# Nuts & Bolts Problem (Lock & Key problem) | Set 2 (Hashmap)

Given a set of n nuts of different sizes and n bolts of different sizes. There is a one-one mapping between nuts and bolts. Match nuts and bolts efficiently.

**Constraint:** Comparison of a nut to another nut or a bolt to another bolt is not allowed. It means nut can only be compared with bolt and bolt can only be compared with nut to see which one is bigger/smaller.

Examples:

Input : nuts[] = {'@', '#', '$', '%', '^', '&'} bolts[] = {'$', '%', '&', '^', '@', '#'} Output : Matched nuts and bolts are- $ % & ^ @ # $ % & ^ @ #

Other way of asking this problem is, given a box with locks and keys where one lock can be opened by one key in the box. We need to match the pair.

We have discussed a sorting based solution in below post.

Nuts & Bolts Problem (Lock & Key problem) | Set 1

In this post, hashmap based approach is discussed.

- Travese the nuts array and create a hashmap
- Traverse the bolts array and search for it in hashmap.
- If it is found in the hashmap of nuts than this means bolts exist for that nut.

`// Hashmap based solution to solve ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// function to match nuts and bolts ` `void` `nutboltmatch(` `char` `nuts[], ` `char` `bolts[], ` `int` `n) ` `{ ` ` ` `unordered_map<` `char` `, ` `int` `> hash; ` ` ` ` ` `// creating a hashmap for nuts ` ` ` `for` `(` `int` `i = 0; i < n; i++) ` ` ` `hash[nuts[i]] = i; ` ` ` ` ` `// searching for nuts for each bolt in hash map ` ` ` `for` `(` `int` `i = 0; i < n; i++) ` ` ` `if` `(hash.find(bolts[i]) != hash.end()) ` ` ` `nuts[i] = bolts[i]; ` ` ` ` ` `// print the result ` ` ` `cout << ` `"matched nuts and bolts are-"` `<< endl; ` ` ` `for` `(` `int` `i = 0; i < n; i++) ` ` ` `cout << nuts[i] << ` `" "` `; ` ` ` `cout << endl; ` ` ` `for` `(` `int` `i = 0; i < n; i++) ` ` ` `cout << bolts[i] << ` `" "` `; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `char` `nuts[] = {` `'@'` `, ` `'#'` `, ` `'$'` `, ` `'%'` `, ` `'^'` `, ` `'&'` `}; ` ` ` `char` `bolts[] = {` `'$'` `, ` `'%'` `, ` `'&'` `, ` `'^'` `, ` `'@'` `, ` `'#'` `}; ` ` ` `int` `n = ` `sizeof` `(nuts) / ` `sizeof` `(nuts[0]); ` ` ` `nutboltmatch(nuts, bolts, n); ` ` ` `return` `0; ` `} ` |

*chevron_right*

*filter_none*

Output:

matched nuts and bolts are- $ % & ^ @ # $ % & ^ @ #

Time complexity for this solution is O(n).

This article is contributed by **Niteesh kumar**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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