Nuts & Bolts Problem (Lock & Key problem) using 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 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.
- Traverse 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 then this means bolts exist for that nut.
Implementation:
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 |
Javascript
<script> // Hashmap based solution to solve // function to match nuts and bolts function nutboltmatch(nuts, bolts, n) { let hash = new Map(); // creating a hashmap for nuts for (let i = 0; i < n; i++) hash.set(nuts[i], i); // searching for nuts for each bolt in hash map for (let i = 0; i < n; i++) if (hash.has(bolts[i])) nuts[i] = bolts[i]; // print the result document.write( "matched nuts and bolts are-<br>" ); for (let i = 0; i < n; i++) document.write(nuts[i] + " " ); document.write( "<br>" ); for (let i = 0; i < n; i++) document.write(bolts[i] + " " ); } // Driver code let nuts = [ '@' , '#' , '$' , '%' , '^' , '&' ]; let bolts = [ '$' , '%' , '&' , '^' , '@' , '#' ]; let n = nuts.length; nutboltmatch(nuts, bolts, n); </script> |
matched nuts and bolts are- $ % & ^ @ # $ % & ^ @ #
The time complexity for this solution is O(n).
Auxiliary space: O(n) because using hashmap
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