The idea is to use an array to store the ugly numbers and use three pointers for 2, 3, and 5 to calculate the next potential ugly numbers. We then choose the minimum of these numbers to be the next ugly number and appropriately move the pointers.
Time Complexity: O(n), where n is the number of ugly numbers to generate.
Space Complexity: O(n), for storing the ugly numbers array.
1#include <iostream>
2#include <vector>
3using namespace std;
4
5int nthUglyNumber(int n) {
6 vector<int> ugly(n);
7 ugly[0] = 1;
8 int i2 = 0, i3 = 0, i5 = 0;
9 int next_2 = 2, next_3 = 3, next_5 = 5;
10
11 for (int i = 1; i < n; ++i) {
12 int next_ugly = min(next_2, min(next_3, next_5));
13 ugly[i] = next_ugly;
14
15 if (next_ugly == next_2) next_2 = ugly[++i2] * 2;
16 if (next_ugly == next_3) next_3 = ugly[++i3] * 3;
17 if (next_ugly == next_5) next_5 = ugly[++i5] * 5;
18 }
19 return ugly[n - 1];
20}
21
22int main() {
23 int n = 10;
24 cout << nthUglyNumber(n) << endl; // Output: 12
25 return 0;
26}This C++ solution follows the dynamic programming approach with three pointers. Here, a vector is used to store the sequence of ugly numbers, and pointers i2, i3, and i5 are used in the same way as described for the C solution, determining the smallest next number and updating corresponding pointers.
This method involves using a min-heap to manage the sequence of potential ugly numbers. We start with 1 in the min-heap and repeatedly extract the smallest element, multiplying it by 2, 3, and 5 to generate new candidates, which are then inserted back into the heap. Duplicate entries are avoided by using a set for tracking which numbers have been added to the heap.
Time Complexity: O(n log n), primarily due to heap operations.
Space Complexity: O(n), for data structures that might store up to n numbers.
1import heapq
2
3def nthUglyNumber(n):
4 min_heap = [1]
5 seen = {1}
6
7 for _ in range(n):
8 curr_ugly = heapq.heappop(min_heap)
9 for factor in [2, 3, 5]:
10 new_ugly = curr_ugly * factor
11 if new_ugly not in seen:
12 seen.add(new_ugly)
13 heapq.heappush(min_heap, new_ugly)
14 return curr_ugly
15
16n = 10
17print(nthUglyNumber(n)) # Output: 12
18The Python solution leverages a min-heap for ordering ugly numbers. The smallest number is removed in each iteration and multiplied by 2, 3, and 5 to yield potential new ugly numbers. Tracking entries in a set prevents duplicates in the heap.