LeetCode #313 — MEDIUM

Super Ugly Number

Move from brute-force thinking to an efficient approach using array strategy.

Solve on LeetCode

The Problem

Problem Statement

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.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Math · Dynamic Programming

Example 1

12
[2,7,13,19]

Example 2

1
[2,3,5]

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #313: Super Ugly Number
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;
    }
}

// Accepted solution for LeetCode #313: Super Ugly Number
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
}

# Accepted solution for LeetCode #313: Super Ugly Number
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

// Accepted solution for LeetCode #313: Super Ugly Number
struct Solution;
use std::cmp::Reverse;
use std::collections::BinaryHeap;
use std::collections::HashSet;

impl Solution {
    fn nth_super_ugly_number(mut n: i32, primes: Vec<i32>) -> i32 {
        let mut queue: BinaryHeap<Reverse<i32>> = BinaryHeap::new();
        let mut visited: HashSet<i32> = HashSet::new();
        queue.push(Reverse(1));
        while n > 1 {
            let min = queue.pop().unwrap().0;
            for &x in &primes {
                if let Some(y) = x.checked_mul(min) {
                    if visited.insert(y) {
                        queue.push(Reverse(y));
                    }
                }
            }
            n -= 1;
        }
        queue.pop().unwrap().0
    }
}

#[test]
fn test() {
    let n = 12;
    let primes = vec![2, 7, 13, 19];
    let res = 32;
    assert_eq!(Solution::nth_super_ugly_number(n, primes), res);
}

// Accepted solution for LeetCode #313: Super Ugly Number
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #313: Super Ugly Number
// 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;
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × m × log (n × m)

Space

O(n × m)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.