<p>A <strong>super ugly number</strong> is a positive integer whose prime factors are in the array <code>primes</code>.</p>

<p>Given an integer <code>n</code> and an array of integers <code>primes</code>, return <em>the</em> <code>n<sup>th</sup></code> <em><strong>super ugly number</strong></em>.</p>

<p>The <code>n<sup>th</sup></code> <strong>super ugly number</strong> is <strong>guaranteed</strong> to fit in a <strong>32-bit</strong> signed integer.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> n = 12, primes = [2,7,13,19]
<strong>Output:</strong> 32
<strong>Explanation:</strong> [1,2,4,7,8,13,14,16,19,26,28,32] is the sequence of the first 12 super ugly numbers given primes = [2,7,13,19].
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 1, primes = [2,3,5]
<strong>Output:</strong> 1
<strong>Explanation:</strong> 1 has no prime factors, therefore all of its prime factors are in the array primes = [2,3,5].
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= primes.length &lt;= 100</code></li>
	<li><code>2 &lt;= primes[i] &lt;= 1000</code></li>
	<li><code>primes[i]</code> is <strong>guaranteed</strong> to be a prime number.</li>
	<li>All the values of <code>primes</code> are <strong>unique</strong> and sorted in <strong>ascending order</strong>.</li>
</ul>

class Solution {
public:
    int nthSuperUglyNumber(int n, vector<int>& primes) {
        vector<long long> dp(n + 1, 0);
        dp[1] = 1;

        // Priority queue to store (value, prime, index)
        priority_queue<vector<long long>, vector<vector<long long>>, greater<vector<long long>>> q;
         
        for (int i = 0; i < primes.size(); i++) {
            q.push({primes[i], primes[i], 2});
        }

        long long last = -1;
        int x = 2;

        for (int i = 2; i <= n; i++) {
            auto it = q.top();
            q.pop();

            if (it[0] == last) {
                i--;
            } else {
                dp[x++] = it[0];
            }
           
            if (it[2] <= n) { 
                long long number = it[1] * dp[it[2]];
                last = it[0];
                q.push({number, it[1], it[2] + 1});
            }
        }
        return dp[n];
    }
};

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