Program to find the Hidden Number

Given an array of n integers, The task is to find another integer x such that, if all the elements of the array are subtracted individually from the number x, then the sum of all the differences should add to 0. If any such integer exists, print the value of x, else print -1.

Example

Input: arr[] = {1, 2, 3}
Output: 2
Explanation: 
Substracting all the elements of arrays from 2,
The sum of difference is:
1 + 0 + (-1) = 0.


Solution: The idea is to calculate the total sum of the array and divide it by the size of the array. If the sum of the array is perfectly divisible by its size then the quotient obtained from this division operation will be the required hidden number.

Below is the implementation of the above idea:

Java

// Java Program to find the
// Hidden Number
import java.lang.*;
import java.util.*;
  
public class GFG {
  
    // Driver Code
    public static void main(String args[])
    {
  
        // Getting the size of array
        int n = 3;
  
        // Getting the array of size n
        int a[] = { 1, 2, 3 };
  
        // Solution
        int i = 0;
  
        // Finding sum of the array elements
        long sum = 0;
        for (i = 0; i < n; i++) {
            sum += a[i];
        }
  
        // Dividing sum by size n
        long x = sum / n;
  
        // Print x, if found
        if (x * n == sum)
            System.out.println(x);
        else
            System.out.println("-1");
    }
}

C#

// C# Program to find the
// Hidden Number
using System;

class GFG
{

// Driver Code
public static void Main()
{

// Getting the size of array
int n = 3;

// Getting the array of size n
int []a = { 1, 2, 3 };

// Solution
int i = 0;

// Finding sum of the
// array elements
long sum = 0;
for (i = 0; i < n; i++) { sum += a[i]; } // Dividing sum by size n long x = sum / n; // Print x, if found if (x * n == sum) Console.WriteLine(x); else Console.WriteLine("-1"); } } // This code is contributed // by inder_verma [tabbyending]

Output:

2


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Improved By : inderDuMCA