Given an array arr[] of n positive integers. There are two operations allowed:
- Operation 1 : Pick any two indexes, increase value at one index by 1 and decrease value at another index by 1. It will cost a.
- Operation 2 : Pick any index and increase its value by 1. It will cost b.
The task is to find the minimum cost to make all the elements equal in the array.
Examples:
Input : n = 4, a = 2, b = 3
arr[] = { 3, 4, 2, 2 }
Output : 5
Perform operation 2 on 3rd index
(0 based indexing). It will cost 2.
Perform operation 1 on index 1 (decrease)
and index 2 (increase). It will cost 3.
Input : n = 3, a = 2, b = 1
arr[] = { 5, 5, 5 }
Output : 0
Approach:
Observe, the final array will not have elements greater than the maximum element of the given array because there is no sense to increase all the elements. Also, they will greater than the minimal element in the original array. Now, iterate over all the possible value of the final array element which needs to be equal and check how many operations of second type must be performed.
For elements to be i (which is one of the possible value of final array element) that number is (n * i – s), where s is the sum of all numbers in the array. The number of operation of the first type for element to be i in the final array can be find by:
for (int j = 0; j < n; j++) op1 += max(0, a[j] - i)
At the end of each iteration simply check whether the new value ans = (n * i – s) * b + op1 * a is less than previous value of ans. If it is less, update the final ans.
Below is the implementation of above approach.
C++
// CPP Program to find the minimum cost to equal// all elements of array using two operation#include <bits/stdc++.h>using namespace std;// Return the minimum cost requiredint minCost(int n, int arr[], int a, int b){ int sum = 0, ans = INT_MAX; int maxval = 0; // finding the maximum element and sum of the array. for (int i = 0; i < n; i++) { sum += arr[i]; maxval = max(maxval, arr[i]); } // for each of the possible value for (int i = 1; i <= maxval; i++) { int op1 = 0; // finding the number of operation 1 required for (int j = 0; j < n; j++) op1 += max(0, arr[j] - i); // finding the minimum cost. if (sum <= n * i) ans = min(ans, (n * i - sum) * b + op1 * a); } return ans;}// Driven Codeint main(){ int n = 4, a = 2, b = 3; int arr[] = { 3, 4, 2, 2 }; cout << minCost(n, arr, a, b) << endl; return 0;} |
Java
// Java Program to find the minimum cost// to equal all elements of array using// two operationimport java.lang.*;class GFG { // Return the minimum cost required static int minCost(int n, int arr[], int a, int b) { int sum = 0, ans = Integer.MAX_VALUE; int maxval = 0; // finding the maximum element and // sum of the array. for (int i = 0; i < n; i++) { sum += arr[i]; maxval = Math.max(maxval, arr[i]); } // for each of the possible value for (int i = 1; i <= maxval; i++) { int op1 = 0; // finding the number of operation // 1 required for (int j = 0; j < n; j++) op1 += Math.max(0, arr[j] - i); // finding the minimum cost. if (sum <= n * i) ans = Math.min(ans, (n * i - sum) * b + op1 * a); } return ans; } // Driven Code public static void main(String [] args) { int n = 4, a = 2, b = 3; int arr[] = { 3, 4, 2, 2 }; System.out.println(minCost(n, arr, a, b)); }}// This code is contributed by Smitha. |
python3
# Python 3 Program to find the minimum # cost to equal all elements of array# using two operationimport sys# Return the minimum cost requireddef minCost(n, arr, a, b): sum = 0 ans = sys.maxsize maxval = 0 # finding the maximum element and # sum of the array. for i in range(0, n) : sum += arr[i] maxval = max(maxval, arr[i]) # for each of the possible value for i in range(0, n) : op1 = 0 # finding the number of operation # 1 required for j in range(0, n) : op1 += max(0, arr[j] - i) # finding the minimum cost. if (sum <= n * i): ans = min(ans, (n * i - sum) * b + op1 * a) return ans# Driven Coden = 4a = 2b = 3arr = [3, 4, 2, 2]print(minCost(n, arr, a, b))# This code is contributed by Smitha |
C#
// C# Program to find the minimum// cost to equal all elements of// array using two operationusing System;class GFG { // Return the minimum cost required static int minCost(int n, int [] arr, int a, int b) { int sum = 0, ans = int.MaxValue; int maxval = 0; // finding the maximum element and // sum of the array. for (int i = 0; i < n; i++) { sum += arr[i]; maxval = Math.Max(maxval, arr[i]); } // for each of the possible value for (int i = 1; i <= maxval; i++) { int op1 = 0; // finding the number of operation // 1 required for (int j = 0; j < n; j++) op1 += Math.Max(0, arr[j] - i); // finding the minimum cost. if (sum <= n * i) ans = Math.Min(ans, (n * i - sum) * b + op1 * a); } return ans; } // Driven Code public static void Main() { int n = 4, a = 2, b = 3; int []arr= { 3, 4, 2, 2 }; Console.Write(minCost(n, arr, a, b)); }}// This code is contributed by Smitha |
PHP
<?php//PHP Program to find the minimum cost// to equal all elements of array using // two operation// Return the minimum cost requiredfunction minCost($n, $arr, $a, $b){ $sum = 0; $ans = PHP_INT_MAX; $maxval = 0; // finding the maximum element and // sum of the array. for ($i = 0; $i < $n; $i++) { $sum += $arr[$i]; $maxval = max($maxval, $arr[$i]); } // for each of the possible value for ($i = 1; $i <= $maxval; $i++) { $op1 = 0; // finding the number of operation // 1 required for ($j = 0; $j < $n; $j++) $op1 += max(0, $arr[$j] - $i); // finding the minimum cost. if ($sum <= $n * $i) $ans = min($ans, ($n * $i - $sum) * $b + $op1 * $a); } return $ans;}// Driven Code $n = 4; $a = 2; $b = 3; $arr= array(3, 4, 2, 2 ); echo minCost($n, $arr, $a, $b) ,"\n";// This code is contributed by ajit?> |
Javascript
<script>// javascript Program to find the minimum cost// to equal all elements of array using// two operation // Return the minimum cost required function minCost(n , arr , a , b) { var sum = 0, ans = Number.MAX_VALUE; var maxval = 0; // finding the maximum element and // sum of the array. for (i = 0; i < n; i++) { sum += arr[i]; maxval = Math.max(maxval, arr[i]); } // for each of the possible value for (i = 1; i <= maxval; i++) { var op1 = 0; // finding the number of operation // 1 required for (j = 0; j < n; j++) op1 += Math.max(0, arr[j] - i); // finding the minimum cost. if (sum <= n * i) ans = Math.min(ans, (n * i - sum) * b + op1 * a); } return ans; } // Driven Code var n = 4, a = 2, b = 3; var arr = [ 3, 4, 2, 2 ]; document.write(minCost(n, arr, a, b));// This code contributed by umadevi9616</script> |
Output
5
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