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For more details, please go through - <a href="https://www.geeksforgeeks.org/python/bisect-algorithm-functions-in-python/">Bisect Algorithm Functions in Python</a>

Bisect Algorithm Functions in Python.

Insert In BST (Python)

For more details, please go through - <a href="https://www.geeksforgeeks.org/dsa/deletion-in-binary-search-tree/">Deletion in Binary Search Tree (BST)</a>

Deletion in BST.

The Trapping Rainwater Problem involves calculating the total amount of water that can be trapped between non-negative integers in an elevation map, where each number represents the height of a bar. The naive approach iterates through every element to compute left and right maximums for each index, with a time complexity of O(n^2). A better solution uses prefix and suffix arrays to store the maximum heights on the left and right, reducing the time complexity to O(n) and requiring O(n) space.The most efficient solution utilizes the two-pointer technique, achieving O(n) time and O(1) space. By comparing the maximum heights on the left and right, the algorithm decides whether to calculate water for the left or right pointer, updating the respective maximums and pointers accordingly. Alternate approaches, such as using a stack, offer O(n) time complexity but require O(n) space, leveraging next and previous greater elements to calculate trapped water efficiently. For more details, please go through - <a href="https://www.geeksforgeeks.org/trapping-rain-water/">Trapping Rain Water Problem – Tutorial with Illustrations</a>

Trapping Rain Water Problem

The task is to find the longest subarray in a binary array with an equal number of 0s and 1s. A naive approach involves using a nested loop to calculate the sum of subarrays while treating 0s as -1. If the cumulative sum of a subarray equals zero, the subarray has an equal number of 0s and 1s, and the length is updated. This approach has a time complexity of O(n^2), which is inefficient for larger arrays.A more efficient approach uses a hash map and prefix sum technique, reducing time complexity to O(n). By treating 0s as -1, the problem is transformed into finding the longest subarray with a sum of 0. The hash map stores the first occurrence of each prefix sum. If a prefix sum repeats, it indicates a subarray with an equal number of 0s and 1s. The difference in indices gives the length of the subarray, and the maximum length is updated accordingly. For more details, please go through - <a href="https://www.geeksforgeeks.org/largest-subarray-with-equal-number-of-0s-and-1s/">Longest Subarray With Equal Number of 0s and 1s</a>

Longest Subarray With Equal Number of 0s and 1s

The problem of finding the length of the longest subarray with a sum equal to k can be solved using two approaches. The naive approach involves iterating over all possible subarrays using nested loops to calculate their sums, and if the sum matches k, the length of the subarray is checked and updated. This approach has a time complexity of O(n^2), making it inefficient for large arrays.A more efficient approach uses a hash map and prefix sums, achieving O(n) time complexity. In this method, the prefix sum is calculated iteratively, and for each prefix sum, the algorithm checks if prefixSum−k exists in the hash map. If it does, the subarray length is calculated and compared to find the maximum. This approach ensures an optimal solution while maintaining O(n) space complexity by storing the first occurrence of each prefix sum in the hash map. For more details, please go through - <a href="https://www.geeksforgeeks.org/longest-sub-array-sum-k/">Longest Subarray With Sum K</a>

Longest Subarray With Sum K

To find the subarray in an array that sums up to a given target, several approaches can be utilized. The naive approach involves iterating through all possible subarrays using nested loops and checking the cumulative sum for each. This approach is simple but has a time complexity of O(n^2). If the target sum is found, the starting and ending indices of the subarray are returned; otherwise, [−1] is returned.The expected approach leverages the sliding window technique for O(n) efficiency. Starting with an empty window, elements are added to the window until the sum equals or exceeds the target. If the sum exceeds the target, elements are removed from the start of the window until it is reduced to the target. This method is efficient for positive integers and ensures minimal computation with constant space usage. For more details, please go through - <a href="https://www.geeksforgeeks.org/find-subarray-with-given-sum/">Subarray with Given Sum</a>

Subarray with Given Sum | DSA

To find the number of times a graph crosses the X-axis, the task involves calculating the cumulative sum of an array where positive values indicate upward movement and negative values indicate downward movement. Starting from the origin, the crossing count is incremented whenever there is a transition from a negative to a non-negative cumulative sum or from a positive to a non-positive cumulative sum. This approach efficiently tracks changes in the graph's position relative to the X-axis.The algorithm iterates through the array while maintaining a running cumulative sum. For each element, the current level is updated, and the crossing condition is checked by comparing the previous and current levels. If the graph transitions across the X-axis, the crossing count is incremented. This method ensures an O(n) time complexity and uses O(1) auxiliary space. For more details, please go through - <a href="https://www.geeksforgeeks.org/count-the-number-of-times-graph-crosses-x-axis/">Count the number of times graph crosses X-axis</a>

Count the number of times graph crosses X-axis

An equilibrium index in an array is an index where the sum of elements on its left equals the sum of elements on its right. To find this, the naive approach uses nested loops to calculate the left and right sums for every index, resulting in O(n^2) time complexity. While simple, this method is inefficient for large arrays.Optimized methods involve computing prefix and suffix sums in O(n) time. The prefix and suffix sum arrays store cumulative sums up to each index and from each index, respectively. A more space-efficient approach uses running prefix and suffix sums, comparing these at each pivot index to find the equilibrium index in O(1) space and O(n) time. For more details, please go through - <a href="https://www.geeksforgeeks.org/equilibrium-index-of-an-array/">Equilibrium Index</a>

Equilibrium Index

To find if a pair of elements in a sorted and rotated array sums up to a given target, a hashing-based approach can be used for O(n) time and space complexity. By iterating through the array, a set is used to store elements, and for each element, its complement with respect to the target is checked. If the complement exists in the set, the pair is found. Otherwise, the element is added to the set. This approach provides simplicity but requires additional space for the hash set.A more efficient approach is the two-pointer technique, which operates in O(n) time and O(1) space. First, the pivot (largest element) and the smallest element in the rotated array are identified. Pointers are initialized at the smallest and largest elements. The sum of the elements at these pointers is compared with the target, and the pointers are adjusted accordingly using modulo operations to handle the circular nature of the array. If the pair is found, the function returns true; otherwise, it returns false after the iteration completes. For more details, please go through -<a href="https://www.geeksforgeeks.org/given-a-sorted-and-rotated-array-find-if-there-is-a-pair-with-a-given-sum/"> Pair Sum in a Sorted and Rotated Array</a>

Pair Sum in a Sorted and Rotated Array

To find the minimum removals required to achieve a target sum k by removing elements from either end of an array, three approaches can be utilized. The naive approach uses recursion to explore all combinations of removals, calculating the sum for every possible path. Though straightforward, this method is highly inefficient with O(2^n) time complexity. A better approach leverages prefix arrays and hash maps to determine the longest subarray with a sum equal to total&nbsp;sum−k, reducing the problem to O(n) time complexity but requiring O(n) space for storing prefix sums.The optimized solution employs the sliding window technique, which uses two pointers to dynamically adjust a window representing the longest subarray with the desired sum. By maintaining a current sum and adjusting the window size, this method achieves O(n) time complexity and O(1) space usage. It ensures efficient computation by incrementing or decrementing the window while tracking the longest valid subarray. The result is derived as the difference between the array's size and the maximum subarray length found. For more details, please go through - <a href="https://www.geeksforgeeks.org/minimum-removals-for-target-sum/">Minimum Removals for Target Sum</a>

Minimum Removals for Target Sum

To find the count of subarrays with exactly k distinct elements, two approaches can be utilized. The naive approach involves iterating over all possible subarrays while using a hash set to track distinct elements. For each subarray, the result is incremented if the size of the hash set equals k. Though simple, this approach has a time complexity of O(n^2), which becomes impractical for large arrays.The optimized approach leverages the sliding window technique and the formula: K=atMostK−atMost(k−1). By calculating subarrays with at most k and k−1 distinct elements, their difference gives the count of subarrays with exactly k distinct elements. This method efficiently processes the array in O(n) time and O(n) space using frequency maps, making it more suitable for large datasets. For more details, please go through - <a href="https://www.geeksforgeeks.org/count-of-subarrays-having-exactly-k-distinct-elements/">Count Subarrays With Exactly K Distinct Elements</a>

Count Subarrays With Exactly K Distinct Elements