<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:
        return self.fun2(m, n, maxMove, startRow, startColumn)

    def fun1(self, m: int, n: int, maxMove: int, startRow: int, startColumn: int) -> int:

        dp = [[[0] * (n + 2) for _ in range(m + 2)] for _ in range(maxMove + 1)]
        for i in range(m + 2):
            dp[0][i][0] = 1
            dp[0][i][-1] = 1
        for i in range(n + 2):
            dp[0][0][i] = 1
            dp[0][-1][i] = 1

        for k in range(1, maxMove + 1):
            for i in range(1, m + 1):
                for j in range(1, n + 1):
                    dp[k][i][j] = (
                        dp[k - 1][i - 1][j]
                        + dp[k - 1][i + 1][j]
                        + dp[k - 1][i][j - 1]
                        + dp[k - 1][i][j + 1]
                    )

        return sum(dp[i][startRow + 1][startColumn + 1] for i in range(1, maxMove + 1)) % (
            10**9 + 7
        )

    def fun2(self, m: int, n: int, maxMove: int, startRow: int, startColumn: int) -> int:
        dp = [[[-1] * (n + 1) for _ in range(m + 1)] for _ in range(maxMove + 1)]

        @functools.lru_cache()
        def dfs(i, j, move):
            if move < 0:
                return 0
            if i < 0 or i >= m or j < 0 or j >= n:
                return 1

            if dp[move][i][j] != -1:
                return dp[move][i][j]

            dp[move][i][j] = (
                dfs(i - 1, j, move - 1)
                + dfs(i + 1, j, move - 1)
                + dfs(i, j - 1, move - 1)
                + dfs(i, j + 1, move - 1)
            )
            return dp[move][i][j]

        return dfs(startRow, startColumn, maxMove) % (10**9 + 7)


19800