g0301_0400.s0313_super_ugly_number.Solution Maven / Gradle / Ivy

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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0301_0400.s0313_super_ugly_number;

// #Medium #Array #Dynamic_Programming #Math #2022_07_10_Time_16_ms_(96.57%)_Space_42.3_MB_(64.43%)

/**
 * 313 - Super Ugly Number\.
 *
 * Medium
 *
 * A **super ugly number** is a positive integer whose prime factors are in the array `primes`.
 *
 * Given an integer `n` and an array of integers `primes`, return _the_ n^th _**super ugly number**_.
 *
 * The n^th **super ugly number** is **guaranteed** to fit in a **32-bit** signed integer.
 *
 * **Example 1:**
 *
 * **Input:** n = 12, primes = [2,7,13,19]
 *
 * **Output:** 32
 *
 * **Explanation:** [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]. 
 *
 * **Example 2:**
 *
 * **Input:** n = 1, primes = [2,3,5]
 *
 * **Output:** 1
 *
 * **Explanation:** 1 has no prime factors, therefore all of its prime factors are in the array primes = [2,3,5]. 
 *
 * **Constraints:**
 *
 * *   1 <= n <= 10⁶
 * *   `1 <= primes.length <= 100`
 * *   `2 <= primes[i] <= 1000`
 * *   `primes[i]` is **guaranteed** to be a prime number.
 * *   All the values of `primes` are **unique** and sorted in **ascending order**.
**/
@SuppressWarnings("java:S3012")
public class Solution {
    public int nthSuperUglyNumber(int n, int[] primes) {
        long[] primes1 = new long[primes.length];
        for (int i = 0; i < primes.length; i++) {
            primes1[i] = primes[i];
        }
        int[] index = new int[primes.length];
        long[] n1 = new long[n];
        n1[0] = 1L;
        for (int i = 1; i < n; i++) {
            long min = Long.MAX_VALUE;
            for (long l : primes1) {
                min = Math.min(min, l);
            }
            n1[i] = min;
            for (int j = 0; j < primes1.length; j++) {
                if (min == primes1[j]) {
                    primes1[j] = primes[j] * (n1[++index[j]]);
                }
            }
        }
        return (int) n1[n - 1];
    }
}