HackerEarth problem : number of occurrences of a number in the array I have written a solution to Memorise me problem on Hacker Earth but it takes a lot of memory space and time limit exceeds in the last test case.

The first line of input will contain \$N\$, an integer, which is the total number of numbers shown to your team.

The second line of input contains \$N\$ space separated integers.

The third line of input contains an integer \$Q\$, denoting the total number of integers.

The Next \$Q\$ lines will contain an integer denoting an integer, \$Bi\$, for which you have to print the number of occurrences of that number (\$Bi\$) in those \$N\$ numbers on a new line.

If the number \$Bi\$ isn’t present then print NOT PRESENT on a new line.

Constraints

\$1 \le N \le 105,
0 \le Bi \le 1000,
1 \le Q \le 1055\$

This is my code : 

public static void main(String args[]) throws Exception {

    // BufferedReader
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    String noOfElements = br.readLine();
    int n = Integer.parseInt(noOfElements);
    int i = 0;

    String[] numberArray = br.readLine().split(" ");

    String noOfQueries = br.readLine();
    int q = Integer.parseInt(noOfQueries);

    String[] queries = new String[q];

    for (i = 0; i &lt; q; i++) {
        String query = br.readLine();
        String queryNumber = query.trim();
        queries[i] = queryNumber;
    }

    for (i = 0; i &lt; queries.length; i++) {
        int count = 0;
        for (int j = 0; j &lt; n; j++) {
            if (queries[i].equals(numberArray[j])) {
                count++;
            }
        }
        if (count == 0) {
            System.out.println("NOT PRESENT");
        } else {
            System.out.println(count);
        }
    }

}
You can maintain a HashMap where key will be the number and value will be its frequency.