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Check if there exist two elements in an array whose sum is equal to the sum of rest of the array
• Difficulty Level : Easy
• Last Updated : 19 Apr, 2021

We have an array of integers and we have to find two such elements in the array such that sum of these two elements is equal to the sum of rest of elements in array.
Examples:

```Input  : arr[] = {2, 11, 5, 1, 4, 7}
Output : Elements are 4 and 11
Note that 4 + 11 = 2 + 5 + 1 + 7

Input  : arr[] = {2, 4, 2, 1, 11, 15}
Output : Elements do not exist```

A simple solution is to consider every pair one by one, find its sum and compare the sum with sum of rest of the elements. If we find a pair whose sum is equal to rest of elements, we print the pair and return true. Time complexity of this solution is O(n3)
An efficient solution is to find sum of all array elements. Let this sum be “sum”. Now the task reduces to finding a pair with sum equals to sum/2.
Another optimization is, a pair can exist only if the sum of whole array is even because we are basically dividing it into two parts with equal sum.
1- Find the sum of whole array. Let this sum be “sum”
2- If sum is odd, return false.
3- Find a pair with sum equals to “sum/2” using hashing based method discussed here as method 2. If a pair is found, print it and return true.
4- If no pair exists, return false.
Below is the implementation of above steps.

C++

 `// C++ program to find whether two elements exist``// whose sum is equal to sum of rest of the elements.``#include``using` `namespace` `std;` `// Function to check whether two elements exist``// whose sum is equal to sum of rest of the elements.``bool` `checkPair(``int` `arr[],``int` `n)``{``    ``// Find sum of whole array``    ``int` `sum = 0;``    ``for` `(``int` `i=0; i s;``    ``for` `(``int` `i=0; i

Java

 `// Java program to find whether two elements exist``// whose sum is equal to sum of rest of the elements.``import` `java.util.*;` `class` `GFG``{``    ` `    ``// Function to check whether two elements exist``    ``// whose sum is equal to sum of rest of the elements.``    ``static` `boolean` `checkPair(``int` `arr[], ``int` `n)``    ``{``        ``// Find sum of whole array``        ``int` `sum = ``0``;``        ``for` `(``int` `i = ``0``; i < n; i++)``        ``{``            ``sum += arr[i];``        ``}` `        ``// If sum of array is not even than we can not``        ``// divide it into two part``        ``if` `(sum % ``2` `!= ``0``)``        ``{``            ``return` `false``;``        ``}` `        ``sum = sum / ``2``;` `        ``// For each element arr[i], see if there is``        ``// another element with vaalue sum - arr[i]``        ``HashSet s = ``new` `HashSet();``        ``for` `(``int` `i = ``0``; i < n; i++)``        ``{``            ``int` `val = sum - arr[i];` `            ``// If element exist than return the pair``            ``if` `(s.contains(val) &&``                ``val == (``int``) s.toArray()[s.size() - ``1``])``            ``{``                ``System.out.printf(``"Pair elements are %d and %d\n"``,``                        ``arr[i], val);``                ``return` `true``;``            ``}``            ``s.add(arr[i]);``        ``}``        ``return` `false``;``    ``}` `    ``// Driver program.``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `arr[] = {``2``, ``11``, ``5``, ``1``, ``4``, ``7``};``        ``int` `n = arr.length;``        ``if` `(checkPair(arr, n) == ``false``)``        ``{``            ``System.out.printf(``"No pair found"``);``        ``}``    ``}``}` `/* This code contributed by PrinciRaj1992 */`

Python3

 `# Python3 program to find whether``# two elements exist whose sum is``# equal to sum of rest of the elements.` `# Function to check whether two``# elements exist whose sum is equal``# to sum of rest of the elements.``def` `checkPair(arr, n):``    ``s ``=` `set``()``    ``sum` `=` `0` `    ``# Find sum of whole array``    ``for` `i ``in` `range``(n):``        ``sum` `+``=` `arr[i]``    ` `    ``# / If sum of array is not``    ``# even than we can not``    ``# divide it into two part``    ``if` `sum` `%` `2` `!``=` `0``:``        ``return` `False``    ``sum` `=` `sum` `/` `2` `    ``# For each element arr[i], see if``    ``# there is another element with``    ``# value sum - arr[i]``    ``for` `i ``in` `range``(n):``        ``val ``=` `sum` `-` `arr[i]``        ``if` `arr[i] ``not` `in` `s:``            ``s.add(arr[i])``            ` `        ``# If element exist than``        ``# return the pair``        ``if` `val ``in` `s:``            ``print``(``"Pair elements are"``,``                   ``arr[i], ``"and"``, ``int``(val))` `# Driver Code``arr ``=` `[``2``, ``11``, ``5``, ``1``, ``4``, ``7``]``n ``=` `len``(arr)``if` `checkPair(arr, n) ``=``=` `False``:``    ``print``(``"No pair found"``)` `# This code is contributed``# by Shrikant13`

C#

 `// C# program to find whether two elements exist``// whose sum is equal to sum of rest of the elements.``using` `System;``using` `System.Collections.Generic;` `class` `GFG``{``    ` `    ``// Function to check whether two elements exist``    ``// whose sum is equal to sum of rest of the elements.``    ``static` `bool` `checkPair(``int` `[]arr, ``int` `n)``    ``{``        ``// Find sum of whole array``        ``int` `sum = 0;``        ``for` `(``int` `i = 0; i < n; i++)``        ``{``            ``sum += arr[i];``        ``}` `        ``// If sum of array is not even than we can not``        ``// divide it into two part``        ``if` `(sum % 2 != 0)``        ``{``            ``return` `false``;``        ``}` `        ``sum = sum / 2;` `        ``// For each element arr[i], see if there is``        ``// another element with vaalue sum - arr[i]``        ``HashSet<``int``> s = ``new` `HashSet<``int``>();``        ``for` `(``int` `i = 0; i < n; i++)``        ``{``            ``int` `val = sum - arr[i];` `            ``// If element exist than return the pair``            ``if` `(s.Contains(val))``            ``{``                ``Console.Write(``"Pair elements are {0} and {1}\n"``,``                        ``arr[i], val);``                ``return` `true``;``            ``}``            ``s.Add(arr[i]);``        ``}``        ``return` `false``;``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main(String[] args)``    ``{``        ``int` `[]arr = {2, 11, 5, 1, 4, 7};``        ``int` `n = arr.Length;``        ``if` `(checkPair(arr, n) == ``false``)``        ``{``            ``Console.Write(``"No pair found"``);``        ``}``    ``}``}` `// This code contributed by Rajput-Ji`

PHP

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Javascript

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Output:

`Pair elements are 4 and 11`

Time complexity : O(n). unordered_set is implemented using hashing. Time complexity hash search and insert is assumed as O(1) here.
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