<p>There is an <code>m x n</code> grid with a ball. The ball is initially at the position <code>[startRow, startColumn]</code>. You are allowed to move the ball to one of the four adjacent cells in the grid (possibly out of the grid crossing the grid boundary). You can apply <strong>at most</strong> <code>maxMove</code> moves to the ball.</p>

<p>Given the five integers <code>m</code>, <code>n</code>, <code>maxMove</code>, <code>startRow</code>, <code>startColumn</code>, return the number of paths to move the ball out of the grid boundary. Since the answer can be very large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/04/28/out_of_boundary_paths_1.png" style="width: 500px; height: 296px;" />
<pre>
<strong>Input:</strong> m = 2, n = 2, maxMove = 2, startRow = 0, startColumn = 0
<strong>Output:</strong> 6
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/04/28/out_of_boundary_paths_2.png" style="width: 500px; height: 293px;" />
<pre>
<strong>Input:</strong> m = 1, n = 3, maxMove = 3, startRow = 0, startColumn = 1
<strong>Output:</strong> 12
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= m, n &lt;= 50</code></li>
	<li><code>0 &lt;= maxMove &lt;= 50</code></li>
	<li><code>0 &lt;= startRow &lt; m</code></li>
	<li><code>0 &lt;= startColumn &lt; n</code></li>
</ul>

class Solution:
    
    def findPaths(self, m: int, n: int, maxMove: int, startRow: int, startColumn: int) -> int:

        self.m, self.n = m, n 
        self.max_move = maxMove 
        self.moves = [(-1, 0), (1, 0), (0, -1), (0, 1)]
        self.cache = {}

        self.MOD = 10**9 + 7
        return self.count_paths(startRow, startColumn, 0)

    def count_paths(self, i: int, j: int, cur_move: int) -> int: 
        if cur_move > self.max_move: 
            return 0 

        if i < 0 or j < 0 or i >= self.m or j >= self.n: 
            return 1

        if (i, j, cur_move) in self.cache: 
            return self.cache[(i, j, cur_move)]

        # start moving 
        paths = 0 
        for (a, b) in self.moves: 
            paths = (paths + self.count_paths(a + i, b + j, cur_move + 1)) % self.MOD

        self.cache[(i, j, cur_move)] = paths 
        
        return paths 
        
        
    

    
18900