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

  • Last Updated : 08 Jun, 2021

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 <bits/stdc++.h>
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




<script>
 
// Javascript program for the above approach
 
// Function to find the sum of values
// removed to make all array elements equal
function minValue(arr, n)
{
    // Stores the minimum of the array
    var minimum = Math.min.apply(Math,arr);
 
    // Stores required sum
    var sum = 0;
     
    var i;
    // Traverse the array
    for (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
 
 var arr = [1, 2, 3];
 var N = arr.length;
 
// Function Call
 document.write(minValue(arr, N));
 
</script>
Output: 
3

 

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

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