Minimize Max Distance to Gas Station
<p>You are given an integer array <code>stations</code> that represents the positions of the gas stations on the <strong>x-axis</strong>. You are also given an integer <code>k</code>.</p>

<p>You should add <code>k</code> new gas stations. You can add the stations anywhere on the <strong>x-axis</strong>, and not necessarily on an integer position.</p>

<p>Let <code>penalty()</code> be the maximum distance between <strong>adjacent</strong> gas stations after adding the <code>k</code> new stations.</p>

<p>Return <em>the smallest possible value of</em> <code>penalty()</code>. Answers within <code>10<sup>-6</sup></code> of the actual answer will be accepted.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> stations = [1,2,3,4,5,6,7,8,9,10], k = 9
<strong>Output:</strong> 0.50000
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> stations = [23,24,36,39,46,56,57,65,84,98], k = 1
<strong>Output:</strong> 14.00000
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>10 &lt;= stations.length &lt;= 2000</code></li>
	<li><code>0 &lt;= stations[i] &lt;= 10<sup>8</sup></code></li>
	<li><code>stations</code> is sorted in a <strong>strictly increasing</strong> order.</li>
	<li><code>1 &lt;= k &lt;= 10<sup>6</sup></code></li>
</ul>
Array; Binary Search
class Solution:
    def minmaxGasDist(self, stations: List[int], k: int) -> float:
        def check(x):
            return sum(int((b - a) / x) for a, b in pairwise(stations)) <= k

        left, right = 0, 1e8
        while right - left > 1e-6:
            mid = (left + right) / 2
            if check(mid):
                right = mid
            else:
                left = mid
        return left
