Minimize the Maximum of Two Arrays
<p>We have two arrays <code>arr1</code> and <code>arr2</code> which are initially empty. You need to add positive integers to them such that they satisfy all the following conditions:</p>

<ul>
	<li><code>arr1</code> contains <code>uniqueCnt1</code> <strong>distinct</strong> positive integers, each of which is <strong>not divisible</strong> by <code>divisor1</code>.</li>
	<li><code>arr2</code> contains <code>uniqueCnt2</code> <strong>distinct</strong> positive integers, each of which is <strong>not divisible</strong> by <code>divisor2</code>.</li>
	<li><strong>No</strong> integer is present in both <code>arr1</code> and <code>arr2</code>.</li>
</ul>

<p>Given <code>divisor1</code>, <code>divisor2</code>, <code>uniqueCnt1</code>, and <code>uniqueCnt2</code>, return <em>the <strong>minimum possible maximum</strong> integer that can be present in either array</em>.</p>

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

<pre>
<strong>Input:</strong> divisor1 = 2, divisor2 = 7, uniqueCnt1 = 1, uniqueCnt2 = 3
<strong>Output:</strong> 4
<strong>Explanation:</strong> 
We can distribute the first 4 natural numbers into arr1 and arr2.
arr1 = [1] and arr2 = [2,3,4].
We can see that both arrays satisfy all the conditions.
Since the maximum value is 4, we return it.
</pre>

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

<pre>
<strong>Input:</strong> divisor1 = 3, divisor2 = 5, uniqueCnt1 = 2, uniqueCnt2 = 1
<strong>Output:</strong> 3
<strong>Explanation:</strong> 
Here arr1 = [1,2], and arr2 = [3] satisfy all conditions.
Since the maximum value is 3, we return it.</pre>

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

<pre>
<strong>Input:</strong> divisor1 = 2, divisor2 = 4, uniqueCnt1 = 8, uniqueCnt2 = 2
<strong>Output:</strong> 15
<strong>Explanation:</strong> 
Here, the final possible arrays can be arr1 = [1,3,5,7,9,11,13,15], and arr2 = [2,6].
It can be shown that it is not possible to obtain a lower maximum satisfying all conditions. 
</pre>

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

<ul>
	<li><code>2 &lt;= divisor1, divisor2 &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= uniqueCnt1, uniqueCnt2 &lt; 10<sup>9</sup></code></li>
	<li><code>2 &lt;= uniqueCnt1 + uniqueCnt2 &lt;= 10<sup>9</sup></code></li>
</ul>
Math; Binary Search; Number Theory
func minimizeSet(divisor1 int, divisor2 int, uniqueCnt1 int, uniqueCnt2 int) int {
	divisor := lcm(divisor1, divisor2)
	left, right := 1, 10000000000
	for left < right {
		mid := (left + right) >> 1
		cnt1 := mid/divisor1*(divisor1-1) + mid%divisor1
		cnt2 := mid/divisor2*(divisor2-1) + mid%divisor2
		cnt := mid/divisor*(divisor-1) + mid%divisor
		if cnt1 >= uniqueCnt1 && cnt2 >= uniqueCnt2 && cnt >= uniqueCnt1+uniqueCnt2 {
			right = mid
		} else {
			left = mid + 1
		}
	}
	return left
}

func lcm(a, b int) int {
	return a * b / gcd(a, b)
}

func gcd(a, b int) int {
	if b == 0 {
		return a
	}
	return gcd(b, a%b)
}