<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>

// TC - O(n logn) time - By channa sir
class Solution1 {
public:
    int nthUglyNumber(int n) {
        set<int> fringeSet;
        fringeSet.insert(1);
        while(n > 1) {
            n--;
            int num = *fringeSet.begin(); // next ugly number
            fringeSet.erase(num);
            vector<int> primes = {2, 3, 5};
            for(int prime : primes) {
                // if(prime*num <= INT_MAX && fringeSet.find(prime*num) == fringeSet.end()) {
                if(num <= INT_MAX/prime && fringeSet.find(prime*num) == fringeSet.end()) {
                    fringeSet.insert(prime*num);
                }
            }
        }
        return *fringeSet.begin();
    }
};

// O(n logn) time - by Channa Sir
class Solution {
public:
    int nthUglyNumber(int n) {
        priority_queue<int, vector<int>, greater<int>> pq;
        pq.push(1);
        while(n > 1) {
            n--;
            int num = pq.top();
            pq.pop();
            if(num % 3 != 0 && num % 5 != 0) {
                if(num <= INT_MAX / 2) pq.push(2 * num);
            }
            if(num % 5 != 0) {
                if(num <= INT_MAX / 3) pq.push(3 * num);
            }
            if(num <= INT_MAX / 5) pq.push(5 * num);
        }
        return pq.top();
    }
};

// TC-O(n) by Mohit Kuk
class Solution3 {
public:
    int nthUglyNumber(int n) {
        vector<long long> arr(n+1);
        arr[1]=1;
        int i=1,j=1,k=1;
        for(int x=2;x<=n;x++){
            long long fi=arr[i]*2;
            long long se=arr[j]*3;
            long long th=arr[k]*5;
            long long mn=min({fi,se,th});
            if(mn==fi)
            {
                arr[x]=fi;
                i++;
            }
            if(mn==se)
            {
                arr[x]=se;
                j++;
            }
            if(mn==th)
            {
                arr[x]=th;
                k++;
            }
        }
        return arr[n];
    }
};
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