**Prerequisite:** Sliding Window Median

Given an array **arr[]** consisting of **N** integers and an integer **K**, the task is to find the minimum cost required to make each element of every subarray of length **K** equal. Cost of replacing any array element by another element is the absolute difference between the two.

**Examples:**

Input:A[] = {1, 2, 3, 4, 6}, K = 3Output:7Explanation:

Subarray 1: Cost to convert subarray {1, 2, 3} to {2, 2, 2} = |1-2| + |2-2| + |3-2| = 2

Subarray 2: Cost to convert subarray {2, 3, 4} to {3, 3, 3} = |2-3| + |3-3| + |4-3| = 2

Subarray 3: Cost to convert subarray {3, 4, 6} to {4, 4, 4} = |3-4| + |4-4| + |6-4| = 3

Minimum Cost = 2 + 2 + 3 = 7/

Input:A[] = {2, 3, 4, 4, 1, 7, 6}, K = 4Output:21

**Approach:**

To find the minimum cost to convert each element of the subarray to a single element, change every element of the subarray to the median of that subarray. Follow the steps below to solve the problem:

- To find the
**median**for each running subarray efficiently, use a multiset to get the sorted order of elements in each subarray. Median will be the**middle element**of this**multiset**. - For the next subarray remove the leftmost element of the previous subarray from the multiset, add the current element to the multiset.
- Keep a pointer
**mid**to efficiently keep track of the middle element of the multiset. - If the newly added element is
**smaller**than the**previous middle element**, move**mid**to its immediate smaller element. Otherwise, move**mid**to its immediate next element. - Calaculate cost of replacing every element of the subarray by the equation
**| A[i] – median**._{each_subarray}| - Finally print the total cost.

Below is the implementation for the above approach:

## C++

`// C++ Program to implement` `// the above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` ` ` `// Function to find the minimum` `// cost to convert each element of` `// every subarray of size K equal` `int` `minimumCost(vector<` `int` `> arr, ` `int` `n,` ` ` `int` `k)` `{` ` ` `// Stores the minimum cost` ` ` `int` `totalcost = 0;` ` ` `int` `i, j;` ` ` ` ` `// Stores the first K elements` ` ` `multiset<` `int` `> mp(arr.begin(),` ` ` `arr.begin() + k);` ` ` ` ` `if` `(k == n) {` ` ` ` ` `// Obtain the middle element of` ` ` `// the multiset` ` ` `auto` `mid = next(mp.begin(),` ` ` `n / 2 - ((k + 1) % 2));` ` ` ` ` `int` `z = *mid;` ` ` ` ` `// Calculate cost for the subarray` ` ` `for` `(i = 0; i < n; i++)` ` ` `totalcost += ` `abs` `(z - arr[i]);` ` ` ` ` `// Return the total cost` ` ` `return` `totalcost;` ` ` `}` ` ` `else` `{` ` ` ` ` `// Obtain the middle element` ` ` `// in multiset` ` ` `auto` `mid = next(mp.begin(),` ` ` `k / 2 - ((k + 1) % 2));` ` ` ` ` `for` `(i = k; i < n; i++) {` ` ` ` ` `int` `zz = *mid;` ` ` `int` `cost = 0;` ` ` `for` `(j = i - k; j < i; j++) {` ` ` ` ` `// Cost for the previous` ` ` `// k length subarray` ` ` `cost += ` `abs` `(arr[j] - zz);` ` ` `}` ` ` `totalcost += cost;` ` ` ` ` `// Insert current element` ` ` `// into multiset` ` ` `mp.insert(arr[i]);` ` ` ` ` `if` `(arr[i] < *mid) {` ` ` ` ` `// New element appears` ` ` `// to the left of mid` ` ` `mid--;` ` ` `}` ` ` ` ` `if` `(arr[i - k] <= *mid) {` ` ` ` ` `// New element appears` ` ` `// to the right of mid` ` ` `mid++;` ` ` `}` ` ` ` ` `// Remove leftmost element` ` ` `// from the window` ` ` `mp.erase(mp.lower_bound(arr[i - k]));` ` ` ` ` `// For last element` ` ` `if` `(i == n - 1) {` ` ` `zz = *mid;` ` ` `cost = 0;` ` ` ` ` `for` `(j = i - k + 1;` ` ` `j <= i; j++) {` ` ` ` ` `// Calculate cost for the subarray` ` ` `cost += ` `abs` `(zz - arr[j]);` ` ` `}` ` ` ` ` `totalcost += cost;` ` ` `}` ` ` `}` ` ` ` ` `// Return the total cost` ` ` `return` `totalcost;` ` ` `}` `}` ` ` `// Driver Code` `int` `main()` `{` ` ` `int` `N = 5, K = 3;` ` ` ` ` `vector<` `int` `> A({ 1, 2, 3, 4, 6 });` ` ` ` ` `cout << minimumCost(A, N, K);` `}` |

**Output:**

7

**Time Complexity:** O(NlogN)**Auxiliary Space:** O(1)

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