Sudo Placement | Beautiful Pairs

Given two arrays of integers where maximum size of first array is big and that of second array is small. Your task is to find if there is a pair in the first array whose sum is present in the second array.

Examples:  

Input: 
4
1 5 10 8
3
2 20 13
Output: 1

Approach: We have x + y = z. This can be rewritten as x = z – y. This means we need to find an element x in array 1 such that it is the result of z(second array) – y(first array). For this, use hashing to keep track of such element x. 

Implementation:

C++

// C++ code for finding required pairs
#include <bits/stdc++.h>
using namespace std;

// The function to check if beautiful pair exists
bool pairExists(int arr1[], int m, int arr2[], int n)
{
    // Set for hashing
    unordered_set<int> s;

    // Traversing the first array
    for (int i = 0; i < m; i++) {

        // Traversing the second array to check for
        // every j corresponding to single i
        for (int j = 0; j < n; j++) {

            // x + y = z => x = y - z
            if (s.find(arr2[j] - arr1[i]) != s.end())

                // if such x exists then we return true
                return true;

        }

       // hash to make use of it next time
       s.insert(arr1[i]);
 }

    // no pair exists
    return false;
}

// Driver Code
int main()
{
    int arr1[] = { 1, 5, 10, 8 };
    int arr2[] = { 2, 20, 13 };

    // If pair exists then 1 else 0
    // 2nd argument as size of first array
    // fourth argument as sizeof 2nd array
    if (pairExists(arr1, 4, arr2, 3))
        cout << 1 << endl;
    else
        cout << 0 << endl;

    return 0;
}

Java

// Java code for finding required pairs
import java.util.*;
class GFG
{
        // The function to check if beautiful pair exists
        static boolean pairExists(int []arr1, int m, int []arr2, int n)
        {
            // Set for hashing
            Set<Integer> s =new HashSet<Integer>();

            // Traversing the first array
            for (int i = 0; i < m; i++) {

                // Traversing the second array to check for
                // every j corresponding to single i
                for (int j = 0; j < n; j++)
                {

                    // x + y = z => x = y - z
                    if (s.contains(arr2[j] - arr1[i]))

                        // if such x exists then we return true
                        return true;

                }

            // hash to make use of it next time
            s.add(arr1[i]);
        }

            // no pair exists
            return false;
        }

        // Driver Code
        public static void main(String []args)
        {
            int []arr1 = { 1, 5, 10, 8 };
            int []arr2 = { 2, 20, 13 };

            // If pair exists then 1 else 0
            // 2nd argument as size of first array
            // fourth argument as sizeof 2nd array
            if (pairExists(arr1, 4, arr2, 3))
                System.out.println(1);
            else
                System.out.println(0);

        }

// This code is contributed by ihritik
}

Python3

# Python3 code for finding required pairs
from typing import List

# The function to check if beautiful pair exists
def pairExists(arr1: List[int], m: int,
               arr2: List[int], n: int) -> bool:

    # Set for hashing
    s = set()

    # Traversing the first array
    for i in range(m):

        # Traversing the second array to check for
        # every j corresponding to single i
        for j in range(n):

            # x + y = z => x = y - z
            if (arr2[2] - arr1[2]) not in s:

                # If such x exists then we
                # return true
                return True

        # Hash to make use of it next time
        s.add(arr1[i])

    # No pair exists
    return False

# Driver Code
if __name__ == "__main__":

    arr1 = [ 1, 5, 10, 8 ]
    arr2 = [ 2, 20, 13 ]

    # If pair exists then 1 else 0
    # 2nd argument as size of first array
    # fourth argument as sizeof 2nd array
    if (pairExists(arr1, 4, arr2, 3)):
        print(1)
    else:
        print(0)

# This code is contributed by sanjeev2552

C#

// C# code for finding required pairs
using System;
using System.Collections.Generic;

class GFG
{
        // The function to check if
        // beautiful pair exists
        static bool pairExists(int []arr1,
                int m, int []arr2, int n)
        {
            // Set for hashing
            HashSet<int> s = new HashSet<int>();

            // Traversing the first array
            for (int i = 0; i < m; i++)
            {

                // Traversing the second array to check for
                // every j corresponding to single i
                for (int j = 0; j < n; j++)
                {

                    // x + y = z => x = y - z
                    if (s.Contains(arr2[j] - arr1[i]))

                        // if such x exists then we return true
                        return true;

                }

            // hash to make use of it next time
            s.Add(arr1[i]);
        }

            // no pair exists
            return false;
        }

        // Driver Code
        public static void Main()
        {
            int []arr1 = { 1, 5, 10, 8 };
            int []arr2 = { 2, 20, 13 };

            // If pair exists then 1 else 0
            // 2nd argument as size of first array
            // fourth argument as sizeof 2nd array
            if (pairExists(arr1, 4, arr2, 3))
                Console.WriteLine(1);
            else
                Console.WriteLine(0);
        }
}

/* This code contributed by PrinciRaj1992 */

Javascript

<script>

// JavaScript code for finding required pairs


// The function to check if beautiful pair exists
function pairExists(arr1, m, arr2, n) {
    // Set for hashing
    let s = new Set();

    // Traversing the first array
    for (let i = 0; i < m; i++) {

        // Traversing the second array to check for
        // every j corresponding to single i
        for (let j = 0; j < n; j++) {

            // x + y = z => x = y - z
            if (s.has(arr2[j] - arr1[i]))

                // if such x exists then we return true
                return true;

        }

        // hash to make use of it next time
        s.add(arr1[i]);
    }

    // no pair exists
    return false;
}

// Driver Code

let arr1 = [1, 5, 10, 8];
let arr2 = [2, 20, 13];

// If pair exists then 1 else 0
// 2nd argument as size of first array
// fourth argument as sizeof 2nd array
if (pairExists(arr1, 4, arr2, 3))
    document.write(1 + "<br>");
else
    document.write(0 + "<br>");

</script>

Output

1

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