Super Ugly Number – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.
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 nth
super ugly number.
The nth
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 <= 105
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.
C++ Super Ugly Number LeetCode Solution
int nthSuperUglyNumber(int n, vector<int>& primes) {
vector<int> index(primes.size(), 0), ugly(n, INT_MAX);
ugly[0]=1;
for(int i=1; i<n; i++){
for(int j=0; j<primes.size(); j++) ugly[i]=min(ugly[i],ugly[index[j]]*primes[j]);
for(int j=0; j<primes.size(); j++) index[j]+=(ugly[i]==ugly[index[j]]*primes[j]);
}
return ugly[n-1];
}
Java Super Ugly Number LeetCode Solution
public int nthSuperUglyNumberI(int n, int[] primes) {
int[] ugly = new int[n];
int[] idx = new int[primes.length];
ugly[0] = 1;
for (int i = 1; i < n; i++) {
//find next
ugly[i] = Integer.MAX_VALUE;
for (int j = 0; j < primes.length; j++)
ugly[i] = Math.min(ugly[i], primes[j] * ugly[idx[j]]);
//slip duplicate
for (int j = 0; j < primes.length; j++) {
while (primes[j] * ugly[idx[j]] <= ugly[i]) idx[j]++;
}
}
return ugly[n - 1];
}
Python 3 Super Ugly Number LeetCode Solution
def nthSuperUglyNumber(self, n, primes):
uglies = [1]
def gen(prime):
for ugly in uglies:
yield ugly * prime
merged = heapq.merge(*map(gen, primes))
while len(uglies) < n:
ugly = next(merged)
if ugly != uglies[-1]:
uglies.append(ugly)
return uglies[-1]
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