Sparse Arrays HackerRank Solution

Sparse Arrays HackerRank Solution
Source : https://www.hackerrank.com/challenges/sparse-arrays

There is a collection of input strings and a collection of query strings. For each query string, determine how many times it occurs in the list of input strings.

For example, given input and , we find instances of ', of '' and of ''. For each query, we add an element to our return array, .

Function Description

Complete the function matchingStrings in the editor below. The function must return an array of integers representing the frequency of occurrence of each query string in strings.

matchingStrings has the following parameters:

• strings - an array of strings to search
• queries - an array of query strings

Input Format

The first line contains and integer , the size of .
Each of the next lines contains a string .
The next line contains , the size of .
Each of the next lines contains a string .

Constraints

.

Output Format

Return an integer array of the results of all queries in order.

Sample Input 1

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4
aba
baba
aba
xzxb
3
aba
xzxb
ab

Sample Output 1

2
1
0

Explanation 1

Here, "aba" occurs twice, in the first and third string. The string "xzxb" occurs once in the fourth string, and "ab" does not occur at all.

Sample Input 2

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3
def
de
fgh
3
de
lmn
fgh

Sample Output 2

1
0
1

Sample Input 3

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13
abcde
sdaklfj
asdjf
na
basdn
sdaklfj
asdjf
na
asdjf
na
basdn
sdaklfj
asdjf
5
abcde
sdaklfj
asdjf
na
basdn

Sample Output 3

1
3
4
3
2

Source : https://www.hackerrank.com/challenges/sparse-arrays

Solution

	// Karthikalapati.blogspot.com
	import java.util.Scanner;
	import java.util.HashMap;
	public class Solution {
	public static void main(String[] args) {
	Scanner scan = new Scanner(System.in);
	int N = scan.nextInt();
	scan.nextLine(); // gets rid of pesky newline
	/* Create and fill HashMap */
	HashMap<String, Integer> map = new HashMap();
	for (int i = 0; i < N; i++) {
	String str = scan.nextLine();
	map.merge(str, 1, Integer::sum);
	}
	/* Query HashMap */
	int Q = scan.nextInt();
	scan.nextLine(); // gets rid of pesky newline
	for (int i = 0; i < Q; i++) {
	String str = scan.nextLine();
	System.out.println(map.getOrDefault(str, 0));
	}
	}
	}