<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 {
public:
    int ways(vector<string>& pizza, int k) {
        const int MOD = 1e9 + 7;
        int rows = pizza.size();
        int cols = pizza[0].size();

        // Preprocess the number of apples in the submatrix from (row,col) to (rows-1, cols-1)
        vector<vector<int>> apples(rows + 1, vector<int>(cols + 1, 0));
        for (int row = rows - 1; row >= 0; --row) {
            for (int col = cols - 1; col >= 0; --col) {
                apples[row][col] = (pizza[row][col] == 'A') + apples[row + 1][col] + apples[row][col + 1] - apples[row + 1][col + 1];
            }
        }

        // Memoization cache
        vector<vector<vector<int>>> dp(rows, vector<vector<int>>(cols, vector<int>(k, -1)));

        // Helper function with memoization
        function<int(int, int, int)> dfs = [&](int row, int col, int cuts_left) -> int {
            if (apples[row][col] == 0) return 0; // No apples in current piece
            if (cuts_left == 0) return 1; // Valid way found
            if (dp[row][col][cuts_left] != -1) return dp[row][col][cuts_left];

            int res = 0;

            // Try horizontal cuts
            for (int new_row = row + 1; new_row < rows; ++new_row) {
                // If upper piece has at least one apple
                if (apples[row][col] - apples[new_row][col] > 0) {
                    res = (res + dfs(new_row, col, cuts_left - 1)) % MOD;
                }
            }

            // Try vertical cuts
            for (int new_col = col + 1; new_col < cols; ++new_col) {
                // If left piece has at least one apple
                if (apples[row][col] - apples[row][new_col] > 0) {
                    res = (res + dfs(row, new_col, cuts_left - 1)) % MOD;
                }
            }

            return dp[row][col][cuts_left] = res;
        };

        return dfs(0, 0, k - 1);
    }
};

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