<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:
        nrow, ncol = len(pizza), len(pizza[0])
        pizza = [x[::-1] for x in pizza]
        pizza = pizza[::-1]
        mod = 10**9 + 7
        @cache
        def countApple(i,j):
            if i == 0 and j == 0:
                return (1 if pizza[i][j] == 'A' else 0)
            if i == 0:
                return countApple(i, j-1) + (1 if pizza[i][j] == 'A' else 0)
            if j == 0:
                return countApple(i-1, j) + (1 if pizza[i][j] == 'A' else 0)
            return countApple(i-1, j) + countApple(i, j-1) - countApple(i-1,j-1) + (1 if pizza[i][j] == 'A' else 0)

        @cache
        def find(row_index, col_index, piece):
            if countApple(row_index, col_index) == 0:
                return 0
            if piece == 1:
                return 1
            ans = 0
            for i in range(row_index):
                if countApple(row_index, col_index) - countApple(i, col_index) > 0:
                    ans += find(i, col_index, piece - 1) % mod
            for j in range(col_index):
                if countApple(row_index, col_index) - countApple(row_index, j) > 0:
                    ans += find(row_index, j, piece - 1) % mod
            return ans % mod
        return find(nrow-1, ncol-1, k)   
    

19000