# 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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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.

## CPP

 `// CPP code for finding required pairs ` `#include ` `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 s =``new` `HashSet(); ` `         `  `            ``// 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 ` `} `

## 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 */`

Output:

```1
```

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Improved By : ihritik, princiraj1992