This program implements an inorder traversal for a Binary Search Tree (BST). Inorder traversal visits nodes in a specific order: left subtree, root, then right subtree. For a BST, this naturally results in visiting the n...

The inorder traversal algorithm for a Binary Search Tree (BST) is typically implemented recursively. The `inorder_helper` function takes a node and a mutable vector to store the traversal result. If the current node is not `None`, it first recursively calls itself on the left child, then pushes the current node's value into the result vector, and finally recursively calls itself on the right child. This order ensures that for a BST, the values are visited in ascending order. The time complexity is O(N), where N is the number of nodes in the tree, because each node is visited exactly once. The space complexity is O(H) in the average case (for a balanced tree) and O(N) in the worst case (for a skewed tree), due to the recursion stack depth, plus O(N) for storing the result vector.

function inorder_traversal(root): result_list = empty list inorder_helper(root, result_list) return result_list function inorder_helper(node, result_list): if node is not null: inorder_helper(node.left, result_list) add...

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

The Longest Common Subsequence (LCS) algorithm uses dynamic programming to find the longest sequence of characters common to two strings. A 2D DP table `dp[i][j]` is constructed, where `dp[i][j]` represents the length of the LCS of the first `i` characters of `text1` and the first `j` characters of `text2`. If the `i`-th character of `text1` matches the `j`-th character of `text2`, then `dp[i][j] = dp[i-1][j-1] + 1`. Otherwise, `dp[i][j]` is the maximum of `dp[i-1][j]` and `dp[i][j-1]`. The time complexity is O(m*n), where m and n are the lengths of the two strings, due to filling the DP table. The space complexity is also O(m*n) for the DP table. After filling the table, the LCS string is reconstructed by backtracking from `dp[m][n]`. Edge cases like empty strings are handled by returning an empty string.

function longest_common_subsequence(text1, text2): m = length of text1 n = length of text2 if m == 0 or n == 0, return "" create 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 text...

This program implements the Merge Sort algorithm, a highly efficient, comparison-based sorting technique. It follows the divide-and-conquer paradigm by recursively dividing the input array into smaller halves, sorting th...

Merge Sort is a recursive sorting algorithm that divides an array into two halves, sorts each half recursively, and then merges the two sorted halves. The base case for the recursion is when the array has 0 or 1 element, as it is already sorted. The `merge` function is crucial; it takes two sorted subarrays and merges them into a single sorted array. It uses three pointers to iterate through the left subarray, right subarray, and the destination array, comparing elements and placing the smaller one into the destination. The time complexity of Merge Sort is O(N log N) in all cases (best, average, and worst), where N is the number of elements, due to the balanced division and linear time merging. The space complexity is O(N) because it requires auxiliary space to store the merged subarrays during the merge step. This implementation uses `clone()` for generic type `T` to handle various data types.

function merge_sort(array): n = length of array if n <= 1, return mid = n / 2 left_half = copy of array from index 0 to mid-1 right_half = copy of array from index mid to n-1 merge_sort(left_half) merge_sort(right_half)...

This function finds any peak element in a given array of integers. A peak element is defined as an element that is strictly greater than its immediate neighbors. This is useful in scenarios where you need to find a local...

The `find_peak_element` function iterates through an array to locate a peak element. It first handles edge cases: an empty array returns `None`, and a single-element array's only element is considered a peak. It then checks the first and last elements against their single neighbor. For the remaining elements, it checks if the current element is greater than both its left and right neighbors. The time complexity is O(n) because, in the worst case, it might iterate through the entire array. The space complexity is O(1) as it uses a constant amount of extra space. The algorithm guarantees finding a peak if one exists because the problem statement implies that `nums[i] != nums[i+1]` for all valid `i`, and we assume `nums[-1] = nums[n] = -infinity`.

function find_peak_element(array): n = length of array if n is 0, return None if n is 1, return index 0 if array[0] > array[1], return index 0 if array[n-1] > array[n-2], return index n-1 for i from 1 to n-2: if array[i]...

This Rust function `find_peak_element` identifies an index of a peak element within a given array of integers. A peak element is defined as an element that is strictly greater than its immediate neighbors. The function h...

The `find_peak_element` function aims to find any peak element in an array. A peak element is greater than its neighbors. The problem statement implies that `nums[-1]` and `nums[n]` are negative infinity, ensuring that at least one peak always exists. The algorithm first handles edge cases: an empty array returns -1, and a single-element array returns index 0. It then checks if the first or last elements are peaks. For arrays with more than two elements, it employs a binary search. In each step, it checks if the middle element `mid` is a peak. If not, it determines which half (left or right) must contain a peak based on the comparison of `nums[mid]` with its neighbors. If `nums[mid-1] > nums[mid]`, a peak must exist in the left half because the array is increasing towards `mid-1` and then decreases at `mid`. Conversely, if `nums[mid+1] > nums[mid]`, a peak must exist in the right half. 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.

function find_peak_element(nums): n = length of nums if n is 0: return -1 if n is 1: return 0 if nums[0] > nums[1]: return 0 if nums[n-1] > nums[n-2]: return n-1 left = 1 right = n-2 while left <= right: mid = left + (ri...

This Rust function `reverse_words` takes a string and reverses the order of words within it. Words are defined as sequences of non-space characters. The function handles multiple spaces between words, as well as leading...

The `reverse_words` function first uses `s.split_whitespace()` to split the input string `s` into an iterator of string slices, where each slice is a word. `split_whitespace()` automatically handles multiple spaces between words and trims leading/trailing whitespace. The `collect()` method then gathers these words into a `Vec<&str>`. An edge case check `words.is_empty()` handles inputs that are empty or contain only whitespace, returning an empty string. The `into_iter().rev().collect()` chain reverses the order of the words in the vector. Finally, `reversed_words.join(" ")` concatenates the reversed words into a new `String`, using a single space as the separator between them. The time complexity is dominated by the splitting and joining operations, which are typically O(n) where n is the length of the string. The space complexity is O(n) in the worst case, as a new string is created to store the result, and intermediate vectors might hold up to n characters.

function reverse_words(s): words = split s by whitespace, filter out empty strings if words is empty: return empty string reverse the order of words join words with a single space return the joined string

This Rust function `sum_array` computes the sum of all integer elements within a given array. It iterates through each number in the array and adds it to a running total. This is a fundamental operation for data analysis...

The `sum_array` function initializes a mutable variable `total` to 0. It then iterates through each element `num` in the input slice `nums`. In each iteration, the current `num` is added to `total`. After the loop finishes, `total` holds the sum of all elements. If the input slice `nums` is empty, the loop will not execute, and the function will correctly return the initial value of `total`, which is 0. The time complexity is O(n), where n is the number of elements in the array, because each element is visited exactly once. The space complexity is O(1) as it only uses a single variable to store the sum.

function sum_array(nums): total = 0 for each number num in nums: total = total + num return total

This Rust function `length_of_lis` calculates the length of the Longest Increasing Subsequence (LIS) within an array of integers. It employs an efficient O(n log n) algorithm using patience sorting principles. For each n...

The `length_of_lis` function finds the length of the Longest Increasing Subsequence (LIS) in O(n log n) time. It maintains a `tails` vector where `tails[i]` stores the smallest ending element of all increasing subsequences of length `i+1`. This `tails` vector is always sorted in increasing order. For each number `num` in the input array `nums`, a binary search is performed on `tails` to find the index `low` where `num` should be placed. If `num` is greater than all elements in `tails` (i.e., `low == tails.len()`), it means `num` can extend the longest subsequence found so far, so it's appended to `tails`. Otherwise, `num` is smaller than or equal to `tails[low]`. In this case, `tails[low]` is replaced by `num`. This replacement is key: it allows for an increasing subsequence of the same length (`low + 1`) but with a smaller ending element, which increases the chances of finding longer subsequences later. The final length of the LIS is simply the length of the `tails` vector. The time complexity is O(n log n) because for each of the n elements, we perform a binary search (log n). The space complexity is O(n) in the worst case, as the `tails` vector can grow up to the size of the input array.

function length_of_lis(nums): if nums is empty: return 0 tails = empty list for each number num in nums: perform binary search on tails to find the smallest element greater than or equal to num let index be the result of...