<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>

#define lli long long int 
#define ld long double
#define vi vector<int>
#define vlli vector<lli>
#define vpii vector<pair<int, int>>
#define pb push_back 

template<typename T> void debug(T _a) {cout << _a << " ";}
template<typename T1, typename T2> void debug(pair<T1, T2> _p) {cout<<"{";debug(_p.first);cout<<": ";debug(_p.second);cout<<"}\n";}
template<typename T> void debug(vector<T> _aa) {for (auto h: _aa) debug(h); cout << endl;}
template<typename T> void debug(deque<T> _aa) {for (auto h: _aa) debug(h); cout << endl;}
template<typename T> void debug(multiset<T> _aa) {for (auto h: _aa) debug(h); cout << endl;}
template<typename T> void debug(set<T> _aa) {for (auto h: _aa) debug(h); cout << endl;}
template<typename T1, typename T2> void debug(map<T1, T2> _mm) {for (auto h: _mm) debug(h);}

class Solution {
public:
    int nthUglyNumber(int n) {
        vector<int> primes = {2,3,5};
        vector<int> dp(n+1);
        dp[1] = 1;
        priority_queue<pair<lli, pair<lli, lli>>> pq;
        for (int i: primes) pq.push({-i, {i, 1}});

        int i = 1;

        while (i < n) {
            auto cur = pq.top(); pq.pop();
            lli val = -cur.first, p = cur.second.first, idx = cur.second.second;
            if (val > dp[i]) {
                dp[++i] = val;
            }
            pq.push({-p * dp[idx+1], {p, idx+1}});
        }
        return dp[n];
    }
};
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