# Solving Leetcode Interviews in Seconds with AI: Out of Boundary Paths


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "576" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> There is an m x n grid with a ball. The ball is initially at the position [startRow, startColumn]. 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 at most maxMove moves to the ball. Given the five integers m, n, maxMove, startRow, startColumn, return the number of paths to move the ball out of the grid boundary. Since the answer can be very large, return it modulo 109 + 7.   Example 1:   Input: m = 2, n = 2, maxMove = 2, startRow = 0, startColumn = 0 Output: 6  Example 2:   Input: m = 1, n = 3, maxMove = 3, startRow = 0, startColumn = 1 Output: 12    Constraints:  1 <= m, n <= 50 0 <= maxMove <= 50 0 <= startRow < m 0 <= startColumn < n  

	# Explanation
	Here's a solution to the problem, along with an explanation of the approach and complexity analysis.

*   **Dynamic Programming:** Use a 3D DP table `dp[moves][row][col]` to store the number of paths to reach cell `(row, col)` in `moves` steps.
*   **Base Case:** For each cell on the boundary, increment the path count when `moves` is equal to 1 from the adjacent internal cell.
*   **Iteration:** Iterate through the number of moves, rows, and columns, updating the DP table based on the number of paths from the adjacent cells in the previous move.

*   **Runtime Complexity:** O(maxMove * m * n), **Storage Complexity:** O(maxMove * m * n)

	
	# Code
	```python
	def findPaths(m: int, n: int, maxMove: int, startRow: int, startColumn: int) -> int:
    """
    Calculates the number of paths to move the ball out of the grid boundary.

    Args:
        m: The number of rows in the grid.
        n: The number of columns in the grid.
        maxMove: The maximum number of moves allowed.
        startRow: The starting row of the ball.
        startColumn: The starting column of the ball.

    Returns:
        The number of paths to move the ball out of the grid boundary, modulo 10^9 + 7.
    """

    dp = [[[0] * n for _ in range(m)] for _ in range(maxMove + 1)]
    MOD = 10**9 + 7
    directions = [(0, 1), (0, -1), (1, 0), (-1, 0)]
    
    # Initialize DP table for 0 moves (base case)
    # dp[0][startRow][startColumn] = 1  # No need for initialization, start from the beginning

    for moves in range(1, maxMove + 1):
        for row in range(m):
            for col in range(n):
                for dr, dc in directions:
                    prev_row, prev_col = row - dr, col - dc
                    if 0 <= prev_row < m and 0 <= prev_col < n:
                        dp[moves][row][col] = (dp[moves][row][col] + dp[moves - 1][prev_row][prev_col]) % MOD
                    else:
                        dp[moves][row][col] = (dp[moves][row][col] + 1) % MOD

    return dp[maxMove][startRow][startColumn]
	```