Leetcode 576. Out of Boundary Paths

There is an m by n grid with a ball. Given the start coordinate (i,j) of the ball, you can move the ball to adjacent cell or cross the grid boundary in four directions (up, down, left, right). However, you can at most move N times. Find out the number of paths to move the ball out of grid boundary. The answer may be very large, return it after mod 109 + 7.

Example 1:

Input:m = 2, n = 2, N = 2, i = 0, j = 0
Output: 6
Explanation:

Example 2:

Input:m = 1, n = 3, N = 3, i = 0, j = 1
Output: 12
Explanation:

Note:

1. Once you move the ball out of boundary, you cannot move it back.
2. The length and height of the grid is in range [1,50].
3. N is in range [0,50].

Analysis:
This problem is a DP problem. Define dp[N + 1][m][n], where dp[k][row][col] means the number of paths to move from (row, col) to out of boundary with k steps. 
So the transist function would be 
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].

And the final state would be dp[k][i][j];

Code (Java):

class Solution {
    public int findPaths(int m, int n, int N, int i, int j) {
        if (N == 0) {
            return 0;
        }
        
        int num = (int)Math.pow(10, 9) + 7;
        
        int[][] prev = new int[m][n];
        
        for (int k = 1; k <= N; k++) {
            int[][] curr = new int[m][n];
            for (int row = 0; row < m; row++) {
                for (int col = 0; col < n; col++) {
                    curr[row][col] = (curr[row][col] + (row == 0 ? 1 : prev[row - 1][col])) % num;
                    curr[row][col] = (curr[row][col] + (row == m - 1 ? 1 : prev[row + 1][col])) % num;
                    curr[row][col] = (curr[row][col] + (col == 0 ? 1 : prev[row][col - 1])) % num;
                    curr[row][col] = (curr[row][col] + (col == n - 1 ? 1 : prev[row][col + 1])) % num;
                }
            }
            
            prev = curr;
        }
        
        return (int)(prev[i][j] % num);
    }
}