Programa para encontrar el Número Oculto

Dada una array de  números enteros, la tarea es encontrar otro número entero  tal que, si todos los elementos de la array se restan individualmente del número  , entonces la suma de todas las diferencias debe sumar 0. Si existe algún número entero, imprima el valor de  , de lo contrario imprimir  . 
Ejemplo 
 

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

Solución : la idea es calcular la suma total de la array y dividirla por el tamaño de la array. Si la suma de la array es perfectamente divisible por su tamaño, el cociente obtenido de esta operación de división será el número oculto requerido.
A continuación se muestra la implementación de la idea anterior: 
 

// 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 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 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# 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 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
?>

<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>

2

Complejidad de tiempo: O(n), donde n es la longitud de la array dada. 
Espacio auxiliar : O(1)