Number of Ways of Cutting a Pizza
<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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1400-1499/1444.Number%20of%20Ways%20of%20Cutting%20a%20Pizza/images/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>
Memoization; Array; Dynamic Programming; Matrix; Prefix Sum
class Solution {
public:
    int ways(vector<string>& pizza, int k) {
        const int mod = 1e9 + 7;
        int m = pizza.size(), n = pizza[0].size();
        vector<vector<vector<int>>> f(m, vector<vector<int>>(n, vector<int>(k, -1)));
        vector<vector<int>> s(m + 1, vector<int>(n + 1));
        for (int i = 1; i <= m; ++i) {
            for (int j = 1; j <= n; ++j) {
                int x = pizza[i - 1][j - 1] == 'A' ? 1 : 0;
                s[i][j] = s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1] + x;
            }
        }
        function<int(int, int, int)> dfs = [&](int i, int j, int k) -> int {
            if (k == 0) {
                return s[m][n] - s[i][n] - s[m][j] + s[i][j] > 0 ? 1 : 0;
            }
            if (f[i][j][k] != -1) {
                return f[i][j][k];
            }
            int ans = 0;
            for (int x = i + 1; x < m; ++x) {
                if (s[x][n] - s[i][n] - s[x][j] + s[i][j] > 0) {
                    ans = (ans + dfs(x, j, k - 1)) % mod;
                }
            }
            for (int y = j + 1; y < n; ++y) {
                if (s[m][y] - s[i][y] - s[m][j] + s[i][j] > 0) {
                    ans = (ans + dfs(i, y, k - 1)) % mod;
                }
            }
            return f[i][j][k] = ans;
        };
        return dfs(0, 0, k - 1);
    }
};