Given an array arr[] of size N and an integer K, the task is to find the maximum possible frequency of any array element by at most K increments.Examples:Input: arr[] = {1, 4, 8, 13}, N = 4, K = 5 Output: 2 Explanation: Incrementing arr[0] twice modifies arr[] to {4, 4, 8, 13}. Maximum frequency = 2. Incrementing arr[1] four times modifies arr[] to {1, 8, 8, 13}. Maximum frequency = 2. Incrementing arr[2] five times modifies arr[] to {1, 4, 13, 13}. Maximum frequency = 2. Therefore, the maximum possible frequency of any array element that can be obtained by at most 5 increments is 2.Input: arr[] = {2, 4, 5}, N = 3, K = 4 Output: 3Approach: This problem can be solved by using Sliding Window Technique and Sorting. Follow the steps to solve this problem.Sort the array arr[].Initialize variables sum = 0, start = 0 and resultant frequency res = 0.Traverse the array over the range of indices [0, N – 1] and perform the following steps:Increment sum by arr[end].Iterate a loop until the value of [(end – start + 1) * arr[end] – sum] is less than K and perform the following operatiosn: Decrement the value of sum by arr[start].Increment the value of start by 1.After completing the above steps, all the elements over the range [start, end] can be made equal by using at most K operations. Therefore, update the value of res as the maximum of res and (end – start + 1).Finally, print the value of res as frequency of most frequent element after performing Koperations.Below is the implementation of the above approach:C++#include using namespace std;void maxFrequency(int arr[], int N, int K){ sort(arr, arr + N); int start = 0, end = 0; int sum = 0, res = 0; for (end = 0; end < N; end++) { sum += arr[end]; while ((end - start + 1) * arr[end] - sum > K) { sum -= arr[start]; start++; } res = max(res, end – start + 1); } cout