Compare la suma de los primeros N-1 elementos con el N-ésimo elemento de una array

Dada una array arr[] de tamaño N , la tarea es comprobar si la suma de los primeros N – 1 elementos de la array es igual al último elemento.
Ejemplos: 
 

Entrada: arr[] = {1, 2, 3, 4, 10} 
Salida: Sí
Entrada: arr[] = {1, 2, 3, 4, 12} 
Salida: No 
 

Enfoque: Encuentre la suma de los primeros N – 1 elementos de la array, es decir , arr[0] + arr[1] + … + arr[N – 2] y compárelo con arr[N – 1] .
A continuación se muestra la implementación del enfoque anterior: 
 

// C++ implementation of the approach
#include <iostream>
using namespace std;
 
// Function that returns true if sum of
// first n-1 elements of the array is
// equal to the last element
bool isSumEqual(int ar[], int n)
{
    int sum = 0;
 
    // Find the sum of first n-1
    // elements of the array
    for (int i = 0; i < n - 1; i++)
        sum += ar[i];
 
    // If sum equals to the last element
    if (sum == ar[n - 1])
        return true;
 
    return false;
}
 
// Driver code
int main()
{
    int arr[] = { 1, 2, 3, 4, 10 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    if (isSumEqual(arr, n))
        cout << "Yes";
    else
        cout << "No";
 
    return 0;
}

// Java implementation of the approach
 
import java.io.*;
 
class GFG {
 
    // Function that returns true if sum of
    // first n-1 elements of the array is
    // equal to the last element
    static boolean isSumEqual(int ar[], int n)
    {
        int sum = 0;
 
        // Find the sum of first n-1
        // elements of the array
        for (int i = 0; i < n - 1; i++)
            sum += ar[i];
 
        // If sum equals to the last element
        if (sum == ar[n - 1])
            return true;
 
        return false;
    }
 
    // Driver code
    public static void main(String[] args)
    {
 
        int arr[] = { 1, 2, 3, 4, 10 };
        int n = arr.length;
 
        if (isSumEqual(arr, n))
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}
 
// This code is contributed by jit_t.

# Python 3 implementation of the approach
 
# Function that returns true if sum of
# first n-1 elements of the array is
# equal to the last element
def isSumEqual(ar, n):
    sum = 0
 
    # Find the sum of first n-1
    # elements of the array
    for i in range(n - 1):
        sum += ar[i]
 
    # If sum equals to the last element
    if (sum == ar[n - 1]):
        return True
 
    return False
 
# Driver code
if __name__ == '__main__':
    arr = [1, 2, 3, 4, 10]
    n = len(arr)
 
    if (isSumEqual(arr, n)):
        print("Yes")
    else:
        print("No")
         
# This code is contributed by
# Surendra_Gangwar

// C# implementation of the approach
using System;
 
class GFG {
 
    // Function that returns true if sum of
    // first n-1 elements of the array is
    // equal to the last element
    static bool isSumEqual(int[] ar, int n)
    {
        int sum = 0;
 
        // Find the sum of first n-1
        // elements of the array
        for (int i = 0; i < n - 1; i++)
            sum += ar[i];
 
        // If sum equals to the last element
        if (sum == ar[n - 1])
            return true;
 
        return false;
    }
 
    // Driver code
    static public void Main()
    {
        int[] arr = { 1, 2, 3, 4, 10 };
        int n = arr.Length;
 
        if (isSumEqual(arr, n))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}
 
// This code is contributed by ajit

<?php
// PHP implementation of the approach
 
// Function that returns true if sum of
// first n-1 elements of the array is
// equal to the last element
function isSumEqual($ar, $n)
{
    $sum = 0;
 
    // Find the sum of first n-1
    // elements of the array
    for ($i = 0; $i < $n - 1; $i++)
        $sum += $ar[$i];
 
    // If sum equals to the last element
    if ($sum == $ar[$n - 1])
        return true;
 
    return false;
}
 
// Driver code
$arr = array( 1, 2, 3, 4, 10 );
$n = count($arr);
 
if (isSumEqual($arr, $n))
    echo "Yes";
else
    echo "No";
     
// This code is contributed by AnkitRai01
 
?>

<script>
    // Javascript implementation of the approach
     
    // Function that returns true if sum of
    // first n-1 elements of the array is
    // equal to the last element
    function isSumEqual(ar, n)
    {
        let sum = 0;
   
        // Find the sum of first n-1
        // elements of the array
        for (let i = 0; i < n - 1; i++)
            sum += ar[i];
   
        // If sum equals to the last element
        if (sum == ar[n - 1])
            return true;
   
        return false;
    }
     
    let arr = [ 1, 2, 3, 4, 10 ];
    let n = arr.length;
 
    if (isSumEqual(arr, n))
      document.write("Yes");
    else
      document.write("No");
     
</script>

Yes

Complejidad de tiempo: O(n)

Espacio Auxiliar: O(1)