313. Super Ugly Number

Description

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 <= 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.

Solutions

Solution 1: Priority Queue (Min Heap)

We use a priority queue (min heap) to maintain all possible super ugly numbers, initially putting $1$ into the queue.

Each time we take the smallest super ugly number $x$ from the queue, multiply $x$ by each number in the array primes, and put the product into the queue. Repeat the above operation $n$ times to get the $n$th super ugly number.

Since the problem guarantees that the $n$th super ugly number is within the range of a 32-bit signed integer, before we put the product into the queue, we can first check whether the product exceeds $2^{31} - 1$. If it does, there is no need to put the product into the queue. In addition, the Euler sieve can be used for optimization.

The time complexity is $O(n \times m \times \log (n \times m))$, and the space complexity is $O(n \times m)$. Where $m$ and $n$ are the length of the array primes and the given integer $n$ respectively.

Python3JavaC++Go

class Solution:
    def nthSuperUglyNumber(self, n: int, primes: List[int]) -> int:
        q = [1]
        x = 0
        mx = (1 << 31) - 1
        for _ in range(n):
            x = heappop(q)
            for k in primes:
                if x <= mx // k:
                    heappush(q, k * x)
                if x % k == 0:
                    break
        return x

class Solution {
    public int nthSuperUglyNumber(int n, int[] primes) {
        PriorityQueue<Integer> q = new PriorityQueue<>();
        q.offer(1);
        int x = 0;
        while (n-- > 0) {
            x = q.poll();
            while (!q.isEmpty() && q.peek() == x) {
                q.poll();
            }
            for (int k : primes) {
                if (k <= Integer.MAX_VALUE / x) {
                    q.offer(k * x);
                }
                if (x % k == 0) {
                    break;
                }
            }
        }
        return x;
    }
}

class Solution {
public:
    int nthSuperUglyNumber(int n, vector<int>& primes) {
        priority_queue<int, vector<int>, greater<int>> q;
        q.push(1);
        int x = 0;
        while (n--) {
            x = q.top();
            q.pop();
            for (int& k : primes) {
                if (x <= INT_MAX / k) {
                    q.push(k * x);
                }
                if (x % k == 0) {
                    break;
                }
            }
        }
        return x;
    }
};

func nthSuperUglyNumber(n int, primes []int) (x int) {
    q := hp{[]int{1}}
    for n > 0 {
        n--
        x = heap.Pop(&q).(int)
        for _, k := range primes {
            if x <= math.MaxInt32/k {
                heap.Push(&q, k*x)
            }
            if x%k == 0 {
                break
            }
        }
    }
    return
}

type hp struct{ sort.IntSlice }

func (h *hp) Push(v any) { h.IntSlice = append(h.IntSlice, v.(int)) }
func (h *hp) Pop() any {
    a := h.IntSlice
    v := a[len(a)-1]
    h.IntSlice = a[:len(a)-1]
    return v
}

Solution 2

Go

type Ugly struct{ value, prime, index int }
type Queue []Ugly

func (u Queue) Len() int           { return len(u) }
func (u Queue) Swap(i, j int)      { u[i], u[j] = u[j], u[i] }
func (u Queue) Less(i, j int) bool { return u[i].value < u[j].value }
func (u *Queue) Push(v any)        { *u = append(*u, v.(Ugly)) }
func (u *Queue) Pop() any {
    old, x := *u, (*u)[len(*u)-1]
    *u = old[:len(old)-1]
    return x
}

func nthSuperUglyNumber(n int, primes []int) int {
    ugly, pq, p := make([]int, n+1), &Queue{}, 2
    ugly[1] = 1
    heap.Init(pq)
    for _, v := range primes {
        heap.Push(pq, Ugly{value: v, prime: v, index: 2})
    }
    for p <= n {
        top := heap.Pop(pq).(Ugly)
        if ugly[p-1] != top.value {
            ugly[p], p = top.value, p+1
        }
        top.value, top.index = ugly[top.index]*top.prime, top.index+1
        heap.Push(pq, top)
    }
    return ugly[n]
}

313. Super Ugly Number

Description

Solutions

Solution 1: Priority Queue (Min Heap)

Solution 2

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