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Nuts & Bolts Problem (Lock & Key problem) | Set 2 (Hashmap)
• Difficulty Level : Medium
• Last Updated : 08 Nov, 2020

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 a nut to see which one is bigger/smaller.
Examples:

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

Another 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 below post.
Nuts & Bolts Problem (Lock & Key problem) | Set 1
In this post, hashmap based approach is discussed.

1. Traverse 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 then this means bolts exist for that nut.

## C++

 `// 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;` `}`

## Python3

 `# Python3 program to implement ` `# above approach` `# Hashmap based solution to ` `# solve`   `# Function to match nuts and ` `# bolts` `def` `nutboltmatch(nuts, ` `                 ``bolts, n):`   `    ``hash1 ``=` `{}`   `    ``# creating a hashmap ` `    ``# for nuts` `    ``for` `i ``in` `range``(n):` `        ``hash1[nuts[i]] ``=` `i`   `    ``# searching for nuts for ` `    ``# each bolt in hash map` `    ``for` `i ``in` `range``(n):` `        ``if` `(bolts[i] ``in` `hash1):` `            ``nuts[i] ``=` `bolts[i]`   `    ``# Print the result` `    ``print``(``"matched nuts and bolts are-"``)` `    ``for` `i ``in` `range``(n):` `        ``print``(nuts[i], ` `              ``end ``=` `" "``)` `    ``print``()` `    ``for` `i ``in` `range``(n):` `        ``print``(bolts[i], ` `              ``end ``=` `" "``)`   `# Driver code` `if` `__name__ ``=``=` `"__main__"``:`   `    ``nuts ``=` `[``'@'``, ``'#'``, ``'\$'``,` `            ``'%'``, ``'^'``, ``'&'``]` `    ``bolts ``=` `[``'\$'``, ``'%'``, ``'&'``,` `             ``'^'``, ``'@'``, ``'#'``]` `    ``n ``=` `len``(nuts)` `    ``nutboltmatch(nuts, bolts, n)`   `# This code is contributed by Chitranayal`

Output:

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

The 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.