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

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 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 codeint main(){    int arr[] = { 10, 4, 8, 13, 5 };    int n = sizeof(arr) / sizeof(arr);    findPair(arr, n);    return 0;}

Java

 // A simple Java program to find// pair whose sum already exists// in arrayimport 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 codeif __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 arrayusing 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.



Javascript



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 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 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 codeint main(){    int arr[] = { 10, 4, 8, 13, 5 };    int n = sizeof(arr) / sizeof(arr);    findPair(arr, n);    return 0;}

Java

 // Java program to find pair whose sum// already exists in arrayimport 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 s = new HashSet();         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 codearr = [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 arrayusing 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 s = new HashSet();         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



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.