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  1. Out of Boundary Paths

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 boundary) in one step. You can move the ball up, down, left, or right. Calculate the number of paths to move the ball out of grid boundary in exactly maxMove steps. Since the answer can be very large, return it modulo 10^9 + 7.

Example

Input: m = 2, n = 2, maxMove = 2, startRow = 0, startColumn = 0
Output: 6
Explanation: There are 6 ways to move the ball out of boundary in 2 moves.

Clarifying Questions

  1. Can the ball move zero steps and still be considered moving out of bounds?
    • No, the ball needs to move; zero steps would mean no movement.
  2. How large can m and n be?
    • Typically, these values can range up to 50.
  3. Should paths that lead to already out-of-bound positions be considered repeatedly?
    • Only if they occur within the exact number of allowed moves.

Strategy

We can solve this problem using dynamic programming. The idea is to use a 3D DP table dp[i][j][k] where:

Each cell will keep track of the number of ways to reach out of the boundary from that point with k moves left. The base condition is when k = 0, indicating no moves left and thus zero out-bound paths from any cell.

We’ll iterate through the grid for each move, updating the DP table by aggregating the possible steps from the current cell to neighboring cells and checking boundary conditions.

Code

def findPaths(m, n, maxMove, startRow, startColumn):
    MODULO = 10**9 + 7
    
    # Initialize the DP table
    dp = [[[0 for _ in range(maxMove + 1)] for _ in range(n + 2)] for _ in range(m + 2)]
    
    # Increment all boundaries
    for i in range(1, m + 1):
        dp[i][0][0] = 1
        dp[i][n + 1][0] = 1
    for j in range(1, n + 1):
        dp[0][j][0] = 1
        dp[m + 1][j][0] = 1

    for moves in range(1, maxMove + 1):
        for i in range(1, m + 1):
            for j in range(1, n + 1):
                dp[i][j][moves] = (dp[i - 1][j][moves - 1] + dp[i + 1][j][moves - 1] + 
                                   dp[i][j - 1][moves - 1] + dp[i][j + 1][moves - 1]) % MODULO
    
    return sum(dp[startRow + 1][startColumn + 1][moves] for moves in range(1, maxMove + 1)) % MODULO

# Example usage:
m = 2
n = 2
maxMove = 2
startRow = 0
startColumn = 0
print(findPaths(m, n, maxMove, startRow, startColumn))  # Output: 6

Strategy Explanation

  1. Initialization:
    • Create a DP table of size (m + 2) x (n + 2) x (maxMove + 1) to store the number of paths for cells including boundaries, initialized to zero.
    • Set the boundaries as base cases (dp table around the grid).
  2. DP Transition:
    • For each position (i, j) and each number of moves remaining k, sum the paths from the four adjacent positions with one less move.
  3. Boundary Inclusion:
    • Ensure boundary conditions are handled by initializing boundaries in the DP array and updating their influence in each move.
  4. Result Calculation:
    • Sum the results for exactly maxMove steps from the start position.

Time Complexity

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