Minimum value to be assigned to the elements so that sum becomes greater than initial sum

Given an array arr[] of N elements, the task is to update all the elements of the given array to some value X such that the sum of all the updated array elements is strictly greater than the sum of all the elements of the initial array and X is the minimum possible.

Examples:

Input: arr[] = {4, 2, 1, 10, 6} 
Output: 5 
Sum of original array = 4 + 2 + 1 + 10 + 6 = 23 
Sum of the modified array = 5 + 5 + 5 + 5 + 5 = 25

Input: arr[] = {9876, 8654, 5470, 3567, 7954} 
Output: 7105 

Approach:

• Find the sum of the original array elements and store it in a variable sumArr
• Calculate X = sumArr / n where n is the number of elements in the array.
• Now, in order to exceed the sum of the original array, every element of the new array has to be at least X + 1.

Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to return the minimum
// required value
int findMinValue(int arr[], int n)
{

    // Find the sum of the
    // array elements
    long sum = 0;
    for (int i = 0; i < n; i++)
        sum += arr[i];

    // Return the required value
    return ((sum / n) + 1);
}

// Driver code
int main()
{
    int arr[] = { 4, 2, 1, 10, 6 };
    int n = sizeof(arr) / sizeof(int);

    cout << findMinValue(arr, n);

    return 0;
}

Java

// Java implementation of the approach
import java.io.*;

public class GFG {

    // Function to return the minimum
    // required value
    static int findMinValue(int arr[], int n)
    {

        // Find the sum of the
        // array elements
        long sum = 0;
        for (int i = 0; i < n; i++)
            sum += arr[i];

        // Return the required value
        return ((int)(sum / n) + 1);
    }

    // Driver code
    public static void main(String args[])
    {
        int arr[] = { 4, 2, 1, 10, 6 };
        int n = arr.length;

        System.out.print(findMinValue(arr, n));
    }
}

Python3

# Python3 implementation of the approach

# Function to return the minimum
# required value
def findMinValue(arr, n):

    # Find the sum of the
    # array elements
    sum = 0
    for i in range(n):
        sum += arr[i]

    # Return the required value
    return (sum // n) + 1

# Driver code
arr = [4, 2, 1, 10, 6]
n = len(arr)
print(findMinValue(arr, n))

C#

// C# implementation of the above approach
using System;

class GFG
{

    // Function to return the minimum
    // required value
    static int findMinValue(int []arr, int n)
    {

        // Find the sum of the
        // array elements
        long sum = 0;
        for (int i = 0; i < n; i++)
            sum += arr[i];

        // Return the required value
        return ((int)(sum / n) + 1);
    }

    // Driver code
    static public void Main ()
    {
        int []arr = { 4, 2, 1, 10, 6 };
        int n = arr.Length;

        Console.WriteLine(findMinValue(arr, n));
    }
}

// This code is contributed by AnkitRai01

Javascript

<script>
// Javascript implementation of the approach

// Function to return the minimum
// required value
function findMinValue(arr, n)
{

    // Find the sum of the
    // array elements
    let sum = 0;
    for (let i = 0; i < n; i++)
        sum += arr[i];

    // Return the required value
    return (parseInt(sum / n) + 1);
}

// Driver code
    let arr = [ 4, 2, 1, 10, 6 ];
    let n = arr.length;

    document.write(findMinValue(arr, n));

</script>

Output:

5

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