<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) {
        if (n==1) return 1;
        priority_queue<long long,vector<long long>,greater<long long> > l1,l2,l3;
        l1.push(2);
        l2.push(3);
        l3.push(5);
        
        for (int i = 1 ; i < n - 1; i++)
        {
            if (l1.top() < l2.top())
            {
                if (l1.top() < l3.top())
                {
                    long long ele = l1.top();
                    l1.pop();
                    l1.push(2*ele);
                    l2.push(3*ele);
                    l3.push(5*ele);
                }
                else 
                {
                    long long ele = l3.top();
                    l3.pop();
                    l3.push(5*ele);
                }
            }
            else
            {
                if (l3.top() < l2.top())
                {
                    long long ele = l3.top();
                    l3.pop();
                    l3.push(5*ele);                    
                }
                else
                {
                    long long ele = l2.top();
                    l2.pop();
                    l2.push(3*ele);
                    l3.push(5*ele);
                }
            }
        }
        return min(l1.top(),min(l2.top(),l3.top()));
    }
};
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