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:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ Program to find the
// hidden number
  
#include <iostream>
using namespace std;
  
    // Driver Code
  
int 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)
            cout<<x<<endl;
        else
            cout<<("-1")<<endl;
      
  
    return 0;
    // This code is contributed
// by shashank
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java Program to find the
// hidden number
  
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");
    }
}

chevron_right


Python 3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python 3 Program to find the
# hidden number
  
# Driver Code
if __name__ == "__main__":
      
    # Getting the size of array
    n = 3
  
    # Getting the array of size n
    a = [ 1, 2, 3 ]
  
    # Solution
    i = 0
  
    # Finding sum of the .
    # array elements
    sum = 0
    for i in range(n):
        sum += a[i]
  
    # Dividing sum by size n
    x = sum // n
  
    # Print x, if found
    if (x * n == sum):
        print(x)
    else:
        print("-1")
  
# This code is contributed 
# by ChitraNayal

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php
// PHP Program to find the
// hidden number
  
// Driver Code
  
// Getting the size of array
$n = 3;
  
// Getting the array of size n
$a = array( 1, 2, 3 );
  
// Solution
$i = 0;
  
// Finding sum of the array elements
$sum = 0;
for ($i = 0; $i < $n; $i++)
{
    $sum += $a[$i];
}
  
// Dividing sum by size n
$x = $sum / $n;
  
// Print x, if found
if ($x * $n == $sum)
echo($x);
else
echo("-1");
  
// This code is contributed
// by inder_verma
?>

chevron_right


Output:

2


My Personal Notes arrow_drop_up

Code for enjoyment

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.