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_264.java
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package com.fishercoder.solutions;
import java.util.TreeSet;
public class _264 {
public static class Solution1 {
/**
* credit: https://discuss.leetcode.com/topic/21791/o-n-java-solution
*/
public int nthUglyNumber(int n) {
int[] ugly = new int[n];
ugly[0] = 1;
int index2 = 0;
int index3 = 0;
int index5 = 0;
int factor2 = 2;
int factor3 = 3;
int factor5 = 5;
for (int i = 1; i < n; i++) {
int min = Math.min(Math.min(factor2, factor3), factor5);
ugly[i] = min;
if (factor2 == min) {
factor2 = 2 * ugly[++index2];
}
if (factor3 == min) {
factor3 = 3 * ugly[++index3];
}
if (factor5 == min) {
factor5 = 5 * ugly[++index5];
}
}
return ugly[n - 1];
}
}
public static class Solution2 {
/**
* My completely original solution on 11/7/2021.
* Although not super robust, as the input increases, I'll have to increase the times (variable n) on line 61 as some smaller numbers might appear later.
*/
public int nthUglyNumber(int n) {
TreeSet<Long> treeSet = new TreeSet<>();
treeSet.add(1L);
int count = 1;
int polled = 0;
int[] primes = new int[]{2, 3, 5};
while (!treeSet.isEmpty()) {
int size = treeSet.size();
for (int i = 0; i < size; i++) {
Long curr = treeSet.pollFirst();
polled++;
if (polled == n) {
return curr.intValue();
}
for (int prime : primes) {
treeSet.add(prime * curr);
count++;
}
}
if (count >= n * 3) {
while (polled < n - 1) {
treeSet.pollFirst();
polled++;
}
return treeSet.pollFirst().intValue();
}
}
return -1;
}
}
}