<p>Given a rectangular pizza represented as a <code>rows x cols</code>&nbsp;matrix containing the following characters: <code>&#39;A&#39;</code> (an apple) and <code>&#39;.&#39;</code> (empty cell) and given the integer <code>k</code>. You have to cut the pizza into <code>k</code> pieces using <code>k-1</code> cuts.&nbsp;</p>

<p>For each cut you choose the direction: vertical or horizontal, then you choose a cut position at the cell boundary and cut the pizza into two pieces. If you cut the pizza vertically, give the left part of the pizza to a person. If you cut the pizza horizontally, give the upper part of the pizza to a person. Give the last piece of pizza to the last person.</p>

<p><em>Return the number of ways of cutting the pizza such that each piece contains <strong>at least</strong> one apple.&nbsp;</em>Since the answer can be a huge number, return this modulo 10^9 + 7.</p>

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

<p><strong><img alt="" src="https://assets.leetcode.com/uploads/2020/04/23/ways_to_cut_apple_1.png" style="width: 500px; height: 378px;" /></strong></p>

<pre>
<strong>Input:</strong> pizza = [&quot;A..&quot;,&quot;AAA&quot;,&quot;...&quot;], k = 3
<strong>Output:</strong> 3 
<strong>Explanation:</strong> The figure above shows the three ways to cut the pizza. Note that pieces must contain at least one apple.
</pre>

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

<pre>
<strong>Input:</strong> pizza = [&quot;A..&quot;,&quot;AA.&quot;,&quot;...&quot;], k = 3
<strong>Output:</strong> 1
</pre>

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

<pre>
<strong>Input:</strong> pizza = [&quot;A..&quot;,&quot;A..&quot;,&quot;...&quot;], k = 1
<strong>Output:</strong> 1
</pre>

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

<ul>
	<li><code>1 &lt;= rows, cols &lt;= 50</code></li>
	<li><code>rows ==&nbsp;pizza.length</code></li>
	<li><code>cols ==&nbsp;pizza[i].length</code></li>
	<li><code>1 &lt;= k &lt;= 10</code></li>
	<li><code>pizza</code> consists of characters <code>&#39;A&#39;</code>&nbsp;and <code>&#39;.&#39;</code> only.</li>
</ul>

class Solution:
    def ways(self, pizza: List[str], k: int) -> int:
        ### recursively cut the pizza, adding up the possible ways and 
        ### caching already computed results

        cache = {}
        def helper(r,c,r2,c2,k):
            # base case, k - 1 cuts already made
            if k == 1 and self.hasApple(r,c,r2,c2, pizza):
                return 1
            elif k == 1:
                return 0
            elif (r,c,r2,c2,k) in cache:
                return cache[(r,c,r2,c2,k)]
            elif not self.hasApple(r,c,r2,c2,pizza):
                return 0

            # make all possible vertical and horizontal cuts, where 
            # the slice that is removed has an apple :)
            # first horizontal cuts, remove rows above the cut
            cuts = 0
            for cutR in range(r+1, r2+1):
                # check if slize cut off has an apple
                if self.hasApple(r, c, cutR-1, c2, pizza):
                    cuts += helper(cutR, c, r2, c2, k-1)
            # now vertical cuts, remove the cols to the left of the cut
            for cutH in range(c+1, c2+1):
                if self.hasApple(r, c, r2, cutH-1, pizza):
                    cuts += helper(r, cutH, r2, c2, k-1)

            cache[(r,c,r2,c2,k)] = cuts
            return cuts
        return helper(0,0,len(pizza)-1, len(pizza[0])-1, k) % (10**9 + 7)
            
    def hasApple(self, r, c, r2, c2, pizza):
        # check if the current slice has an apple or not
        for row in pizza[r:r2+1]:
            if 'A' in row[c:c2+1]:
                return True
        return False

18600