Minimum cost to make every Kth element in Array equal

Given an array arr[] of integers and an integer K, the task is to find the minimum number of operations required to make every K^th element in the array equal. While performing one operation you can either increase a number by one or decrease the number by one.

Examples:

Input: arr[] = {1, 2, 3, 4, 4, 6}, K = 3
Output: 8
Explanation: For the given array, value of K = 3 which means every 3rd element in the array should be equal. So we have to make 1 = 4, 2 = 4 and 3 = 6. So by performing 3 operations on 1 we can make it equal to 4 (or vice versa it will take same number of operations) and 2 operations for making 2 and 4 equal and 3 operations to make 6 and 3 equal .So total number of operations is 8 which is the minimum cost to make every 3rd element in array equal.

Input: {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, K = 3
Output: 24
Explanation: Value of K = 3, so every 3rd element should be equal which means we have to make 1 = 4 = 7 = 10, 2 = 5 = 8 and 3 = 6 = 9. For making 1 = 4 = 7 = 10, the minimum cost will be 12 and for making 2 = 5 = 8, the minimum cost will be 6 and for making 3 = 6 = 9, the minimum cost will be 6. So for making every 3rd element of array equal we have to perform minimum 24 operations.

Approach: To solve the problem follow the below idea:

We need to find average of every K^th element and then find difference between the K^th element and average.

This could be implemented by following below mentioned steps:

• Initialize an array say average of size k for calculating the average of every K^th element.
• Iterate over the input array and then calculate the average of every k^th element.
• Initialize a variable say, minCost. Iterate over the input array and add the absolute difference of arr[i] and average[i%k], minCost += abs(arr[i] - average[i % k]).

Below is the implementation of the above approach:

#include <bits/stdc++.h>
using namespace std;
int findCost(vector<int> arr, int k)
{
    int n = arr.size();
    vector<int> average(k);
    for (int i = 0; i < k; i++) {
        average[i] = arr[i];
        int total_ele = 1;
        for (int j = i + k; j < n; j += k) {
            average[i] = (average[i] + arr[j]);
            total_ele++;
        }

        // Calculating average
        average[i] /= total_ele;
    }
    int minCost = 0;
    for (int i = 0; i < n; i++) {

        // Adding the absolute difference
        // to minCost variable
        minCost += abs(arr[i] - average[i % k]);
    }
    return minCost;
}

// Drivers code
int main()
{

    // Input array
    vector<int> arr = { 1, 2, 3, 4, 4, 6 };
    int k = 3;

    // Function call
    cout << findCost(arr, k);

    return 0;
}

// Java implementation of the above approach:

import java.io.*;

class GFG {

    static int findCost(int[] arr, int k)
    {
        int n = arr.length;
        int[] average = new int[k];
        for (int i = 0; i < k; i++) {
            average[i] = arr[i];
            int total_ele = 1;
            for (int j = i + k; j < n; j += k) {
                average[i] += arr[j];
                total_ele++;
            }

            // Calculating average
            average[i] /= total_ele;
        }
        int minCost = 0;
        for (int i = 0; i < n; i++) {

            // Adding the absolute difference
            // to minCost variable
            minCost += Math.abs(arr[i] - average[i % k]);
        }
        return minCost;
    }

    public static void main(String[] args)
    {
        // Input array
        int[] arr = { 1, 2, 3, 4, 4, 6 };
        int k = 3;

        // Function call
        System.out.print(findCost(arr, k));
    }
}

// This code is contributed by karthik.

# Python3 implementation of the above approach:

# Function to find the minimum number of operations 
# required to make every Kth element in the array equal.
def findCost(arr, k):
    n = len(arr)
    average = [0] * k
    for i in range(k):
        average[i] = arr[i]
        total_ele = 1
        for j in range(i + k, n, k):
            average[i] += arr[j]
            total_ele += 1

        # Calculating average
        average[i] //= total_ele

    minCost = 0
    for i in range(n):
        # Adding the absolute difference
        # to minCost variable
        minCost += abs(arr[i] - average[i % k])
    return minCost

# Driver code
arr = [1, 2, 3, 4, 4, 6]
k = 3

# Function call
print(findCost(arr, k))

// C# implementation for the above approach:

using System;

public class GFG {

    static int findCost(int[] arr, int k)
    {
        int n = arr.Length;
        int[] average = new int[k];
        for (int i = 0; i < k; i++) {
            average[i] = arr[i];
            int total_ele = 1;
            for (int j = i + k; j < n; j += k) {
                average[i] += arr[j];
                total_ele++;
            }
            // Calculating average
            average[i] /= total_ele;
        }
        int minCost = 0;
        for (int i = 0; i < n; i++) {
            // Adding the absolute difference to minCost
            // variable
            minCost += Math.Abs(arr[i] - average[i % k]);
        }
        return minCost;
    }

    static public void Main()
    {

        // Code
        // Input array
        int[] arr = { 1, 2, 3, 4, 4, 6 };
        int k = 3;

        // Function call
        Console.Write(findCost(arr, k));
    }
}

// This code is contributed by lokesh.

// JavaScript implementation of the above approach:

// Function to find the minimum number of operations 
// required to make every Kth element in the array equal.
function findCost(arr, k) {
    let n = arr.length;
    let average = new Array(k).fill(0);
    for (let i = 0; i < k; i++) {
        average[i] = arr[i];
        let total_ele = 1;
        for (let j = i + k; j < n; j += k) {
            average[i] += arr[j];
            total_ele += 1;
        }

        // Calculating average
        average[i] = Math.floor(average[i] / total_ele);
    }
    let minCost = 0;
    for (let i = 0; i < n; i++) {
        // Adding the absolute difference
        // to minCost variable
        minCost += Math.abs(arr[i] - average[i % k]);
    }
    return minCost;
}

// Driver code
let arr = [1, 2, 3, 4, 4, 6];
let k = 3;

// Function call
console.log(findCost(arr, k));


// This code is contributed by Tapesh(tapeshdua420)

8

Time Complexity: O(N)
Auxiliary Space: O(k) //For the array which was initialized for calculating the average.