Compare sum of first N-1 elements to Nth element of an array
Last Updated :
07 Jun, 2022
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++
#include <iostream>
using namespace std;
bool isSumEqual( int ar[], int n)
{
int sum = 0;
for ( int i = 0; i < n - 1; i++)
sum += ar[i];
if (sum == ar[n - 1])
return true ;
return false ;
}
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
import java.io.*;
class GFG {
static boolean isSumEqual( int ar[], int n)
{
int sum = 0 ;
for ( int i = 0 ; i < n - 1 ; i++)
sum += ar[i];
if (sum == ar[n - 1 ])
return true ;
return false ;
}
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" );
}
}
|
Python3
def isSumEqual(ar, n):
sum = 0
for i in range (n - 1 ):
sum + = ar[i]
if ( sum = = ar[n - 1 ]):
return True
return False
if __name__ = = '__main__' :
arr = [ 1 , 2 , 3 , 4 , 10 ]
n = len (arr)
if (isSumEqual(arr, n)):
print ( "Yes" )
else :
print ( "No" )
|
C#
using System;
class GFG {
static bool isSumEqual( int [] ar, int n)
{
int sum = 0;
for ( int i = 0; i < n - 1; i++)
sum += ar[i];
if (sum == ar[n - 1])
return true ;
return false ;
}
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" );
}
}
|
PHP
<?php
function isSumEqual( $ar , $n )
{
$sum = 0;
for ( $i = 0; $i < $n - 1; $i ++)
$sum += $ar [ $i ];
if ( $sum == $ar [ $n - 1])
return true;
return false;
}
$arr = array ( 1, 2, 3, 4, 10 );
$n = count ( $arr );
if (isSumEqual( $arr , $n ))
echo "Yes" ;
else
echo "No" ;
?>
|
Javascript
<script>
function isSumEqual(ar, n)
{
let sum = 0;
for (let i = 0; i < n - 1; i++)
sum += ar[i];
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>
|
Time Complexity: O(n)
Auxiliary Space: O(1)
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