Out of Boundary Paths
<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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0500-0599/0576.Out%20of%20Boundary%20Paths/images/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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0500-0599/0576.Out%20of%20Boundary%20Paths/images/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>
Dynamic Programming
class Solution {
    private int m, n;
    private Integer[][][] f;
    private final int mod = (int) 1e9 + 7;

    public int findPaths(int m, int n, int maxMove, int startRow, int startColumn) {
        this.m = m;
        this.n = n;
        f = new Integer[m][n][maxMove + 1];
        return dfs(startRow, startColumn, maxMove);
    }

    private int dfs(int i, int j, int k) {
        if (i < 0 || i >= m || j < 0 || j >= n) {
            return k >= 0 ? 1 : 0;
        }
        if (k <= 0) {
            return 0;
        }
        if (f[i][j][k] != null) {
            return f[i][j][k];
        }
        int ans = 0;
        final int[] dirs = {-1, 0, 1, 0, -1};
        for (int d = 0; d < 4; ++d) {
            int x = i + dirs[d], y = j + dirs[d + 1];
            ans = (ans + dfs(x, y, k - 1)) % mod;
        }
        return f[i][j][k] = ans;
    }
}
