Given an array of N positive elements. The task is to find an element which is equal to the sum of all elements of array except itself.
Examples:
Input: arr[] = {1, 2, 3, 6}
Output: 6
6 is the element which is equal to the sum of all
remaining elements i.e. 1 + 2+ 3 = 6
Input: arr[] = {2, 2, 2, 2}
Output: -1
Approach: First of all, find the sum of all elements of an array. Then iterate over each element and check the condition that if (a[i] == sum-a[i] ). If true then print that a[i], else print “-1” if no such element is found.
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach#include <bits/stdc++.h>#define ll long long intusing namespace std;// Function to find the elementint findEle(int arr[], int n){ // sum is use to store // sum of all elements // of array ll sum = 0; for (int i = 0; i < n; i++) sum += arr[i]; // iterate over all elements for (int i = 0; i < n; i++) if (arr[i] == sum - arr[i]) return arr[i]; return -1;}// Driver codeint main(){ int arr[] = { 1, 2, 3, 6 }; int n = sizeof(arr) / sizeof(arr[0]); cout << findEle(arr, n); return 0;} |
Java
// Java implementation of the above approachimport java.io.*;class GFG {// Function to find the elementstatic int findEle(int arr[], int n){ // sum is use to store // sum of all elements // of array int sum = 0; for (int i = 0; i < n; i++) sum += arr[i]; // iterate over all elements for (int i = 0; i < n; i++) if (arr[i] == sum - arr[i]) return arr[i]; return -1;}// Driver code public static void main (String[] args) { int arr[] = { 1, 2, 3, 6 }; int n = arr.length; System.out.print(findEle(arr, n)); }}// This code is contributed by shs.. |
Python3
# Python 3 implementation of # the above approach# Function to find the elementdef findEle(arr, n) : # sum is use to store # sum of all elements # of array sum = 0 for i in range(n) : sum += arr[i] # iterate over all elements for i in range(n) : if arr[i] == sum - arr[i] : return arr[i] return -1# Driver Codeif __name__ == "__main__" : arr = [1, 2, 3, 6 ] n = len(arr) print(findEle(arr, n))# This code is contributed by Ryuga |
C#
// C# implementation of the // above approach using System; class GFG { // Function to find the element static int findEle(int []arr, int n) { // sum is use to store // sum of all elements // of array int sum = 0; for (int i = 0; i < n; i++) sum += arr[i]; // iterate over all elements for (int i = 0; i < n; i++) if (arr[i] == sum - arr[i]) return arr[i]; return -1; } // Driver code static public void Main (String []args){ int []arr = { 1, 2, 3, 6 }; int n = arr.Length; Console.WriteLine(findEle(arr, n)); } } // This code is contributed // by Arnab Kundu |
PHP
<?php// PHP implementation of the above approach// Function to find the elementfunction findEle($arr, $n){ // sum is use to store // sum of all elements // of array $sum = 0; for ($i = 0; $i < $n; $i++) $sum += $arr[$i]; // iterate over all elements for ($i = 0; $i < $n; $i++) if ($arr[$i] == $sum - $arr[$i]) return $arr[$i]; return -1;}// Driver code$arr= array(1, 2, 3, 6 );$n = sizeof($arr);echo findEle($arr, $n);// This code is contributed // by Akanksha Rai?> |
Javascript
<script>// javascript implementation of the above approach// Function to find the elementfunction findEle(arr, n){ // sum is use to store // sum of all elements // of array var sum = 0; for (var i = 0; i < n; i++) sum += arr[i]; // iterate over all elements for (var i = 0; i < n; i++) if (arr[i] == sum - arr[i]) return arr[i]; return -1;}// Driver code var arr = [1, 2, 3, 6]; var n = arr.length; document.write(findEle(arr, n));// This code is contributed by ipg016107.</script> |
Output
6
Complexity Analysis:
- Time Complexity: O(n)
- Auxiliary Space: O(1)
Note: Above problem can be solved with the concept used in Check if the array has an element which is equal to sum of all the remaining elements.
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