<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>

# LeetCode Solution - Approach 2: Dynamic Programming w/ Optimized Space Complexity.

class Solution:
    def ways(self, pizza: List[str], k: int) -> int:
        rows = len(pizza)
        cols = len(pizza[0])
        apples = [[0] * (cols + 1) for row in range(rows + 1)]
        for row in range(rows - 1, -1, -1):
            for col in range(cols - 1, -1, -1):
                apples[row][col] = ((pizza[row][col] == 'A')
                                    + apples[row + 1][col]
                                    + apples[row][col + 1]
                                    - apples[row + 1][col + 1])
        f = [[int(apples[row][col] > 0) for col in range(cols)]
             for row in range(rows)]
        mod = 1000000007
        for remain in range(1, k):
            g = [[0 for col in range(cols)] for row in range(rows)]
            for row in range(rows):
                for col in range(cols):
                    for next_row in range(row + 1, rows):
                        if apples[row][col] - apples[next_row][col] > 0:
                            g[row][col] += f[next_row][col]
                    for next_col in range(col + 1, cols):
                        if apples[row][col] - apples[row][next_col] > 0:
                            g[row][col] += f[row][next_col]
                    g[row][col] %= mod
            f = g
        return f[0][0]
17100