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++
// 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 elementbool 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 codeint 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
// Java implementation of the approachimport 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. |
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 elementdef 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 codeif __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#
// C# implementation of the approachusing 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// 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?> |
Javascript
<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> |
Output:
Yes
Time Complexity: O(n)
Auxiliary Space: O(1)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
