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Program to find the Hidden Number

  • Difficulty Level : Easy
  • Last Updated : 20 May, 2021

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++




// 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
}

Java




// 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");
    }
}

Python 3




# 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

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

PHP




<?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
?>

Javascript




<script>
 
// JavaScript Program to find the
// hidden number
     
    // Driver Code
    // Getting the size of array
    let n = 3;
    // Getting the array of size n
    let a=[1, 2, 3];
     
    // Solution
    let i = 0;
     
    // Finding sum of the array elements
    let sum = 0;
    for (i = 0; i < n; i++) {
            sum += a[i];
        }
   
        // Dividing sum by size n
        let x = sum / n;
   
        // Print x, if found
        if (x * n == sum)
            document.write(x);
        else
            document.write("-1");
 
 
// This code is contributed by rag2127
 
</script>
Output: 
2

 

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