TELEPHONE BOOK USING HASHMAP

 

Sample Input

3
sam 99912222
tom 11122222
harry 12299933
sam
edward
harry

Sample Output

sam=99912222
Not found
harry=12299933

Explanation

We add the following  (Key,Value) pairs to our map so it looks like this:

We then process each query and print key=value if the queried  is found in the map; otherwise, we print Not found.

Query 0: sam
Sam is one of the keys in our dictionary, so we print sam=99912222.

Query 1: edward
Edward is not one of the keys in our dictionary, so we print Not found.

Query 2: harry
Harry is one of the keys in our dictionary, so we print harry=12299933.

SOLUTION:

package HackerRank_problems.practice_problems;

import java.util.*;

import java.io.*;

public class TelephoneBook{

    public static void main(String []argh){

        Scanner in = new Scanner(System.in);

        int n = in.nextInt();

        HashMap<String,Integer> hm=new HashMap();

        for(int i = 0; i < n; i++){

            String name = in.next();

            int phone = in.nextInt();

            hm.put(name,phone);

            

        }

        while(in.hasNext()){

            String s = in.next();

            if(hm.get(s)==null)

            {

                System.out.println("Not found");

            }

            else{

                System.out.println(s+"="+hm.get(s));

            }

        }

        in.close();

    }

}