<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 {
    const int MOD = 1e9 + 7;
public:
    int findPaths(int m, int n, int maxMove, int startRow, int startColumn) {
        vector<pair<int,int>> dirs = {{1,0},{-1,0},{0,1},{0,-1}};
        vector<vector<int>> dp_curr(m, vector<int>(n, 0));
        vector<vector<int>> dp_prev(m, vector<int>(n, 0));

        int result = 0;
        for (int k = 1; k <= maxMove; ++k) {
            for (int i = 0; i < m; ++i) {
                for (int j = 0; j < n; ++j) {
                    long long sum = 0;
                    for (auto& dir : dirs) {
                        int X = i + dir.first;
                        int Y = j + dir.second;
                        if (X < 0 || X >= m || Y < 0 || Y >= n) {
                            sum = (sum + 1) % MOD;
                        } else {
                            sum = (sum + dp_prev[X][Y]) % MOD;
                        }
                    }
                    dp_curr[i][j] = sum;
                }
            }
            dp_prev=dp_curr ; 
        }
        
        return dp_prev[startRow][startColumn];
    }
};

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