Knight Probability in Chessboard

Tested Python solution for LeetCode 688 with 20 pytest cases. Generate a practice environment with lcpy.

LeetCode 688, Medium. Topics: Dynamic Programming. View on LeetCode.

Generate this problem as a practice environment: tested reference solution, 20 parametrized pytest cases, and a playground notebook:

lcpy gen -n 688   # by problem number
lcpy gen -s knight_probability_in_chessboard   # by problem name

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Problem

On an n x n chessboard, a knight starts at the cell (row, column) and attempts to make exactly k moves. The rows and columns are 0-indexed, so the top-left cell is (0, 0), and the bottom-right cell is (n - 1, n - 1).

A chess knight has eight possible moves it can make, as illustrated below. Each move is two cells in a cardinal direction, then one cell in an orthogonal direction.

Each time the knight is to move, it chooses one of eight possible moves uniformly at random (even if the piece would go off the chessboard) and moves there.

The knight continues moving until it has made exactly k moves or has moved off the chessboard.

Return the probability that the knight remains on the board after it has stopped moving.

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Examples

Input: n = 3, k = 2, row = 0, column = 0
Output: 0.06250
Explanation: There are two moves (to (1,2), (2,1)) that will keep the knight on the board.
From each of those positions, there are also two moves that will keep the knight on the board.
The total probability the knight stays on the board is 0.0625.

Input: n = 1, k = 0, row = 0, column = 0
Output: 1.00000

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Constraints

• 1 <= n <= 25
• 0 <= k <= 100
• 0 <= row, column <= n - 1

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Solution

Reference implementation from solution.py on GitHub, full suite in test_solution.py:

class Solution:
    # Time: O(k * n^2)
    # Space: O(n^2)
    def knight_probability(self, n: int, k: int, row: int, column: int) -> float:
        moves = ((1, 2), (2, 1), (2, -1), (1, -2), (-1, -2), (-2, -1), (-2, 1), (-1, 2))
        prob = [[0.0] * n for _ in range(n)]
        prob[row][column] = 1.0
        for _ in range(k):
            nxt = [[0.0] * n for _ in range(n)]
            for r in range(n):
                for c in range(n):
                    if prob[r][c] == 0.0:
                        continue
                    share = prob[r][c] / 8.0
                    for dr, dc in moves:
                        nr, nc = r + dr, c + dc
                        if 0 <= nr < n and 0 <= nc < n:
                            nxt[nr][nc] += share
            prob = nxt
        return sum(map(sum, prob))

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Complexity

Time	Space
O(k * n^2)	O(n^2)

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