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Nuts & Bolts Problem (Lock & Key problem) | Set 2 (Hashmap)
  • Difficulty Level : Medium
  • Last Updated : 10 Mar, 2021

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 <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;
}

Java




// Hashmap based solution to solve
import java.util.HashMap;
class GFG
{
 
  // function to match nuts and bolts
  static void nutboltmatch(char nuts[], char bolts[], int n)
  {
    HashMap<Character, Integer> hash = new HashMap<>();
 
    // creating a hashmap for nuts
    for (int i = 0; i < n; i++)
      hash.put(nuts[i], i);
 
    // searching for nuts for each bolt in hash map
    for (int i = 0; i < n; i++)
      if (hash.containsKey(bolts[i]))
        nuts[i] = bolts[i];
 
    // print the result
    System.out.println("matched nuts and bolts are-");
    for (int i = 0; i < n; i++)
      System.out.print(nuts[i] + " ");
    System.out.println();
    for (int i = 0; i < n; i++)
      System.out.print(bolts[i] + " ");
  }
 
  // Driver code
  public static void main(String[] args)
  {
    char nuts[] = { '@', '#', '$', '%', '^', '&' };
    char bolts[] = { '$', '%', '&', '^', '@', '#' };
    int n = nuts.length;
    nutboltmatch(nuts, bolts, n);
  }
}
 
// This code is contributed by sanjeev2552

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

C#




// Hashmap based solution to solve
using System;
using System.Collections.Generic;
 
public class GFG
{
 
  // function to match nuts and bolts
  static void nutboltmatch(char[] nuts, char[] bolts, int n)
  {
    Dictionary<char,int> hash = new Dictionary<char,int>();
 
    // creating a hashmap for nuts
    for (int i = 0; i < n; i++)
    {
      hash.Add(nuts[i], i);
    }
 
    // searching for nuts for each bolt in hash map
    for (int i = 0; i < n; i++)
      if (hash.ContainsKey(bolts[i]))
        nuts[i] = bolts[i];
 
    // print the result
    Console.WriteLine("matched nuts and bolts are-");
    for (int i = 0; i < n; i++)
      Console.Write(nuts[i] + " ");
    Console.WriteLine();
    for (int i = 0; i < n; i++)
      Console.Write(bolts[i] + " ");
  }
 
  // Driver code
  static public void Main ()
  {
 
    char[] nuts = { '@', '#', '$', '%', '^', '&' };
    char[] bolts = { '$', '%', '&', '^', '@', '#' };
    int n = nuts.Length;
    nutboltmatch(nuts, bolts, n);
 
  }
}
 
// This code is contributed by avanitrachhadiya2155

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.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 

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