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 existsbool 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 Codeint 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 pairsimport 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 pairsfrom typing import List
# The function to check if beautiful pair existsdef 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 Codeif __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 pairsusing 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 existsfunction 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 Codelet 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 arrayif (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)
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