Leetcode #2664: The Knight’s Tour
In this guide, we solve Leetcode #2664 The Knight’s Tour 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
Given two positive integers m and n which are the height and width of a 0-indexed 2D-array board, a pair of positive integers (r, c) which is the starting position of the knight on the board. Your task is to find an order of movements for the knight, in a manner that every cell of the board gets visited exactly once (the starting cell is considered visited and you shouldn't visit it again).
Quick Facts
- Difficulty: Medium
- Premium: Yes
- Tags: Array, Backtracking, Matrix
Intuition
We must explore combinations of choices, but many branches can be pruned early.
Backtracking enumerates valid candidates while keeping the search space under control.
Approach
Use DFS to build candidates step by step, and backtrack when constraints are violated.
Pruning keeps the exploration practical for typical constraints.
Steps:
- Define the decision tree.
- DFS through choices and backtrack.
- Prune invalid paths early.
Example
Input: m = 1, n = 1, r = 0, c = 0
Output: [[0]]
Explanation: There is only 1 cell and the knight is initially on it so there is only a 0 inside the 1x1 grid.
Python Solution
class Solution:
def tourOfKnight(self, m: int, n: int, r: int, c: int) -> List[List[int]]:
def dfs(i: int, j: int):
nonlocal ok
if g[i][j] == m * n - 1:
ok = True
return
for a, b in pairwise((-2, -1, 2, 1, -2, 1, 2, -1, -2)):
x, y = i + a, j + b
if 0 <= x < m and 0 <= y < n and g[x][y] == -1:
g[x][y] = g[i][j] + 1
dfs(x, y)
if ok:
return
g[x][y] = -1
g = [[-1] * n for _ in range(m)]
g[r][c] = 0
ok = False
dfs(r, c)
return g
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.