<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:
        if maxMove == 0:
            return 0
        ## Number of N(k, i, j) = N(k-1, i+1, j) + N(k-1, i-1, j) + N(k-1, i, j+1) + 
        ## N(k-1, i, j-1)

        ## We can hence proceed bottom up 1 --> 2 ---> .... ---> K

        ## To avoid painfull boundary condidions, once we go outside the border we place there a zero, as once I am out I have no other options
        dp = [[1 for _ in range(n+2)] for _ in range(m+2)]
        dp_aux = [[1 for _ in range(n+2)] for _ in range(m+2)] ## need an auxiliary solution

        moves = [(1,0), (0,1), (-1, 0), (0,-1)] 

        for r in range(1, m+1):
            for c in range(1, n+1):
                aggr = 4
                for move in moves:
                    raux = r + move[0]
                    caux = c + move[1]
                    if 1 <= raux < m+1 and 1<= caux < n+1:
                        aggr += (-1)
                dp[r][c] = aggr
                dp_aux[r][c] = aggr
        
        print(dp)
        if maxMove == 1:
            return dp[startRow+1][startColumn+1]

        for k in range(2, maxMove+1):
            for r in range(1, m+1):
                for c in range(1, n+1):
                    dp[r][c] = dp_aux[r+1][c] + dp_aux[r-1][c] + dp_aux[r][c+1] + dp_aux[r][c-1]
            
            ## Update the auxiliary matrix
            for r in range(1,m+1):
                for c in range(1, n+1):
                    dp_aux[r][c] = dp[r][c]

        print(dp)
        
        return (dp[startRow+1][startColumn+1])% (10**9 + 7)
                    


        
17300