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)
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