Check if there exist two elements in an array whose sum is equal to the sum of rest of the array - GeeksforGeeks

27 Sep, 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++
Java
Python3
C#
PHP
Javascript
// C++ program to find whether two elements exist// whose sum is equal to sum of rest of the elements.#include<bits/stdc++.h>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<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 value sum - arr[i]    unordered_set<int> s;    for (int i=0; i<n; i++)    {        int val = sum-arr[i];         // If element exist than return the pair        if (s.find(val) != s.end())        {            printf("Pair elements are %d and %d\n",                                    arr[i], val);            return true;        }         s.insert(arr[i]);    }     return false;} // Driver program.int main(){    int arr[] = {2, 11, 5, 1, 4, 7};    int n = sizeof(arr)/sizeof(arr[0]);    if (checkPair(arr, n) == false)       printf("No pair found");    return 0;}
// 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 value sum - arr[i]        HashSet<Integer> s = new HashSet<Integer>();        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 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 Codearr = [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# 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 value 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// PHP 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.function checkPair(&$arr, $n){    // Find sum of whole array    $sum = 0;    for ($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 value sum - arr[i]    $s = array();    for ($i = 0; $i < $n; $i++)    {        $val = $sum - $arr[$i];         // If element exist than return the pair        if (array_search($val, $s))        {            echo "Pair elements are " . $arr[$i] .                 " and " . $val . "\n";                                                 return true;        }         array_push($s, $arr[$i]);    }     return false;} // Driver Code$arr = array(2, 11, 5, 1, 4, 7);$n = sizeof($arr);if (checkPair($arr, $n) == false)    echo "No pair found"; // This code is contributed by ita_c?>
<script> // Javascript 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.    function checkPair(arr,n)    {        // Find sum of whole array        let sum = 0;        for (let 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 = Math.floor(sum / 2);           // For each element arr[i], see if there is        // another element with value sum - arr[i]        let s = new Set();        for (let i = 0; i < n; i++)        {            let val = sum - arr[i];               // If element exist than return the pair                         if(!s.has(arr[i]))            {                s.add(arr[i])            }                         if (s.has(val) )            {                document.write("Pair elements are "+                        arr[i]+" and "+ val+"<br>");                return true;            }            s.add(arr[i]);        }        return false;    }    // Driver program.    let arr=[2, 11, 5, 1, 4, 7];    let n = arr.length;    if (checkPair(arr, n) == false)    {        document.write("No pair found");    }         // This code is contributed by rag2127     </script>
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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shrikanth13
ukasp
princiraj1992
Rajput-Ji
rag2127
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