# Minimum cost required to convert all Subarrays of size K to a single element

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 = 3
Output: 7
Explanation:
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 = 4
Output: 21

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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] – medianeach_subarray |.
• Finally print the total cost.

Below is the implementation for the above approach:

 `// C++ Program to implement ` `// the above approach ` `#include ` `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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