package com.fishercoder.solutions;

/**
 * 313. Super Ugly Number
 *
 * Write a program to find the nth super ugly number.
 * Super ugly numbers are positive numbers whose all prime factors
 * are in the given prime list primes of size k.
 * For example, [1, 2, 4, 7, 8, 13, 14, 16, 19, 26, 28, 32]
 * is the sequence of the first 12 super ugly numbers given primes = [2, 7, 13, 19] of size 4.
 * Note:
 * (1) 1 is a super ugly number for any given primes.
 * (2) The given numbers in primes are in ascending order.
 * (3) 0 < k ≤ 100, 0 < n ≤ 106, 0 < primes[i] < 1000.
 * (4) The nth super ugly number is guaranteed to fit in a 32-bit signed integer.
 */
public class _313 {

    public int nthSuperUglyNumber(int n, int[] primes) {
        int[] ret = new int[n];
        ret[0] = 1;

        int[] indexes = new int[primes.length];

        for (int i = 1; i < n; i++) {
            ret[i] = Integer.MAX_VALUE;

            for (int j = 0; j < primes.length; j++) {
                ret[i] = Math.min(ret[i], primes[j] * ret[indexes[j]]);
            }

            for (int j = 0; j < indexes.length; j++) {
                if (ret[i] == primes[j] * ret[indexes[j]]) {
                    indexes[j]++;
                }
            }
        }

        return ret[n - 1];
    }

}

package com.fishercoder.solutions;

/**
 * 313. Super Ugly Number
 *
 * Write a program to find the nth super ugly number.
 * Super ugly numbers are positive numbers whose all prime factors
 * are in the given prime list primes of size k.
 * For example, [1, 2, 4, 7, 8, 13, 14, 16, 19, 26, 28, 32]
 * is the sequence of the first 12 super ugly numbers given primes = [2, 7, 13, 19] of size 4.
 *
 * Note:
 * (1) 1 is a super ugly number for any given primes.
 * (2) The given numbers in primes are in ascending order.
 * (3) 0 < k ≤ 100, 0 < n ≤ 106, 0 < primes[i] < 1000.
 * (4) The nth super ugly number is guaranteed to fit in a 32-bit signed integer.
 */
public class _313 {

    public static class Solution1 {
        public int nthSuperUglyNumber(int n, int[] primes) {
            int[] ret = new int[n];
            ret[0] = 1;

            int[] indexes = new int[primes.length];

            for (int i = 1; i < n; i++) {
                ret[i] = Integer.MAX_VALUE;

                for (int j = 0; j < primes.length; j++) {
                    ret[i] = Math.min(ret[i], primes[j] * ret[indexes[j]]);
                }

                for (int j = 0; j < indexes.length; j++) {
                    if (ret[i] == primes[j] * ret[indexes[j]]) {
                        indexes[j]++;
                    }
                }
            }
            return ret[n - 1];
        }
    }

}

refactor 313