Skip to content
Related Articles

Related Articles

Improve Article
Save Article
Like Article

Find pairs in array whose sums already exist in array

  • Difficulty Level : Basic
  • Last Updated : 14 May, 2021

Given an array of n distinct and positive elements, the task is to find pair whose sum already exists in the given array. 
Examples : 
 

Input : arr[] = {2, 8, 7, 1, 5};
Output : 2 5
         7 1    
     
Input : arr[] = {7, 8, 5, 9, 11};
Output : Not Exist

 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

A Naive Approach is to run three loops to find pair whose sum exists in an array. 
 



C++




// A simple C++ program to find pair whose sum
// already exists in array
#include <bits/stdc++.h>
using namespace std;
 
// Function to find pair whose sum exists in arr[]
void findPair(int arr[], int n)
{
    bool found = false;
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
            for (int k = 0; k < n; k++) {
                if (arr[i] + arr[j] == arr[k]) {
                    cout << arr[i] << " " << arr[j] << endl;
                    found = true;
                }
            }
        }
    }
 
    if (found == false)
        cout << "Not exist" << endl;
}
 
// Driven code
int main()
{
    int arr[] = { 10, 4, 8, 13, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);
    findPair(arr, n);
    return 0;
}

Java




// A simple Java program to find
// pair whose sum already exists
// in array
import java.io.*;
 
public class GFG {
 
    // Function to find pair whose
    // sum exists in arr[]
    static void findPair(int[] arr, int n)
    {
        boolean found = false;
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                for (int k = 0; k < n; k++) {
                    if (arr[i] + arr[j] == arr[k]) {
                        System.out.println(arr[i] +
                                      " " + arr[j]);
                        found = true;
                    }
                }
            }
        }
 
        if (found == false)
            System.out.println("Not exist");
    }
 
    // Driver code
    static public void main(String[] args)
    {
        int[] arr = {10, 4, 8, 13, 5};
        int n = arr.length;
        findPair(arr, n);
    }
}
 
// This code is contributed by vt_m.

Python3




# A simple python program to find pair
# whose sum already exists in array
 
# Function to find pair whose sum
# exists in arr[]
def findPair(arr, n):
    found = False
    for i in range(0, n):
        for j in range(i + 1, n):
            for k in range(0, n):
                if (arr[i] + arr[j] == arr[k]):
                    print(arr[i], arr[j])
                    found = True
 
    if (found == False):
        print("Not exist")
 
# Driver code
if __name__ == '__main__':
    arr = [ 10, 4, 8, 13, 5 ]
    n = len(arr)
    findPair(arr, n)
     
# This code contributed by 29AjayKumar

C#




// A simple C# program to find
// pair whose sum already exists
// in array
using System;
 
public class GFG {
 
    // Function to find pair whose
    // sum exists in arr[]
    static void findPair(int[] arr, int n)
    {
        bool found = false;
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                for (int k = 0; k < n; k++) {
                    if (arr[i] + arr[j] == arr[k]) {
                        Console.WriteLine(arr[i] +
                                      " " + arr[j]);
                        found = true;
                    }
                }
            }
        }
 
        if (found == false)
            Console.WriteLine("Not exist");
    }
 
    // Driver code
    static public void Main(String []args)
    {
        int[] arr = {10, 4, 8, 13, 5};
        int n = arr.Length;
        findPair(arr, n);
    }
}
 
// This code is contributed by vt_m.

PHP




<?php
// A simple php program to
// find pair whose sum
// already exists in array
 
// Function to find pair whose
// sum exists in arr[]
function findPair($arr, $n)
{
    $found = false;
    for ($i = 0; $i < $n; $i++)
    {
        for ($j = $i + 1; $j < $n; $j++)
        {
            for ($k = 0; $k < $n; $k++)
            {
                if ($arr[$i] + $arr[$j] == $arr[$k])
                {
                    echo $arr[$i] , " " , $arr[$j] ;
                    $found = true;
                }
            }
        }
    }
 
    if ($found == false)
        echo "Not exist" ;
}
 
// Driver code
$arr = array( 10, 4, 8, 13, 5 );
$n = sizeof($arr) / sizeof($arr[0]);
findPair($arr, $n);
     
// This code is contributed by nitin mittal.
?>

Javascript




<script>
// A simple Javascript program to find
// pair whose sum already exists
// in array
     
    // Function to find pair whose
    // sum exists in arr[]
    function findPair(arr,n)
    {
        let found = false;
        for (let i = 0; i < n; i++) {
            for (let j = i + 1; j < n; j++) {
                for (let k = 0; k < n; k++) {
                    if (arr[i] + arr[j] == arr[k]) {
                        document.write(arr[i] +
                                      " " + arr[j]+"<br>");
                        found = true;
                    }
                }
            }
        }
   
        if (found == false)
            document.write("Not exist");
    }
     
    // Driver code
    let arr=[10, 4, 8, 13, 5];
    let n = arr.length;
    findPair(arr, n);
     
     
 
// This code is contributed by patel2127
</script>

Output : 

8 5

An Efficient solution is to store all elements in a hash table (unordered_set in C++) and check one by one all pairs and check its sum exists in set or not. If it exists in the set then print pair. If no pair found in the array then print not exists.
 

 

C++




// C++ program to find pair whose sum already
// exists in array
#include <bits/stdc++.h>
using namespace std;
 
// Function to find pair whose sum exists in arr[]
void findPair(int arr[], int n)
{
    // Hash to store all element of array
    unordered_set<int> s;
    for (int i = 0; i < n; i++)
        s.insert(arr[i]);
 
    bool found = false;
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
            // Check sum already exists or not
            if (s.find(arr[i] + arr[j]) != s.end()) {
                cout << arr[i] << " " << arr[j] << endl;
                found = true;
            }
        }
    }
 
    if (found == false)
        cout << "Not exist" << endl;
}
 
// Driven code
int main()
{
    int arr[] = { 10, 4, 8, 13, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);
    findPair(arr, n);
    return 0;
}

Java




// Java program to find pair whose sum
// already exists in array
import java.util.*;
import java.lang.*;
import java.io.*;
 
class Getpairs {
    // Function to find pair whose sum
    // exists in arr[]
    public static void findPair(int[] arr, int n)
    {
        /* store all the array elements as a
        Hash value*/
        HashSet<Integer> s = new HashSet<Integer>();
 
        for (Integer i : arr) {
            s.add(i);
        }
 
        /* Run two loop and check for the sum
    in the Hashset*/
        /* if not a single pair exist then found
    will be false else true*/
        boolean found = false;
 
        for (int i = 0; i < n - 1; i++) {
            for (int j = i + 1; j < n; j++) {
                int sum = arr[i] + arr[j];
                if (s.contains(sum)) {
                    /* if the sum is present in
                 hashset then found become
                true*/
                    found = true;
 
                    System.out.println(arr[i] + " "
                                       + arr[j]);
                }
            }
        }
 
        if (found == false)
            System.out.println("Not Exist ");
    }
 
    // driver function
    public static void main(String args[])
    {
        int[] arr = { 10, 4, 8, 13, 5 };
        int n = arr.length;
        findPair(arr, n);
    }
}
 
// This code is contributed by Smarak Chopdar

Python3




# Python3 program to find pair whose
# sum already exist in arrar
 
# Function to find pair whose
# sum sxists in arr[]
def findPair(arr, n):
     
    # hash to store all element of array
    s = {i : 1 for i in arr}
     
    found = False
     
    for i in range(n):
        for j in range(i + 1, n):
             
            # check if sum already exists or not
            if arr[i] + arr[j] in s.keys():
                print(arr[i], arr[j])
                found = True
    if found == False:
        print("Not exist")
         
# Driver code
arr = [10, 4, 8, 13, 5]
 
n = len(arr)
 
findPair(arr, n)
     
# This code is contributed
# by Mohit Kumar

C#




// C# program to find pair whose sum
// already exists in array
using System;
using System.Collections.Generic;
 
class Getpairs
{
    // Function to find pair whose sum
    // exists in arr[]
    public static void findPair(int[] arr, int n)
    {
        /* store all the array elements as a
        Hash value*/
        HashSet<int> s = new HashSet<int>();
 
        foreach (int i in arr)
        {
            s.Add(i);
        }
 
        /* Run two loop and check for the sum
    in the Hashset*/
        /* if not a single pair exist then found
    will be false else true*/
        bool found = false;
 
        for (int i = 0; i < n - 1; i++)
        {
            for (int j = i + 1; j < n; j++)
            {
                int sum = arr[i] + arr[j];
                if (s.Contains(sum))
                {
                    /* if the sum is present in
                    hashset then found become
                    true*/
                    found = true;
 
                    Console.WriteLine(arr[i] + " "
                                    + arr[j]);
                }
            }
        }
 
        if (found == false)
            Console.WriteLine("Not Exist ");
    }
 
    // Driver code
    public static void Main()
    {
        int[] arr = { 10, 4, 8, 13, 5 };
        int n = arr.Length;
        findPair(arr, n);
    }
}
 
// This code contributed by Rajput-Ji

Javascript




<script>
 
// Javascript program to find pair whose sum
// already exists in array   
     
    // Function to find pair whose sum
    // exists in arr[]
    function findPair(arr,n)
    {
        /* store all the array elements as a
        Hash value*/
        let s = new Set();
  
        for (let i=0;i<arr.length;i++) {
            s.add(arr[i]);
        }
  
        /* Run two loop and check for the sum
    in the Hashset*/
        /* if not a single pair exist then found
    will be false else true*/
        let found = false;
  
        for (let i = 0; i < n - 1; i++) {
            for (let j = i + 1; j < n; j++) {
                let sum = arr[i] + arr[j];
                if (s.has(sum)) {
                    /* if the sum is present in
                 hashset then found become
                true*/
                    found = true;
  
                    document.write(arr[i] + " "
                                  + arr[j]+"<br>");
                }
            }
        }
  
        if (found == false)
            document.write("Not Exist ");
    }
     
    // driver function
    let arr=[10, 4, 8, 13, 5 ];
    let n = arr.length;
    findPair(arr, n);
     
// This code is contributed by unknown2108
 
</script>

Output:  

8 5

Time Complexity : O(n2)
This article is contributed by DANISH_RAZA. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 




My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!