<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 {
public:
    int nthUglyNumber(int n) {
        std::vector<int> ugly = {1};
        std::unordered_set<int> ugly_set = {1};
        std::vector<int> primes = {2, 3, 5};
        std::vector<int> new_ugly;
        std::vector<int> pointers = {0, 0, 0};
        while (ugly.size() != n) {
            new_ugly.clear();
            for (int i = 0; i < 3; i++) {
                new_ugly.push_back(primes[i] * ugly[pointers[i]]);
            }
            int p = 0;
            int curr_min = new_ugly[0];
            for (int i = 0; i < 3; i++) {
                if (new_ugly[i] < curr_min) {
                    curr_min = new_ugly[i];
                    p = i;
                }
            }
            if (!ugly_set.contains(curr_min)) {
                ugly_set.insert(curr_min);
                ugly.push_back(curr_min);
            }
            pointers[p]++;
        }
        return ugly[n-1];
    }
};
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