Leetcode #688: Knight Probability in Chessboard
In this guide, we solve Leetcode #688 Knight Probability in Chessboard in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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).
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Dynamic Programming
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
Example
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.
Python Solution
class Solution:
def knightProbability(self, n: int, k: int, row: int, column: int) -> float:
f = [[[0] * n for _ in range(n)] for _ in range(k + 1)]
for i in range(n):
for j in range(n):
f[0][i][j] = 1
for h in range(1, k + 1):
for i in range(n):
for j in range(n):
for a, b in pairwise((-2, -1, 2, 1, -2, 1, 2, -1, -2)):
x, y = i + a, j + b
if 0 <= x < n and 0 <= y < n:
f[h][i][j] += f[h - 1][x][y] / 8
return f[k][row][column]
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.