<p>An <strong>ugly number</strong> is a positive integer whose prime factors are limited to <code>2</code>, <code>3</code>, and <code>5</code>.</p>

<p>Given an integer <code>n</code>, return <em>the</em> <code>n<sup>th</sup></code> <em><strong>ugly number</strong></em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> n = 10
<strong>Output:</strong> 12
<strong>Explanation:</strong> [1, 2, 3, 4, 5, 6, 8, 9, 10, 12] is the sequence of the first 10 ugly numbers.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 1
<strong>Output:</strong> 1
<strong>Explanation:</strong> 1 has no prime factors, therefore all of its prime factors are limited to 2, 3, and 5.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 1690</code></li>
</ul>

class Solution:
    def nthUglyNumber(self, n: int) -> int:
        ugly_numbers = [1]  # The first ugly number is 1
        i2, i3, i5 = 0, 0, 0  # Initialize pointers for 2, 3, and 5
        
        for _ in range(1, n):
            # Compute the next possible ugly numbers from 2, 3, and 5
            next_ugly_2 = ugly_numbers[i2] * 2
            next_ugly_3 = ugly_numbers[i3] * 3
            next_ugly_5 = ugly_numbers[i5] * 5
            
            # The next ugly number is the minimum of these
            next_ugly = min(next_ugly_2, next_ugly_3, next_ugly_5)
            ugly_numbers.append(next_ugly)
            
            # Move the pointer that corresponds to the chosen number
            if next_ugly == next_ugly_2:
                i2 += 1
            if next_ugly == next_ugly_3:
                i3 += 1
            if next_ugly == next_ugly_5:
                i5 += 1
        
        return ugly_numbers[-1]

16700