<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:
        self.memo = {}
        self.pizza = [["A" == j for j in i] for i in pizza]
        self.rows = len(self.pizza)
        self.cols = len(self.pizza[0])
        self.waysHelper(k-1,0,0)%(10**9 + 7)
        print(self.memo)
        return self.waysHelper(k-1,0,0)%(10**9 + 7)
    
    def waysHelper(self,k,cutRow,cutCol):
        if (k,cutRow,cutCol) in self.memo:
            return self.memo[k,cutRow,cutCol]
        applesRemaining = 0
        for i in range(cutRow,self.rows):
            for j in range(cutCol,self.cols):
                applesRemaining += self.pizza[i][j]
        if applesRemaining == 0:
            self.memo[k,cutRow,cutCol] = 0
            return 0
        if k == 0:
            self.memo[k,cutRow,cutCol] = 1
            return 1
        
        
        
        
        seenAtLeastOneApple = False
        totRowCuts = 0
        for row in range(cutRow, self.rows):
            if not seenAtLeastOneApple:
                for testCol in range(cutCol,self.cols):
                    if self.pizza[row][testCol]:
                        seenAtLeastOneApple = True
                        break
                
            if seenAtLeastOneApple:
                totRowCuts = (totRowCuts + self.waysHelper(k-1,row+1,cutCol)) % (10**9 + 7)

        seenAtLeastOneApple = False
        totColCuts = 0
        for col in range(cutCol, self.cols):
            if not seenAtLeastOneApple:
                for testRow in range(cutRow,self.rows):
                    if self.pizza[testRow][col]:
                        seenAtLeastOneApple = True
                        break
            if seenAtLeastOneApple:
                totColCuts = (totColCuts + self.waysHelper(k-1,cutRow,col+1)) % (10**9 + 7)
        
        self.memo[k,cutRow,cutCol] = totRowCuts + totColCuts
        return totRowCuts + totColCuts
    
        
18900