<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>

int mod = 1e9+7;
class Solution {
public:
    vector<vector<vector<int>>> dp;

    bool hasApple(vector<string>& pizza, int row, int col, int n, int m) {
        for (int i = row; i < n; i++) {
            for (int j = col; j < m; j++) {
                if (pizza[i][j] == 'A') return true;
            }
        }
        return false;
    }

    int solve(vector<string>& pizza, int k, int row, int col, int n, int m) {
        if (k == 1) {
            return hasApple(pizza, row, col, n, m) ? 1 : 0;
        }

        if (dp[row][col][k] != -1) return dp[row][col][k];

        int res = 0;
        // bool appleFound = false;

        for (int i = row; i < n - 1; i++) {
            if (hasApple(pizza, row, col, i + 1, m)) {
                res = (res + solve(pizza, k - 1, i + 1, col, n, m)) % mod;
            }
        }

        // Vertical cuts
        for (int j = col; j < m - 1; j++) {
            if (hasApple(pizza, row, col, n, j + 1)) {
                res = (res + solve(pizza, k - 1, row, j + 1, n, m)) % mod;
            }
        }

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

    int ways(vector<string>& pizza, int k) {
        int n = pizza.size();
        int m = pizza[0].size();

        dp = vector<vector<vector<int>>>(n, vector<vector<int>>(m, vector<int>(k + 1, -1)));
        return solve(pizza, k, 0, 0, n, m);
    }
};

10000