<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 typing import List

class Solution:
    def ways(self, pizza: List[str], k: int) -> int:
        MOD = 10**9 + 7
        rows, cols = len(pizza), len(pizza[0])
        
        # Precompute the number of apples in submatrices using a prefix sum array
        apples = [[0] * (cols + 1) for _ in range(rows + 1)]
        for r in range(rows - 1, -1, -1):
            for c in range(cols - 1, -1, -1):
                apples[r][c] = (1 if pizza[r][c] == 'A' else 0) + apples[r + 1][c] + apples[r][c + 1] - apples[r + 1][c + 1]
        
        # Memoization table
        dp = [[[-1] * k for _ in range(cols)] for _ in range(rows)]
        
        def has_apple(r1, c1, r2, c2):
            return apples[r1][c1] - apples[r2][c1] - apples[r1][c2] + apples[r2][c2] > 0
        
        def dfs(r, c, cuts):
            if cuts == 0:
                return 1 if has_apple(r, c, rows, cols) else 0
            if dp[r][c][cuts] != -1:
                return dp[r][c][cuts]
            
            ways = 0
            # Try horizontal cuts
            for nr in range(r + 1, rows):
                if has_apple(r, c, nr, cols):
                    ways = (ways + dfs(nr, c, cuts - 1)) % MOD
            # Try vertical cuts
            for nc in range(c + 1, cols):
                if has_apple(r, c, rows, nc):
                    ways = (ways + dfs(r, nc, cuts - 1)) % MOD
            
            dp[r][c][cuts] = ways
            return ways
        
        return dfs(0, 0, k - 1)
17500