Project Euler #38: Pandigital multiples

Question

Take the number 192 and multiply it by each of 1, 2, and 3:

192 × 1 = 192
192 × 2 = 384
192 × 3 = 576

By concatenating each product we get the 1 to 9 pandigital, 192384576. We will call 192384576 the concatenated product of 192 and (1,2,3)
The same can be achieved by starting with 9 and multiplying by 1, 2, 3, 4, and 5, giving the pandigital, 918273645, which is the concatenated product of 9 and (1,2,3,4,5).
What is the largest 1 to 9 pandigital 9-digit number that can be formed as the concatenated product of an integer with (1,2, ... , n) where n > 1?

Answer : 932718654

Hacker Rank Problem

Solution

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import java.util.*; public class Solution { public static void main(String[] args) { /* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */ Scanner in = new Scanner(System.in); int n = in.nextInt(); int k = in.nextInt(); for(int i=2;i<n;i++) { StringBuffer sb = new StringBuffer(); for(int j=1;j<=k;j++) { long product = i*j; sb.append(product); if(sb.length() == k && isKPanDigital(sb, k)) System.out.println(i); } } } public static boolean isKPanDigital(StringBuffer sb, int k) { HashMap<Integer, Integer> map = new HashMap<Integer, Integer>(); String s = sb.toString(); for(int i=0;i<s.length();i++) { if(!map.containsKey(Character.getNumericValue(s.charAt(i)))) map.put(Character.getNumericValue(s.charAt(i)), 1); else return false; } for(int i=1;i<=k;i++) { if(!map.containsKey(i)) return false; } return true; } }