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++
#include <bits/stdc++.h>
using namespace std;
int minValue( int arr[], int n)
{
int minimum = *min_element(
arr, arr + n);
int sum = 0;
for ( int i = 0; i < n; i++) {
sum = sum + (arr[i] - minimum);
}
return sum;
}
int main()
{
int arr[] = { 1, 2, 3 };
int N = sizeof (arr) / sizeof (arr[0]);
cout << minValue(arr, N);
return 0;
}
|
Java
import java.util.Arrays;
class GFG
{
static int minValue( int []arr, int n)
{
Arrays.sort(arr);
int minimum = arr[ 0 ];
int sum = 0 ;
for ( int i = 0 ; i < n; i++)
{
sum = sum + (arr[i] - minimum);
}
return sum;
}
static public void main(String args[])
{
int []arr = { 1 , 2 , 3 };
int N = arr.length;
System.out.println(minValue(arr, N));
}
}
|
Python3
def minValue(arr, n):
minimum = min (arr)
sum = 0
for i in range (n):
sum = sum + (arr[i] - minimum)
return sum
if __name__ = = '__main__' :
arr = [ 1 , 2 , 3 ]
N = len (arr)
print (minValue(arr, N))
|
C#
using System;
class GFG{
static int minValue( int []arr, int n)
{
Array.Sort(arr);
int minimum = arr[0];
int sum = 0;
for ( int i = 0; i < n; i++)
{
sum = sum + (arr[i] - minimum);
}
return sum;
}
static public void Main ()
{
int []arr = { 1, 2, 3 };
int N = arr.Length;
Console.WriteLine(minValue(arr, N));
}
}
|
Javascript
<script>
function minValue(arr, n)
{
var minimum = Math.min.apply(Math,arr);
var sum = 0;
var i;
for (i = 0; i < n; i++) {
sum = sum + (arr[i] - minimum);
}
return sum;
}
var arr = [1, 2, 3];
var N = arr.length;
document.write(minValue(arr, N));
</script>
|
Time Complexity: O(N)
Auxiliary Space: O(1)
Last Updated :
08 Jun, 2021
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