<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:
        
        rows, cols = len(pizza), len(pizza[0])
        MOD = 10**9 + 7
        
        def apple_in_submatrix(r1, c1, r2, c2):
            # Make sure we don't go out of bounds
            for i, j in product(range(r1, r2 + 1), range(c1, c2 + 1)):
                if pizza[i][j] == 'A': return True
            return False

        @cache
        def count_ways(r1, c1, cuts):
            # Ensure entire area has at least one 'A' before considering valid cuts
            if not apple_in_submatrix(r1, c1, rows - 1, cols - 1): return 0
            if cuts == 0: return 1  # No more cuts left, this is a valid piece
                
            count = 0
            # Cut horizontally and ensure the subpiece up to the cut has at least one 'A'
            for i in range(r1 + 1, rows):
                if apple_in_submatrix(r1, c1, i - 1, cols - 1):
                    count += count_ways(i, c1, cuts - 1) % MOD

            # Cut vertically and ensure the subpiece up to the cut has at least one 'A'
            for j in range(c1 + 1, cols):
                if apple_in_submatrix(r1, c1, rows - 1, j - 1):
                    count = (count + count_ways(r1, j, cuts - 1)) % MOD

            return count

        return count_ways(0, 0, k - 1)  # Start from the whole pizza and make k-1 additional cuts

19000