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You are given an n x n chessboard and a knight. Initially, the knight is positioned at coordinates (row, column). You need to find the probability that the knight remains on the chessboard after making exactly k moves.

The knight moves in an L-shape: it can move two squares in one direction (horizontal or vertical) and then one square in the perpendicular direction. Given the initial position of the knight, determine the probability that after k moves, the knight will still be on the chessboard.

The movements can be in any of the following directions:

Clarifying Questions

  1. What are the constraints on n, k, row, and column?
    • Constraints:
      • 1 <= n <= 25
      • 0 <= k <= 100
      • 0 <= row, column < n
  2. Are there any constraints on the type of input (all integers)?
    • Yes, all inputs are integers.
  3. Are we supposed to consider the chessboard as 0-indexed or 1-indexed?
    • The chessboard is considered 0-indexed.

Strategy

  1. Define Movement: All possible moves of a knight can be captured in a list of tuples.
  2. Dynamic Programming Approach:
    • We’ll use a DP table where dp[i][j][m] defines the probability of the knight being at position (i, j) after m moves.
    • Initialize dp[row][column][0] = 1 because the knight starts at the initial position with probability 1.
    • For each move from 1 to k, update the possible positions based on previous moves while ensuring the new positions are within the board.
  3. Sum Probabilities: After k moves, sum up the probabilities of all positions to get the result.

Code

def knightProbability(n: int, k: int, row: int, column: int) -> float:
    # Directions a knight can move
    directions = [(2, 1), (1, 2), (-1, 2), (-2, 1), (-2, -1), (-1, -2), (1, -2), (2, -1)]
    
    # DP table initialized with zeros
    dp = [[[0 for _ in range(k + 1)] for _ in range(n)] for _ in range(n)]
    
    # The initial position with 0 moves: probability is 1
    dp[row][column][0] = 1
    
    # Iterate over number of moves
    for move in range(1, k + 1):
        for r in range(n):
            for c in range(n):
                # If there's a probability to be at (r, c) with move-1
                if dp[r][c][move - 1] > 0:
                    for dr, dc in directions:
                        new_r, new_c = r + dr, c + dc
                        # Check if the new position is within bounds
                        if 0 <= new_r < n and 0 <= new_c < n:
                            dp[new_r][new_c][move] += dp[r][c][move - 1] / 8.0
    
    # Calculate the total probability of the knight being on the board after k moves.
    result = 0
    for r in range(n):
        for c in range(n):
            result += dp[r][c][k]
    
    return result

# Example Usage
n = 3
k = 2
row = 0
column = 0
print(knightProbability(n, k, row, column))  # Output: The probability the knight is on the board after 2 moves

Time Complexity

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