Compare sum of first N-1 elements to Nth element of an array

Given an array arr[] of size N, the task is to check whether the sum of first N – 1 element of the array is equal to the last element.

Examples:

Input: arr[] = {1, 2, 3, 4, 10}
Output: Yes

Input: arr[] = {1, 2, 3, 4, 12}
Output: No

Approach: Find the sum of the first N – 1 elements from the array i.e. arr[0] + arr[1] + … + arr[N – 2] and compare it with arr[N – 1].

Below is the implementation of the above approach:

C++

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// 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;
}

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Java

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// 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.

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Python3

# 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#

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// 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

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PHP

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<?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
  
?>

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Output:

Yes


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