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 = 25Input: 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).
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