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 <stdio.h>
2
3int nthUglyNumber(int n) {
4 int ugly[n];
5 ugly[0] = 1;
6 int i2 = 0, i3 = 0, i5 = 0;
7 int next_2 = 2, next_3 = 3, next_5 = 5;
8
9 for (int i = 1; i < n; i++) {
10 int next_ugly = next_2 < next_3 ? (next_2 < next_5 ? next_2 : next_5) : (next_3 < next_5 ? next_3 : next_5);
11 ugly[i] = next_ugly;
12
13 if (next_ugly == next_2) next_2 = ugly[++i2] * 2;
14 if (next_ugly == next_3) next_3 = ugly[++i3] * 3;
15 if (next_ugly == next_5) next_5 = ugly[++i5] * 5;
16 }
17 return ugly[n-1];
18}
19
20int main() {
21 int n = 10;
22 printf("%d\n", nthUglyNumber(n)); // Output: 12
23 return 0;
24}The code defines a function nthUglyNumber that employs dynamic programming with three pointers to generate ugly numbers in sequence and return the nth ugly number. We use an auxiliary array ugly to store ugly numbers, and we maintain pointers i2, i3, and i5 to track the current position of multiplication for 2, 3, and 5 respectively. The next_2, next_3, and next_5 variables hold the next potential candidates for ugly numbers, and we compute the next ugly number as the minimum of these three values.
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.