Elemento igual a la suma de todos los elementos restantes

Dada una array de N elementos positivos. La tarea es encontrar un elemento que sea igual a la suma de todos los elementos de la array excepto él mismo.
Ejemplos: 
 

Input: arr[] = {1, 2, 3, 6}
Output: 6
6 is the element which is equal to the sum of all 
remaining elements i.e. 1 + 2+ 3 = 6

Input: arr[] = {2, 2, 2, 2}
Output: -1

Enfoque: en primer lugar, encuentre la suma de todos los elementos de una array. Luego itere sobre cada elemento y verifique la condición de que if (a[i] == sum-a[i] ) . Si es verdadero, imprima ese a[i] , de lo contrario, imprima «-1» si no se encuentra dicho elemento.
A continuación se muestra la implementación del enfoque anterior: 
 

// C++ implementation of the above approach
#include <bits/stdc++.h>
#define ll long long int
using namespace std;
 
// Function to find the element
int findEle(int arr[], int n)
{
    // sum is use to store
    // sum of all elements
    // of array
    ll sum = 0;
 
    for (int i = 0; i < n; i++)
        sum += arr[i];
 
    // iterate over all elements
    for (int i = 0; i < n; i++)
        if (arr[i] == sum - arr[i])
            return arr[i];
 
    return -1;
}
 
// Driver code
int main()
{
    int arr[] = { 1, 2, 3, 6 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << findEle(arr, n);
    return 0;
}

// Java implementation of the above approach
 
import java.io.*;
 
class GFG {
 
 
// Function to find the element
static int findEle(int arr[], int n)
{
    // sum is use to store
    // sum of all elements
    // of array
    int sum = 0;
 
    for (int i = 0; i < n; i++)
        sum += arr[i];
 
    // iterate over all elements
    for (int i = 0; i < n; i++)
        if (arr[i] == sum - arr[i])
            return arr[i];
 
    return -1;
}
 
// Driver code
 
    public static void main (String[] args) {
        int arr[] = { 1, 2, 3, 6 };
    int n = arr.length;
    System.out.print(findEle(arr, n));
    }
}
// This code is contributed by shs..

# Python 3 implementation of
# the above approach
 
# Function to find the element
def findEle(arr, n) :
     
    # sum is use to store
    # sum of all elements
    # of array
    sum = 0
     
    for i in range(n) :
        sum += arr[i]
     
    # iterate over all elements
    for i in range(n) :
        if arr[i] == sum - arr[i] :
            return arr[i]
     
    return -1
 
# Driver Code
if __name__ == "__main__" :
     
    arr = [1, 2, 3, 6 ]
    n = len(arr)
    print(findEle(arr, n))
 
# This code is contributed by Ryuga

// C# implementation of the
// above approach
using System;
 
class GFG
{
 
// Function to find the element
static int findEle(int []arr, int n)
{
    // sum is use to store
    // sum of all elements
    // of array
    int sum = 0;
 
    for (int i = 0; i < n; i++)
        sum += arr[i];
 
    // iterate over all elements
    for (int i = 0; i < n; i++)
        if (arr[i] == sum - arr[i])
            return arr[i];
 
    return -1;
}
 
// Driver code
static public void Main (String []args)
{
    int []arr = { 1, 2, 3, 6 };
    int n = arr.Length;
    Console.WriteLine(findEle(arr, n));
}
}
 
// This code is contributed
// by Arnab Kundu

<?php
// PHP implementation of the above approach
 
// Function to find the element
function findEle($arr, $n)
{
    // sum is use to store
    // sum of all elements
    // of array
    $sum = 0;
 
    for ($i = 0; $i < $n; $i++)
        $sum += $arr[$i];
 
    // iterate over all elements
    for ($i = 0; $i < $n; $i++)
        if ($arr[$i] == $sum - $arr[$i])
            return $arr[$i];
 
    return -1;
}
 
// Driver code
$arr= array(1, 2, 3, 6 );
$n = sizeof($arr);
echo findEle($arr, $n);
 
// This code is contributed
// by Akanksha Rai
?>

<script>
 
// javascript implementation of the above approach
 
// Function to find the element
function findEle(arr, n)
{
    // sum is use to store
    // sum of all elements
    // of array
    var sum = 0;
 
    for (var i = 0; i < n; i++)
        sum += arr[i];
 
    // iterate over all elements
    for (var i = 0; i < n; i++)
        if (arr[i] == sum - arr[i])
            return arr[i];
 
    return -1;
}
 
// Driver code
    var arr = [1, 2, 3, 6];
    var n = arr.length;
    document.write(findEle(arr, n));
 
// This code is contributed by ipg016107.
</script>

6

Complejidad de tiempo: O(n)

Espacio Auxiliar: O(1)

Nota: el problema anterior se puede resolver con el concepto utilizado en Comprobar si la array tiene un elemento que es igual a la suma de todos los elementos restantes.