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.

Example:

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] of size 4.

Note:

• 1 is a super ugly number for any given primes.

• The given numbers in primes are in ascending order.

• 0 < k ≤ 100, 0 < n ≤ 106, 0 < primes[i] < 1000.

• The nth super ugly number is guaranteed to fit in a 32-bit signed integer.

int nthSuperUglyNumber(int n, vector<int>& primes) { // time: O(n * # of primes); space: O(n + # of primes)
    vector<int> dp(n, INT_MAX), idx(primes.size(), 0);
    dp[0] = 1;
    for (int i = 1; i < n; ++i) {
        for (int j = 0; j < primes.size(); ++j) {
            dp[i] = min(dp[i], primes[j] * dp[idx[j]]);
        }
        for (int j = 0; j < primes.size(); ++j) {
            if (primes[j] * dp[idx[j]] == dp[i]) ++idx[j];
        }
    }
    return dp.back();
}

int nthSuperUglyNumber(int n, vector<int>& primes) { // time: O(n * # of primes); space: O(n + 2 * # of primes)
    vector<int> dp(n, 0), idx(primes.size(), 0), val(primes.size(), 1);
    int next = 1;
    for (int i = 0; i < n; ++i) {
        dp[i] = next;
        next = INT_MAX;
        for (int j = 0; j < primes.size(); ++j) {
            if (val[j] == dp[i]) {
                val[j] = dp[idx[j]++] * primes[j];
            }
            next = min(next, val[j]);
        }
    }
    return dp.back();
}