# 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.

## Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

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.

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

 `// Hashmap based solution to solve ` `#include ` `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); ` `    ``nutboltmatch(nuts, bolts, n); ` `    ``return` `0; ` `} `

Output:

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

Time complexity for this solution is O(n).

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