<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:
        mod = 10**9 + 7
        m, n = len(pizza), len(pizza[0])
        
        # Create a prefix sum array to quickly check if there is at least one apple in a sub-pizza
        pre_sum = [[0] * (n + 1) for _ in range(m + 1)]
        for i in range(m - 1, -1, -1):
            for j in range(n - 1, -1, -1):
                pre_sum[i][j] = pre_sum[i + 1][j] + pre_sum[i][j + 1] - pre_sum[i + 1][j + 1] + (pizza[i][j] == 'A')
        
        # DP table to store the number of ways to cut the pizza
        dp = [[[0] * (k + 1) for _ in range(n)] for _ in range(m)]
        
        # Initialize the DP table for the base case when there is only one piece left
        for i in range(m):
            for j in range(n):
                dp[i][j][1] = 1 if pre_sum[i][j] > 0 else 0
        
        # Fill the DP table for k > 1
        for remain in range(2, k + 1):
            for i in range(m):
                for j in range(n):
                    # Cut horizontally
                    for n_i in range(i + 1, m):
                        if pre_sum[i][j] - pre_sum[n_i][j] > 0:
                            dp[i][j][remain] += dp[n_i][j][remain - 1]
                            dp[i][j][remain] %= mod
                    
                    # Cut vertically
                    for n_j in range(j + 1, n):
                        if pre_sum[i][j] - pre_sum[i][n_j] > 0:
                            dp[i][j][remain] += dp[i][n_j][remain - 1]
                            dp[i][j][remain] %= mod
        
        return dp[0][0][k]

# Example usage:
solution = Solution()
pizza = ["A..","AAA","..."]
k = 3
print(solution.ways(pizza, k))  # Output: 3
17700