<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>

MOD = 10**9 + 7

class Solution:
    def ways(self, pizza: List[str], k: int) -> int:
        rows, cols = len(pizza), len(pizza[0])
        # Precompute the prefix sum of apples
        preSum = [[0] * (cols + 1) for _ in range(rows + 1)]
        for i in range(rows - 1, -1, -1):
            for j in range(cols - 1, -1, -1):
                preSum[i][j] = (1 if pizza[i][j] == 'A' else 0) \
                               + preSum[i + 1][j] + preSum[i][j + 1] - preSum[i + 1][j + 1]

        # Initialize the dp array with -1
        dp = [[[-1] * (k + 1) for _ in range(cols)] for _ in range(rows)]

        def hasApple(i1, j1, i2, j2):
            # Check if the sub-pizza from (i1, j1) to (i2, j2) contains at least one apple
            return preSum[i1][j1] - preSum[i2][j1] - preSum[i1][j2] + preSum[i2][j2] > 0

        def dfs(i, j, k):
            # If there are no apples in the current sub-pizza, return 0
            if preSum[i][j] == 0:
                return 0
            # If only one piece is needed
            if k == 1:
                return 1
            # Return the cached result if already computed
            if dp[i][j][k] != -1:
                return dp[i][j][k]
            ways = 0
            # Try horizontal cuts
            for row in range(i + 1, rows):
                if hasApple(i, j, row, cols):
                    ways = (ways + dfs(row, j, k - 1)) % MOD
            # Try vertical cuts
            for col in range(j + 1, cols):
                if hasApple(i, j, rows, col):
                    ways = (ways + dfs(i, col, k - 1)) % MOD
            dp[i][j][k] = ways
            return ways

        return dfs(0, 0, k)
17300