# Minimum sum of values subtracted from array elements to make all array elements equal

Given an array arr[] consisting of N positive integers, the task is to find the sum of all the array elements required to be subtracted from each array element such that remaining array elements are all equal.

Examples:

Input: arr[] = {1, 2}
Output: 1
Explanation: Subtracting 1 from arr[1] modifies arr[] to {1, 1}. Therefore, the required sum is 1.

Input: arr[] = {1, 2, 3}
Output: 3
Explanation: Subtracting 1 and 2 from arr[1] and arr[2] modifies arr[] to {1, 1, 1}. Therefore, the required sum = 1 + 2 = 3.

Approach: The idea is to reduce all array elements to the minimum element present in the array. Follow the below steps to solve the problem:

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach` `#include ` `using` `namespace` `std;`   `// Function to find the sum of values` `// removed to make all array elements equal` `int` `minValue(``int` `arr[], ``int` `n)` `{` `    ``// Stores the minimum of the array` `    ``int` `minimum = *min_element(` `        ``arr, arr + n);`   `    ``// Stores required sum` `    ``int` `sum = 0;`   `    ``// Traverse the array` `    ``for` `(``int` `i = 0; i < n; i++) {`   `        ``// Add the value subtracted` `        ``// from the current element` `        ``sum = sum + (arr[i] - minimum);` `    ``}`   `    ``// Return the total sum` `    ``return` `sum;` `}`   `// Driver Code` `int` `main()` `{` `    ``int` `arr[] = { 1, 2, 3 };` `    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr[0]);`   `    ``// Function Call` `    ``cout << minValue(arr, N);` `    ``return` `0;` `}`

## Java

 `// Java program for the above approach` `import` `java.util.Arrays;` `class` `GFG` `{` `    `  `// Function to find the sum of values` `// removed to make all array elements equal` `static` `int` `minValue(``int` `[]arr, ``int` `n)` `{` `    ``Arrays.sort(arr);` `    `  `    ``// Stores the minimum of the array` `    ``int` `minimum = arr[``0``];`   `    ``// Stores required sum` `    ``int` `sum = ``0``;`   `    ``// Traverse the array` `    ``for``(``int` `i = ``0``; i < n; i++)` `    ``{` `        `  `        ``// Add the value subtracted` `        ``// from the current element` `        ``sum = sum + (arr[i] - minimum);` `    ``}` `    `  `    ``// Return the total sum` `    ``return` `sum;` `}`   `// Driver Code` `static` `public` `void` `main(String args[])` `{` `    ``int` `[]arr = { ``1``, ``2``, ``3` `};` `    ``int` `N = arr.length;` `    `  `    ``// Function Call` `    ``System.out.println(minValue(arr, N));` `}` `}`   `// This code is contributed by AnkThon`

## Python3

 `# Python3 program for the above approach`   `# Function to find the sum of values` `# removed to make all array elements equal` `def` `minValue(arr, n):` `    `  `    ``# Stores the minimum of the array` `    ``minimum ``=` `min``(arr)`   `    ``# Stores required sum` `    ``sum` `=` `0`   `    ``# Traverse the array` `    ``for` `i ``in` `range``(n):` `        `  `        ``# Add the value subtracted` `        ``# from the current element` `        ``sum` `=` `sum` `+` `(arr[i] ``-` `minimum)` `        `  `    ``# Return the total sum` `    ``return` `sum`   `# Driver Code` `if` `__name__ ``=``=` `'__main__'``:` `    `  `    ``arr ``=` `[ ``1``, ``2``, ``3` `]` `    ``N ``=` `len``(arr)` `    `  `    ``# Function Call` `    ``print``(minValue(arr, N))`   `# This code is contributed by mohit kumar 29`

## C#

 `// C# program for the above approach` `using` `System;`   `class` `GFG{` `    `  `// Function to find the sum of values` `// removed to make all array elements equal` `static` `int` `minValue(``int` `[]arr, ``int` `n)` `{` `    ``Array.Sort(arr);` `    `  `    ``// Stores the minimum of the array` `    ``int` `minimum = arr[0];`   `    ``// Stores required sum` `    ``int` `sum = 0;`   `    ``// Traverse the array` `    ``for``(``int` `i = 0; i < n; i++)` `    ``{` `        `  `        ``// Add the value subtracted` `        ``// from the current element` `        ``sum = sum + (arr[i] - minimum);` `    ``}` `    `  `    ``// Return the total sum` `    ``return` `sum;` `}`   `// Driver Code` `static` `public` `void` `Main ()` `{` `    ``int` `[]arr = { 1, 2, 3 };` `    ``int` `N = arr.Length;` `    `  `    ``// Function Call` `    ``Console.WriteLine(minValue(arr, N));` `}` `}`   `// This code is contributed by AnkThon`

## Javascript

 ``

Output:

`3`

Time Complexity: O(N)
Auxiliary Space: O(1)

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