<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:
    #define ll long long
    int nthUglyNumber(int n) {
        vector<int> primes = {2,3,5};
        vector<ll> v(n, 0);
        priority_queue<pair<ll, pair<ll, ll>> ,vector<pair<ll, pair<ll, ll>>>,  greater<pair<ll, pair<ll, ll>>>> pq;
        v[0] = 1;
        for(auto it:primes)pq.push({it, {it, 0}});
        for(ll i = 1; i<n; i++){
            auto t = pq.top(); pq.pop();
            v[i] = t.first;
            if(v[i] == v[i-1])i--;
            t.first = t.second.first * v[t.second.second+1];
            t.second.second++;
            if(t.second.second+1<n)pq.push(t);
            // cout<<t.first<<" "<<t.second<<endl;
        }
        // for(ll i = 0; i<n; i++)cout<<v[i]<<" ";
        return v[n-1];
    }
};
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