<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) {
        map<int , int > m;
        m[2]=0;
        m[3]=0;
        m[5]=0;
        vector<int > sub;
        sub.push_back(1);
        int c=1;
        
        int ans=1;
        if (n==c) return c;
        while(n!=c){
            if (2*sub[m[2]] <= 3*sub[m[3]] && 2*sub[m[2]] <= 5*sub[m[5]]){
                if (sub[sub.size()-1]<2*sub[m[2]]){
                    
                    c+=1;
                    ans=2*sub[m[2]];
                    sub.push_back(2*sub[m[2]]);

                }
                
                m[2]++;
            }
            else if(3*sub[m[3]] <=2*sub[m[2]]  && 3*sub[m[3]] <= 5*sub[m[5]]){
                if (sub[sub.size()-1]<3*sub[m[3]]){
                    
                    c+=1;
                    ans=3*sub[m[3]];
                    sub.push_back(3*sub[m[3]]);

                }
                m[3]++;
            }
            else if(5*sub[m[5]] <=2*sub[m[2]]  && 5*sub[m[5]] <= 3*sub[m[3]]){
                if (sub[sub.size()-1]<5*sub[m[5]]){
                    
                    c+=1;
                    ans=5*sub[m[5]];
                    sub.push_back(5*sub[m[5]]);

                }
                m[5]++;
            }
        }
        return ans;
    }
};
13378