<p>An <strong>ugly number</strong> is a positive integer that is divisible by <code>a</code>, <code>b</code>, or <code>c</code>.</p>

<p>Given four integers <code>n</code>, <code>a</code>, <code>b</code>, and <code>c</code>, return the <code>n<sup>th</sup></code> <strong>ugly number</strong>.</p>

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

<pre>
<strong>Input:</strong> n = 3, a = 2, b = 3, c = 5
<strong>Output:</strong> 4
<strong>Explanation:</strong> The ugly numbers are 2, 3, 4, 5, 6, 8, 9, 10... The 3<sup>rd</sup> is 4.
</pre>

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

<pre>
<strong>Input:</strong> n = 4, a = 2, b = 3, c = 4
<strong>Output:</strong> 6
<strong>Explanation:</strong> The ugly numbers are 2, 3, 4, 6, 8, 9, 10, 12... The 4<sup>th</sup> is 6.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> n = 5, a = 2, b = 11, c = 13
<strong>Output:</strong> 10
<strong>Explanation:</strong> The ugly numbers are 2, 4, 6, 8, 10, 11, 12, 13... The 5<sup>th</sup> is 10.
</pre>

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

<ul>
	<li><code>1 &lt;= n, a, b, c &lt;= 10<sup>9</sup></code></li>
	<li><code>1 &lt;= a * b * c &lt;= 10<sup>18</sup></code></li>
	<li>It is guaranteed that the result will be in range <code>[1, 2 * 10<sup>9</sup>]</code>.</li>
</ul>

class Solution:
    def nthUglyNumber(self, n: int, a: int, b: int, c: int) -> int:
        # enough function: given an input num, determine whether there are at least n ugly number less than or equal to num
        # since a might be a multiple of b or c, or the other way round, we need the help of the greatest common divisor to avoid counting duplicate numbers
        def enough(num: int) -> bool:
            total = num//a + num//b + num//c - num//ab - num//bc - num//ac + num//abc
            return total >= n

        ab = a*b//math.gcd(a,b)
        bc = b*c//math.gcd(b,c)
        ac = a*c//math.gcd(a,c)
        abc = a*bc//math.gcd(a, bc)

        l, r = 1, 10**9 * 2
        while l < r:
            mid = l + (r-l) // 2
            if enough(mid):
                r = mid
            else:
                l = mid + 1
        return r

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