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();
}