<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>

from functools import lru_cache
from typing import List

class Solution:
    def ways(self, pizza: List[str], k: int) -> int:
        n, m = len(pizza), len(pizza[0])
        MOD = 10 ** 9 + 7

        # Precompute the number of apples in the rectangle from (i, j) to the bottom-right corner
        apples = [[0] * (m + 1) for _ in range(n + 1)]
        for i in reversed(range(n)):
            for j in reversed(range(m)):
                apples[i][j] = (
                    (1 if pizza[i][j] == 'A' else 0)
                    + apples[i + 1][j]
                    + apples[i][j + 1]
                    - apples[i + 1][j + 1]
                )

        @cache
        def dfs(r: int, c: int, cuts: int) -> int:
            # Check if there are no apples in the current rectangle
            if apples[r][c] == 0:
                return 0
            # If no more cuts are needed and there's at least one apple
            if cuts == 1:
                return 1

            total_ways = 0

            # Try horizontal cuts
            for nr in range(r + 1, n):
                # Ensure the upper piece has at least one apple
                if apples[r][c] - apples[nr][c] > 0:
                    total_ways = (total_ways + dfs(nr, c, cuts - 1)) % MOD

            # Try vertical cuts
            for nc in range(c + 1, m):
                # Ensure the left piece has at least one apple
                if apples[r][c] - apples[r][nc] > 0:
                    total_ways = (total_ways + dfs(r, nc, cuts - 1)) % MOD

            return total_ways

        return dfs(0, 0, k)

18800