Find Digits | HackerRank Solutions By CodingHumans | Problem Solving Algorithms

 

Find Digits 

An integer d  is a divisor of an integer n  if the remainder of n ÷ d = 0.

Given an integer, for each digit that makes up the integer determine whether it is a divisor. Count the number of divisors occurring within the integer.

Note: Each digit is considered to be unique, so each occurrence of the same digit should be counted (e.g. for n=111, 1  is a divisor of  111 each time it occurs so the answer is 3).

Function Description

Complete the findDigits function in the editor below. It should return an integer representing the number of digits of  that are divisors of .

findDigits has the following parameter(s):

n: an integer to analyze

Input Format

The first line is an integer, , indicating the number of test cases.

The  subsequent lines each contain an integer, .

Output Format

For every test case, count the number of digits in  that are divisors of n. Print each answer on a new line.

Sample Input

2
12
1012

Sample Output

2
3

Explanation

The number 12  is broken into two digits, 1 and 2 . When 12  is divided by either of those two digits, the remainder is 0  so they are both divisors.

The number 1012 is broken into four digits 1,0 ,1 , ,and  2. 1012  is evenly divisible by its digits 1, 1 , and 2, but it is not divisible by 0 as division by zero is undefined.

solution

(java)

import java.io.*;
import java.util.*;

public class Solution 
{
   
    static int findDigits(int n) 
    {
        int fix=n;
        int digit=0;
        int count=0;
        while(n!=0)
        {
            digit=n%10;
            if(digit!=0 && fix%digit==0)
            {
                count++;
            }
            n=n/10; 
        }
        return count;
    }

    private static final Scanner scanner = new Scanner(System.in);

    public static void main(String[] args) throws IOException {
        BufferedWriter bufferedWriter = new BufferedWriter(new FileWriter(System.getenv("OUTPUT_PATH")));

        int t = scanner.nextInt();
        scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");

        for (int tItr = 0; tItr < t; tItr++) {
            int n = scanner.nextInt();
            scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");

            int result = findDigits(n);

            bufferedWriter.write(String.valueOf(result));
            bufferedWriter.newLine();
        }
        bufferedWriter.close();
        scanner.close();
    }
}