Java Map | HackerRank Solution

Hello coders, today we are going to solve Java Map HackerRank Solution.

Problem

You are given a phone book that consists of people’s names and their phone number. After that you will be given some person’s name as query. For each query, print the phone number of that person.

Input Format

The first line will have an integer n denoting the number of entries in the phone book. Each entry consists of two lines: a name and the corresponding phone number.

After these, there will be some queries. Each query will contain a person’s name. Read the queries until end-of-file.

Constraints:
A person’s name consists of only lower-case English letters and it may be in the format ‘first-name last-name’ or in the format ‘first-name’. Each phone number has exactly 8 digits without any leading zeros.

1 <= n <= 1000

1 <= Query <= 100000

Output Format

For each case, print “Not found” if the person has no entry in the phone book. Otherwise, print the person’s name and phone number. See sample output for the exact format.

To make the problem easier, we provided a portion of the code in the editor. You can either complete that code or write completely on your own.

Sample Input

 3
 uncle sam
 99912222
 tom
 11122222
 harry
 12299933
 uncle sam
 uncle tom
 harry

Sample Output

 uncle sam=99912222
 Not found
 harry=12299933

Solution – Java Map

//Complete this code or write your own from scratch
import java.util.*;
import java.io.*;

class Solution {
    public static void main(String [] args) throws Exception {
        /* Read input and save as entries in a HashMap */
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int n = Integer.valueOf(br.readLine());
        HashMap<String, Integer> map = new HashMap<>();
        while (n-- > 0) {
            String name = br.readLine();
            int phone   = Integer.valueOf(br.readLine());
            map.put(name, phone);
        }
        
        /* Read each query and check if its in our HashMap */
        String s;
        while((s = br.readLine()) != null) {
            if (map.containsKey(s)) {
                System.out.println(s + "=" + map.get(s));
            } else {
                System.out.println("Not found");
            }
        }
        
        br.close();
    }
}

Disclaimer: The above Problem ( Java Map ) is generated by Hacker Rank but the Solution is Provided by CodingBroz. This tutorial is only for Educational and Learning Purpose.