<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:
    void helper(int twos, int threes, int fives, int n, long curr, vector<long>& nums) {
        switch (n) {
            case -1:
                return;
                break;
            case 2:
                for (int i = 0; i <= twos; ++i) {
                    nums.push_back(curr);
                    helper(i, threes, fives, 3, curr, nums);
                    curr *= 2;
                }
                break;
            case 3:
                helper(twos, 0, fives, 5, curr, nums);
                for (int i = 1; i <= threes - (twos / 1.58); ++i) {
                    curr *= 3;
                    nums.push_back(curr);
                    helper(twos, i, fives, 5, curr, nums);
                }
                break;
            case 5:
                helper(twos, threes, 0, -1, curr, nums);
                for (int i = 1; i <= fives - (twos / 2.32) - (threes / 1.465); ++i) {
                    curr *= 5;
                    nums.push_back(curr);
                    helper(twos, threes, i, -1, curr, nums);
                }
                break;
        }
    }

    int nthUglyNumber(int n) {
        vector<long> nums;

        double num = 4.7 * pow(2.34, log(n)/log(10)-1);
        int num_threes = ceil(num / 1.58);
        int num_fives =  ceil(num / 2.32);
        int num_twos = num;

        helper(num_twos, num_threes, num_fives, 2, 1, nums);
        sort(nums.begin(), nums.end());
        return nums[n-1];
    }
};
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