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.
1public class Main {
2
3 public static int nthUglyNumber(int n) {
4 int[] ugly = new int[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 = Math.min(next_2, Math.min(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
20 public static void main(String[] args) {
21 int n = 10;
22 System.out.println(nthUglyNumber(n)); // Output: 12
23 }
24}
25This Java solution uses an integer array to store the sequence of ugly numbers, with the same logic and three pointers as in the C/C++ solutions. It sequentially generates ugly numbers by multiplying previous numbers by 2, 3, and 5 and selecting the smallest candidate in each iteration.
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.
1#include <iostream>
2#include <queue>
3#include <unordered_set>
4using namespace std;
5
6int nthUglyNumber(int n) {
7 priority_queue<long, vector<long>, greater<long>> minHeap;
8 unordered_set<long> seen;
9 minHeap.push(1);
10 seen.insert(1);
11 long currUgly = 1;
12
13 for (int i = 0; i < n; ++i) {
14 currUgly = minHeap.top();
15 minHeap.pop();
16 if (!seen.count(currUgly * 2)) {
17 minHeap.push(currUgly * 2);
18 seen.insert(currUgly * 2);
19 }
20 if (!seen.count(currUgly * 3)) {
21 minHeap.push(currUgly * 3);
22 seen.insert(currUgly * 3);
23 }
24 if (!seen.count(currUgly * 5)) {
25 minHeap.push(currUgly * 5);
26 seen.insert(currUgly * 5);
27 }
28 }
29 return (int)currUgly;
30}
31
32int main() {
33 int n = 10;
34 cout << nthUglyNumber(n) << endl; // Output: 12
35 return 0;
36}The solution uses a min-heap to keep track of and retrieve the smallest ugly number, avoiding duplicates by using a set. It generates ugly numbers by multiplying the current smallest number by 2, 3, and 5, inserting valid new numbers back into the heap.