOccurrence(array, target): low = 0 high = length of array - 1 result = -1 if array is empty: return -1 while low <= high: mid = low + floor((high - low) / 2) if array[mid] == target: result = mid high =...

This function finds a peak element in a 2D grid. A peak element is defined as an element that is strictly greater than all its adjacent neighbors (up, down, left, right). The algorithm uses a binary search approach on co...

The `findPeakElementII` function employs a binary search strategy on the columns of the 2D grid. In each step of the binary search, it identifies the row containing the maximum element within the current middle column. This maximum element is then compared with its left and right neighbors. If the element is greater than both horizontal neighbors, it is a peak element and its indices are returned. If the left neighbor is greater, the search space for columns is reduced to the left half; otherwise, it's reduced to the right half. This process guarantees finding a peak because at each step, we are moving towards a potentially larger element. The time complexity is O(M log N) or O(N log M), where M is the number of rows and N is the number of columns, depending on whether we binary search rows or columns. The space complexity is O(1) as it uses a constant amount of extra space.

function findPeakElementII(matrix): rows = number of rows in matrix cols = number of columns in matrix low_col = 0 high_col = cols - 1 while low_col <= high_col: mid_col = floor((low_col + high_col) / 2) max_row = find_m...