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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

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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 <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 then 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




// 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 then 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




# 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 then 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 then 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
// 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 then 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
?>


Javascript




<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 then 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.
Auxiliary Space: O(n)

Another Efficient Approach ( Space Optimization): First we will sort the array for Binary search . Then we will iterate whole array and check if a index exist in the array that pair with i such that arr[index] + a[i] == Rest sum of the array . We can use Binary search  to  find a index in the array by modifying the Binary search program . If a pair exist then print that pair . else print no pair exists.

Below is the implementation of the above approach :

C++




// C++ program for the above approach
 
#include<bits/stdc++.h>
using namespace std;
 
// Function to Find if a index exist in array such that
// arr[index] + a[i] == Rest sum of the array
int binarysearch(int arr[], int n, int i ,int Totalsum)
{
     int l = 0, r = n - 1 , index = -1;//initialize as -1
 
     while (l <= r)
     {   int mid = (l + r) / 2;
      
         int Pairsum = arr[mid] + arr[i];//pair sum
         int Restsum = Totalsum - Pairsum;//Rest sum
         
        if ( Pairsum == Restsum )
         {
            if( index != i )// checking a pair has same position or not
            {  index = mid; }//Then update index -1 to mid
               
            // Checking for adjacent element
            else if(index == i && mid>0 && arr[mid-1]==arr[i])
              {  index = mid-1; }//Then update index -1 to mid-1
           
            else if(index == i && mid<n-1 && arr[mid+1]==arr[i])
               {   index = mid+1; } //Then update index-1 to mid+1   
            break;
        }
        else if (Pairsum > Restsum)
        { // If pair sum is greater than rest sum , our index will
          // be in the Range [mid+1,R]
            l = mid + 1;
        }
        else {
          // If pair sum is smaller than rest sum , our index will
          // be in the Range [L,mid-1]
            r = mid - 1;
        }
    }
    // return index=-1 if a pair not exist with arr[i]
    // else return index such that arr[i]+arr[index] == sum of rest of arr
    return index;
}
// Function to check if a pair exist such their sum
// equal to rest of the array or not
bool checkPair(int arr[],int n)
{   int Totalsum=0;
    sort(arr , arr + n);//sort arr for Binary search
  
    for(int i=0;i<n;i++)
    {  Totalsum+=arr[i]; } //Finding total sum of the arr
     
    for(int i=0;i<n;i++)
    { // If index is -1 , Means arr[i] can't pair with any element
      // else arr[i]+a[index] == sum of rest of the arr
      int index = binarysearch(arr, n, i,Totalsum) ;
       
      if(index != -1) {
         cout<<"Pair elements are "<< arr[i]<<" and "<< arr[index];
         return true;
      }
    }
    return false;//Return false if a pair not exist
}
// Driver Code
int main()
{
    int arr[] = {2, 11, 5, 1, 4, 7};
    int n = sizeof(arr)/sizeof(arr[0]);
   
    //Function call
    if (checkPair(arr, n) == false)
    { cout<<"No pair found";  }
    return 0;
}
 
// This Approach is contributed by nikhilsainiofficial546


Java




// Java program for the above approach
 
import java.util.*;
 
class GFG {
    // Function to Find if a index exist in array such that
    // arr[index] + a[i] == Rest sum of the array
    static int binarysearch(int arr[], int n, int i,
                            int Totalsum)
    {
        int l = 0, r = n - 1, index = -1; // initialize as -1
 
        while (l <= r) {
            int mid = (l + r) / 2;
 
            int Pairsum = arr[mid] + arr[i]; // pair sum
            int Restsum = Totalsum - Pairsum; // Rest sum
 
            if (Pairsum == Restsum) {
                if (index != i) // checking a pair has same
                                // position or not
                {
                    index = mid;
                } // Then update index -1 to mid
 
                // Checking for adjacent element
                else if (index == i && mid > 0
                         && arr[mid - 1] == arr[i]) {
                    index = mid - 1;
                } // Then update index -1 to mid-1
 
                else if (index == i && mid < n - 1
                         && arr[mid + 1] == arr[i]) {
                    index = mid + 1;
                } // Then update index-1 to mid+1
                break;
            }
            else if (Pairsum > Restsum) {
                // If pair sum is greater than rest sum ,
                // our index will be in the Range [mid+1,R]
                l = mid + 1;
            }
            else {
                // If pair sum is smaller than rest sum ,
                // our index will be in the Range [L,mid-1]
                r = mid - 1;
            }
        }
        // return index=-1 if a pair not exist with arr[i]
        // else return index such that arr[i]+arr[index] ==
        // sum of rest of arr
        return index;
    }
 
    // Function to check if a pair exist such their sum
    // equal to rest of the array or not
    static boolean checkPair(int arr[], int n)
    {
        int Totalsum = 0;
        Arrays.sort(arr); // sort arr for Binary search
 
        for (int i = 0; i < n; i++) {
            Totalsum += arr[i];
        } // Finding total sum of the arr
 
        for (int i = 0; i < n; i++) {
            // If index is -1 , Means arr[i] can't pair with
            // any element else arr[i]+a[index] == sum of
            // rest of the arr
            int index = binarysearch(arr, n, i, Totalsum);
 
            if (index != -1) {
                System.out.println("Pair elements are "
                                   + arr[i] + " and "
                                   + arr[index]);
                return true;
            }
        }
        return false; // Return false if a pair not exist
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int arr[] = { 2, 11, 5, 1, 4, 7 };
        int n = arr.length;
 
        // Function call
        if (checkPair(arr, n) == false) {
            System.out.println("No pair found");
        }
    }
}


Python3




# Python program for the above approach
 
# Function to find if a index exist in array such that
# arr[index] + a[i] == Rest sum of the array
 
 
def binarysearch(arr, n, i, Totalsum):
    l = 0
    r = n - 1
    index = -1  # Initialize as -1
 
    while l <= r:
        mid = (l + r) // 2
        Pairsum = arr[mid] + arr[i]  # Pair sum
        Restsum = Totalsum - Pairsum  # Rest sum
 
        if Pairsum == Restsum:
            if index != i:  # Checking if a pair has the same position or not
                index = mid  # Then update index -1 to mid
 
            # Checking for adjacent element
            elif index == i and mid > 0 and arr[mid - 1] == arr[i]:
                index = mid - 1  # Then update index -1 to mid-1
 
            elif index == i and mid < n - 1 and arr[mid + 1] == arr[i]:
                index = mid + 1  # Then update index-1 to mid+1
 
            break
        elif Pairsum > Restsum:
            # If pair sum is greater than rest sum, our index will
            # be in the Range [mid+1,R]
            l = mid + 1
        else:
            # If pair sum is smaller than rest sum, our index will
            # be in the Range [L,mid-1]
            r = mid - 1
 
    # Return index=-1 if a pair not exist with arr[i]
    # else return index such that arr[i]+arr[index] == sum of rest of arr
    return index
 
# Function to check if a pair exists such that their sum
# equals to rest of the array or not
 
 
def checkPair(arr, n):
    Totalsum = 0
    arr = sorted(arr)  # Sort arr for Binary search
 
    for i in range(n):
        Totalsum += arr[i]  # Finding total sum of the arr
 
    for i in range(n):
        # If index is -1, means arr[i] can't pair with any element
        # else arr[i]+a[index] == sum of rest of the arr
        index = binarysearch(arr, n, i, Totalsum)
 
        if index != -1:
            print("Pair elements are", arr[i], "and", arr[index])
            return True
 
    return False  # Return false if a pair not exist
 
 
# Driver Code
arr = [2, 11, 5, 1, 4, 7]
n = len(arr)
 
# Function call
if checkPair(arr, n) == False:
    print("No pair found")


C#




using System;
 
class GFG
{
  // Function to Find if a index exist in array such that
  // arr[index] + a[i] == Rest sum of the array
  static int BinarySearch(int[] arr, int n, int i, int totalSum)
  {
    int l = 0, r = n - 1, index = -1; // initialize as -1
 
    while (l <= r)
    {
      int mid = (l + r) / 2;
 
      int pairSum = arr[mid] + arr[i]; // pair sum
      int restSum = totalSum - pairSum; // rest sum
 
      if (pairSum == restSum)
      {
        if (index != i) // checking a pair has same
          // position or not
        {
          index = mid;
        } // Then update index -1 to mid
 
        // Checking for adjacent element
        else if (index == i && mid > 0
                 && arr[mid - 1] == arr[i])
        {
          index = mid - 1;
        } // Then update index -1 to mid-1
 
        else if (index == i && mid < n - 1
                 && arr[mid + 1] == arr[i])
        {
          index = mid + 1;
        } // Then update index-1 to mid+1
        break;
      }
      else if (pairSum > restSum)
      {
        // If pair sum is greater than rest sum ,
        // our index will be in the Range [mid+1,R]
        l = mid + 1;
      }
      else
      {
        // If pair sum is smaller than rest sum ,
        // our index will be in the Range [L,mid-1]
        r = mid - 1;
      }
    }
    // return index=-1 if a pair not exist with arr[i]
    // else return index such that arr[i]+arr[index] ==
    // sum of rest of arr
    return index;
  }
 
  // Function to check if a pair exist such their sum
  // equal to rest of the array or not
  static bool CheckPair(int[] arr, int n)
  {
    int totalSum = 0;
    Array.Sort(arr); // sort arr for Binary search
 
    for (int i = 0; i < n; i++)
    {
      totalSum += arr[i];
    } // Finding total sum of the arr
 
    for (int i = 0; i < n; i++)
    {
      // If index is -1 , Means arr[i] can't pair with
      // any element else arr[i]+a[index] == sum of
      // rest of the arr
      int index = BinarySearch(arr, n, i, totalSum);
 
      if (index != -1)
      {
        Console.WriteLine("Pair elements are " + arr[i] + " and " + arr[index]);
        return true;
      }
    }
    return false; // Return false if a pair not exist
  }
 
  // Driver Code
  static void Main(string[] args)
  {
    int[] arr = { 2, 11, 5, 1, 4, 7 };
    int n = arr.Length;
 
    // Function call
    if (!CheckPair(arr, n))
    {
      Console.WriteLine("No pair found");
    }
  }
}


Javascript




// JavaScript program for the above approach
// function to find if a index exist in array such that
// arr[index] + a[i] == Rest sum of the array
 
function binarysearch(arr, n, i, TotalSum){
    let l = 0;
    let r = n-1;
    let index = -1;
     
    while(l <= r){
        let mid = parseInt((l+r)/2);
        let Pairsum = arr[mid] + arr[i];
        let Restsum = TotalSum - Pairsum;
         
        if ( Pairsum == Restsum )
        {
            if( index != i )// checking a pair has same position or not
            {  index = mid; }//Then update index -1 to mid
                
            // Checking for adjacent element
            else if(index == i && mid>0 && arr[mid-1]==arr[i])
            {  index = mid-1; }//Then update index -1 to mid-1
             
            else if(index == i && mid<n-1 && arr[mid+1]==arr[i])
            {   index = mid+1; } //Then update index-1 to mid+1  
            break;
        }
        else if (Pairsum > Restsum)
        { // If pair sum is greater than rest sum , our index will
          // be in the Range [mid+1,R]
            l = mid + 1;
        }
        else {
          // If pair sum is smaller than rest sum , our index will
          // be in the Range [L,mid-1]
            r = mid - 1;
        }
    }
    // return index=-1 if a pair not exist with arr[i]
    // else return index such that arr[i]+arr[index] == sum of rest of arr
    return index;
}
 
// Function to check if a pair exist such their sum
// equal to rest of the array or not
function checkPair(arr, n){
    let Totalsum = 0;
    arr.sort(function(a, b){return a - b});
     
    for(let i=0;i<n;i++)
    {  Totalsum+=arr[i]; } //Finding total sum of the arr
      
    for(let i=0;i<n;i++)
    { // If index is -1 , Means arr[i] can't pair with any element
      // else arr[i]+a[index] == sum of rest of the arr
      let index = binarysearch(arr, n, i,Totalsum) ;
        
      if(index != -1) {
         console.log("Pair elements are " + arr[i] + " and " + arr[index]);
         return true;
      }
    }
    return false;//Return false if a pair not exist
}
 
// driver code to test above function
let arr = [2, 11, 5, 1, 4, 7];
let n = arr.length;
 
// function call
if(checkPair(arr, n) == false
    console.log("No Pair Found")
     
    // THIS CODE IS CONTRIBUTED BY YASH AGARWAL(YASHAGARWAL2852002)


Output

Pair elements are 11 and 4

Time complexity : O(n * logn) 
Auxiliary Space: O(1)

 



Last Updated : 24 Apr, 2023
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