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

12 Jun, 2021
Prerequisite: Sliding Window MedianGiven 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  
 
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.
Calculate 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++
// 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 equalint 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 Codeint main(){    int N = 5, K = 3;     vector<int> A({ 1, 2, 3, 4, 6 });     cout << minimumCost(A, N, K);}
7
 
Time Complexity: O(NlogN) Auxiliary Space: O(1) 
simmytarika5
cpp-multiset
median-finding
sliding-window
subarray
Arrays
Greedy
Mathematical
Sorting
sliding-window
Arrays
Greedy
Mathematical
Sorting
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