Apple Redistribution into Boxes
<p>You are given an array <code>apple</code> of size <code>n</code> and an array <code>capacity</code> of size <code>m</code>.</p>

<p>There are <code>n</code> packs where the <code>i<sup>th</sup></code> pack contains <code>apple[i]</code> apples. There are <code>m</code> boxes as well, and the <code>i<sup>th</sup></code> box has a capacity of <code>capacity[i]</code> apples.</p>

<p>Return <em>the <strong>minimum</strong> number of boxes you need to select to redistribute these </em><code>n</code><em> packs of apples into boxes</em>.</p>

<p><strong>Note</strong> that, apples from the same pack can be distributed into different boxes.</p>

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

<pre>
<strong>Input:</strong> apple = [1,3,2], capacity = [4,3,1,5,2]
<strong>Output:</strong> 2
<strong>Explanation:</strong> We will use boxes with capacities 4 and 5.
It is possible to distribute the apples as the total capacity is greater than or equal to the total number of apples.
</pre>

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

<pre>
<strong>Input:</strong> apple = [5,5,5], capacity = [2,4,2,7]
<strong>Output:</strong> 4
<strong>Explanation:</strong> We will need to use all the boxes.
</pre>

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

<ul>
	<li><code>1 &lt;= n == apple.length &lt;= 50</code></li>
	<li><code>1 &lt;= m == capacity.length &lt;= 50</code></li>
	<li><code>1 &lt;= apple[i], capacity[i] &lt;= 50</code></li>
	<li>The input is generated such that it&#39;s possible to redistribute packs of apples into boxes.</li>
</ul>
Greedy; Array; Sorting
class Solution:
    def minimumBoxes(self, apple: List[int], capacity: List[int]) -> int:
        capacity.sort(reverse=True)
        s = sum(apple)
        for i, c in enumerate(capacity, 1):
            s -= c
            if s <= 0:
                return i
