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Find all pairs with a given sum in two unsorted arrays

Last Updated : 15 Sep, 2024
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Given two unsorted arrays of distinct elements, the task is to find all pairs from both arrays whose sum is equal to a given value X.

Examples: 

Input: arr1[] = {-1, -2, 4, -6, 5, 7}, arr2[] = {6, 3, 4, 0} , x = 8
Output: 4 4 5 3

Input: arr1[] = {1, 2, 4, 5, 7}, arr2[] = {5, 6, 3, 4, 8}, x = 9
Output: 1 8 4 5 5 4

Table of Content

[Naive Approach] Using Two Nested Loops - O(n2) time and O(1) auxiliary space

The very basic idea is to use two nested loops to iterate over each element in arr1 and for each element in arr1, iterate over each element in arr2. Check if the sum of the current elements from arr1 and arr2 equals X. If yes then print the pair.

Code Implementation:

C++ 
// C++ program to find all pairs in both arrays
// whose sum is equal to given value x
#include <bits/stdc++.h>
using namespace std;

// Function to print all pairs in both arrays
// whose sum is equal to given value x
void findPairs(int arr1[], int arr2[], int n, int m, int x)
{
    for (int i = 0; i < n; i++)
        for (int j = 0; j < m; j++)
            if (arr1[i] + arr2[j] == x)
                cout << arr1[i] << " " << arr2[j] << endl;
}

// Driver code
int main()
{
    int arr1[] = { 1, 2, 3, 7, 5, 4 };
    int arr2[] = { 0, 7, 4, 3, 2, 1 };
    int n = sizeof(arr1) / sizeof(int);
    int m = sizeof(arr2) / sizeof(int);
    int x = 8;
    findPairs(arr1, arr2, n, m, x);
    return 0;
}
C 
// C program to find all pairs in both arrays
// whose sum is equal to given value x
#include <stdio.h>

// Function to print all pairs in both arrays
// whose sum is equal to given value x
void findPairs(int arr1[], int arr2[], int n, int m, int x)
{
    for (int i = 0; i < n; i++)
        for (int j = 0; j < m; j++)
            if (arr1[i] + arr2[j] == x)
                printf("%d %d\n", arr1[i], arr2[j]);
}

// Driver code
int main()
{
    int arr1[] = { 1, 2, 3, 7, 5, 4 };
    int arr2[] = { 0, 7, 4, 3, 2, 1 };
    int n = sizeof(arr1) / sizeof(int);
    int m = sizeof(arr2) / sizeof(int);
    int x = 8;
    findPairs(arr1, arr2, n, m, x);
    return 0;
}
Java 
// Java program to find all pairs in both arrays
// whose sum is equal to given value x
import java.io.*;

class GFG {

    // Function to print all pairs in both arrays
    // whose sum is equal to given value x
    static void findPairs(int arr1[], int arr2[], int n,
                          int m, int x)
    {
        for (int i = 0; i < n; i++)
            for (int j = 0; j < m; j++)
                if (arr1[i] + arr2[j] == x)
                    System.out.println(arr1[i] + " "
                                       + arr2[j]);
    }

    // Driver code
    public static void main(String[] args)
    {
        int arr1[] = { 1, 2, 3, 7, 5, 4 };
        int arr2[] = { 0, 7, 4, 3, 2, 1 };
        int x = 8;
        findPairs(arr1, arr2, arr1.length, arr2.length, x);
    }
}
Python 
# Python 3 program to find all 
# pairs in both arrays whose 
# sum is equal to given value x

# Function to print all pairs 
# in both arrays whose sum is
# equal to given value x
def findPairs(arr1, arr2, n, m, x):

    for i in range(0, n):
        for j in range(0, m):
            if (arr1[i] + arr2[j] == x):
                print(arr1[i], arr2[j])

# Driver code
arr1 = [1, 2, 3, 7, 5, 4]
arr2 = [0, 7, 4, 3, 2, 1]
n = len(arr1)
m = len(arr2)
x = 8
findPairs(arr1, arr2, n, m, x)
C# 
// C# program to find all
// pairs in both arrays
// whose sum is equal to
// given value x
using System;

class GFG {

    // Function to print all
    // pairs in both arrays
    // whose sum is equal to
    // given value x
    static void findPairs(int[] arr1, int[] arr2,
                          int n, int m, int x)
    {
        for (int i = 0; i < n; i++)
            for (int j = 0; j < m; j++)
                if (arr1[i] + arr2[j] == x)
                    Console.WriteLine(arr1[i] + " " + arr2[j]);
    }

    // Driver code
    static void Main()
    {
        int[] arr1 = { 1, 2, 3, 7, 5, 4 };
        int[] arr2 = { 0, 7, 4, 3, 2, 1 };
        int x = 8;
        findPairs(arr1, arr2,
                  arr1.Length,
                  arr2.Length, x);
    }
}
JavaScript 
// Function to find all pairs in both arrays whose sum is equal to given value x
function findPairs(arr1, arr2, n, m, x) {
  for (let i = 0; i < n; i++) {
    for (let j = 0; j < m; j++) {
      if (arr1[i] + arr2[j] === x) {
        console.log(arr1[i], arr2[j]);
      }
    }
  }
}

// Example usage
const arr1 = [1, 2, 3, 7, 5, 4];
const arr2 = [0, 7, 4, 3, 2, 1];
const n = arr1.length;
const m = arr2.length;
const x = 8;

findPairs(arr1, arr2, n, m, x);
PHP 
<?php
// PHP program to find all pairs 
// in both arrays whose sum is 
// equal to given value x

// Function to print all pairs 
// in both arrays whose sum is
// equal to given value x
function findPairs($arr1, $arr2, 
                   $n, $m, $x)
{
    for ($i = 0; $i < $n; $i++)
        for ($j = 0; $j < $m; $j++)
            if ($arr1[$i] + $arr2[$j] == $x)
                echo $arr1[$i] . " " . 
                     $arr2[$j] . "\n";
} 

// Driver code
$arr1 = array(1, 2, 3, 7, 5, 4);
$arr2 = array(0, 7, 4, 3, 2, 1);
$n = count($arr1);
$m = count($arr2);
$x = 8;
findPairs($arr1, $arr2, 
          $n, $m, $x);

// This code is contributed 
// by Sam007
?>

Output
1 7
7 1
5 3
4 4

Time Complexity : O(n^2) 
Auxiliary Space : O(1)

[Optimized Approach] Using Sorting and Binary Search - O(n log(n) + m log(n)) time and O(1) auxiliary space

Sort one of the arrays and use binary search to find the complement of each element in the second array.

Code Implementation:

Java 
import java.util.Arrays;

public class FindPairs {

    // Binary search function
    public static int binarySearch(int[] arr, int l, int r,
                                   int x)
    {
        while (l <= r) {
            int m = l + (r - l) / 2;
            if (arr[m] == x)
                return m;
            if (arr[m] < x)
                l = m + 1;
            else
                r = m - 1;
        }
        return -1;
    }

    // Function to find pairs in two arrays whose sum equals
    // a given value
    public static void findPairs(int[] arr1, int[] arr2,
                                 int n, int m, int x)
    {
        // Sort the first array in ascending order
        Arrays.sort(arr1);
      
        // Iterate through the second array
        for (int i = 0; i < m; i++) {
          
            // Calculate the complement value needed to
            // reach the target sum
            int complement = x - arr2[i];
          
            // Perform binary search on the sorted first
            // array to find the complement
            if (binarySearch(arr1, 0, n - 1, complement)
                != -1) {
              
                // If the complement is found, print the
                // pair
                System.out.println(complement + " "
                                   + arr2[i]);
            }
        }
    }

    public static void main(String[] args)
    {
        // Example arrays and target value
        int[] arr1 = { 1, 2, 3, 7, 5, 4 };
        int[] arr2 = { 0, 7, 4, 3, 2, 1 };
        int n = arr1.length;
        int m = arr2.length;
        int x = 8;
      
        // Find and print pairs with sum equal to x
        findPairs(arr1, arr2, n, m, x);
    }
}
Python 
def binary_search(arr, l, r, x):

    while l <= r:
        m = l + (r - l) // 2
        if arr[m] == x:
            return m
        elif arr[m] < x:
            l = m + 1
        else:
            r = m - 1
    return -1


def find_pairs(arr1, arr2, n, m, x):

    # Sort the first array in ascending order
    arr1.sort()
    # Iterate through the second array
    for i in range(m):
        # Calculate the complement value needed to reach the target sum
        complement = x - arr2[i]
        # Perform binary search on the sorted first array to find the complement
        index = binary_search(arr1, 0, n - 1, complement)
        if index != -1:
            # If the complement is found, print the pair
            print(complement, arr2[i])


# Example arrays and target value
arr1 = [1, 2, 3, 7, 5, 4]
arr2 = [0, 7, 4, 3, 2, 1]
n = len(arr1)
m = len(arr2)
x = 8
# Find and print pairs with sum equal to x
find_pairs(arr1, arr2, n, m, x)
JavaScript 
function findPairs(arr1, arr2, n, m, x) {

  // Sort arr1 in ascending order
  arr1.sort((a, b) => a - b); 

  for (let i = 0; i < m; i++) {
  
    // Calculate the complement needed
    const complement = x - arr2[i]; 
    
    // Search for the complement
    const index = binarySearch(arr1, 0, n - 1, complement); 

    if (index !== -1) {
      console.log(complement, arr2[i]); // Print the pair if found
    }
  }
}

function binarySearch(arr, l, r, x) {
  while (l <= r) {
    const m = Math.floor((l + r) / 2);
    if (arr[m] === x) return m;
    if (arr[m] < x) l = m + 1;
    else r = m - 1;
  }
  return -1; // Element not found
}

// Example usage
const arr1 = [1, 2, 3, 7, 5, 4];
const arr2 = [0, 7, 4, 3, 2, 1];
const n = arr1.length;
const m = arr2.length;
const x = 8;

findPairs(arr1, arr2, n, m, x);

Output
1 7
4 4
5 3
7 1

Time Complexity: O(n log(n) + m log(n))
Auxiliary Space: O(1)

[Expected Approach] Using Hashing Approach - O(n + m) time and O(n + m) auxiliary space

The idea is to use a hash set to store elements of one array and for each element in the second array, calculate the required complement that would sum to X and check if this complement exists in the hash set. If such pair is present, we print the pairs.

Code Implementation:

C++ 
// C++ program to find all pair in both arrays
// whose sum is equal to given value x
#include <bits/stdc++.h>
using namespace std;

// Function to find all pairs in both arrays
// whose sum is equal to given value x
void findPairs(int arr1[], int arr2[], int n,
               int m, int x)
{
    // Insert all elements of first array in a set
    unordered_set<int> s;
    for (int i = 0; i < n; i++)
        s.insert(arr1[i]);

    // Subtract sum from second array elements one
    // by one and check it's present in array first
    // or not
    for (int j = 0; j < m; j++)
        if (s.find(x - arr2[j]) != s.end())
            cout << x - arr2[j] << " "
                 << arr2[j] << endl;
}

// Driver code
int main()
{
    int arr1[] = { 1, 0, -4, 7, 6, 4 };
    int arr2[] = { 0, 2, 4, -3, 2, 1 };
    int x = 8;
    int n = sizeof(arr1) / sizeof(int);
    int m = sizeof(arr2) / sizeof(int);
    findPairs(arr1, arr2, n, m, x);
    return 0;
}
Java 
// JAVA Code for Given two unsorted arrays,
// find all pairs whose sum is x
import java.util.*;

class GFG {

    // Function to find all pairs in both arrays
    // whose sum is equal to given value x
    public static void findPairs(int arr1[], int arr2[],
                                 int n, int m, int x)
    {
        // Insert all elements of first array in a hash
        HashMap<Integer, Integer> s = new HashMap<Integer, Integer>();

        for (int i = 0; i < n; i++)
            s.put(arr1[i], 0);

        // Subtract sum from second array elements one
        // by one and check it's present in array first
        // or not
        for (int j = 0; j < m; j++)
            if (s.containsKey(x - arr2[j]))
                System.out.println(x - arr2[j] + " " + arr2[j]);
    }

    /* Driver program to test above function */
    public static void main(String[] args)
    {
        int arr1[] = { 1, 0, -4, 7, 6, 4 };
        int arr2[] = { 0, 2, 4, -3, 2, 1 };
        int x = 8;

        findPairs(arr1, arr2, arr1.length, arr2.length, x);
    }
}
Python 
# Python3 program to find all 
# pair in both arrays whose 
# sum is equal to given value x

# Function to find all pairs 
# in both arrays whose sum is
# equal to given value x
def findPairs(arr1, arr2, n, m, x):

    # Insert all elements of 
    # first array in a hash
    s = set()
    for i in range (0, n):
        s.add(arr1[i])

    # Subtract sum from second 
    # array elements one by one 
    # and check it's present in
    # array first or not
    for j in range(0, m):
        if ((x - arr2[j]) in s):
            print((x - arr2[j]), '', arr2[j])

# Driver code
arr1 = [1, 0, -4, 7, 6, 4]
arr2 = [0, 2, 4, -3, 2, 1]
x = 8

n = len(arr1)
m = len(arr2)
findPairs(arr1, arr2, n, m, x)
C# 
// C# Code for Given two unsorted arrays,
// find all pairs whose sum is x
using System;
using System.Collections.Generic;

class GFG {

    // Function to find all pairs in
    // both arrays whose sum is equal
    // to given value x
    public static void findPairs(int[] arr1, int[] arr2,
                                 int n, int m, int x)
    {
        // Insert all elements of first
        // array in a hash
        Dictionary<int,
                   int>
            s = new Dictionary<int,
                               int>();

        for (int i = 0; i < n; i++) {
            s[arr1[i]] = 0;
        }

        // Subtract sum from second array
        // elements one by one and check
        // it's present in array first
        // or not
        for (int j = 0; j < m; j++) {
            if (s.ContainsKey(x - arr2[j])) {
                Console.WriteLine(x - arr2[j] + " " + arr2[j]);
            }
        }
    }

    // Driver Code
    public static void Main(string[] args)
    {
        int[] arr1 = new int[] { 1, 0, -4, 7, 6, 4 };
        int[] arr2 = new int[] { 0, 2, 4, -3, 2, 1 };
        int x = 8;

        findPairs(arr1, arr2, arr1.Length,
                  arr2.Length, x);
    }
}
JavaScript 
function findPairs(arr1, arr2, n, m, x) {

  // Create a set to store elements of the first array
  const set = new Set(arr1);

  // Iterate through the second array
  for (let j = 0; j < m; j++) {
  
    // Calculate the complement needed
    const complement = x - arr2[j];

    // Check if the complement exists in the set
    if (set.has(complement)) {
    
      // Print the pair if found
      console.log(complement, arr2[j]);
    }
  }
}

// Example usage
const arr1 = [1, 0, -4, 7, 6, 4];
const arr2 = [0, 2, 4, -3, 2, 1];
const n = arr1.length;
const m = arr2.length;
const x = 8;

findPairs(arr1, arr2, n, m, x);

Output
6 2
4 4
6 2
7 1

Time Complexity: O(n + m)
Auxiliary Space: O(n + m)

Please refer 2Sum - Complete Tutorial for all variations of 2SUM problem.


 


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