# Minimum cost to equal all elements of array using two operation

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
```

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

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 ` `using` `namespace` `std; ` ` `  `// Return the minimum cost required ` `int` `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 Code ` `int` `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 operation ` `import` `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 operation ` `import` `sys ` ` `  `# Return the minimum cost required ` `def` `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 Code ` `n ``=` `4` `a ``=` `2` `b ``=` `3` `arr ``=` `[``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 operation ` `using` `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

 ` `

Output:

```5
```

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