<p>Koko loves to eat bananas. There are <code>n</code> piles of bananas, the <code>i<sup>th</sup></code> pile has <code>piles[i]</code> bananas. The guards have gone and will come back in <code>h</code> hours.</p>

<p>Koko can decide her bananas-per-hour eating speed of <code>k</code>. Each hour, she chooses some pile of bananas and eats <code>k</code> bananas from that pile. If the pile has less than <code>k</code> bananas, she eats all of them instead and will not eat any more bananas during this hour.</p>

<p>Koko likes to eat slowly but still wants to finish eating all the bananas before the guards return.</p>

<p>Return <em>the minimum integer</em> <code>k</code> <em>such that she can eat all the bananas within</em> <code>h</code> <em>hours</em>.</p>

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

<pre>
<strong>Input:</strong> piles = [3,6,7,11], h = 8
<strong>Output:</strong> 4
</pre>

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

<pre>
<strong>Input:</strong> piles = [30,11,23,4,20], h = 5
<strong>Output:</strong> 30
</pre>

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

<pre>
<strong>Input:</strong> piles = [30,11,23,4,20], h = 6
<strong>Output:</strong> 23
</pre>

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

<ul>
	<li><code>1 &lt;= piles.length &lt;= 10<sup>4</sup></code></li>
	<li><code>piles.length &lt;= h &lt;= 10<sup>9</sup></code></li>
	<li><code>1 &lt;= piles[i] &lt;= 10<sup>9</sup></code></li>
</ul>

from typing import List
from math import ceil
import ast


class Solution:
    def minEatingSpeed(self, piles: List[int], h: int) -> int:
        def k_works(k: int) -> bool:
            hours = 0
            for p in piles:
                hours += ceil(p / k)
            return hours <= h

        l = 1
        r = max(piles)
        while l <= r:
            middle = (l + r) // 2
            if k_works(middle):
                r = middle - 1
            else:
                l = middle + 1

        return l
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