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Minimum value to be assigned to the elements so that sum becomes greater than initial sum
  • Last Updated : 11 Oct, 2019

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 sum of all the updated array elements is strictly greater than the sum of all the elements of the initial array and X is 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++

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// 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;
}

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Java

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// Java implementation of the approach
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));
    }
}

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Python3

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# 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))

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C#

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// 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

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Output:

5

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