How do you solve Check Knight Tour Configuration on LeetCode?

Check Knight Tour Configuration (LeetCode #2596) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n²) time complexity and O(n) space complexity. Key insight: Understanding knight moves is crucial.

What is the brute force solution for Check Knight Tour Configuration?

The brute force approach for Check Knight Tour Configuration is Brute Force, with O(n²) time complexity and O(1) space complexity. We will check each move of the knight one by one to see if it follows the valid knight move rules. This involves iterating through the grid and verifying the position of the knight at each step. The optimal Optimal Solution approach improves this to O(n²).

What are the interview tips for Check Knight Tour Configuration?

Clarify the knight's move rules before starting. Think about edge cases, like the smallest grid. Explain your thought process as you code.

What are common mistakes when solving Check Knight Tour Configuration?

Not checking all possible knight moves. Confusing row and column indices.

#2596

Check Knight Tour Configuration

Medium

ArrayDepth-First SearchBreadth-First SearchMatrixSimulationHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n²)
Space	O(1)	O(n)

💡

Intuition

Time O(n²)Space O(n)

We can optimize the brute-force approach by directly validating the knight's moves using a set of pre-computed valid positions. This allows us to quickly check if each move is valid without iterating through all possible moves.

⚙️

Algorithm

2 steps

1Step 1: Create a mapping of each number to its position on the grid.
2Step 2: For each number from 0 to n*n - 1, check if the move to the next number is valid using the pre-computed positions.

solution.py10 lines

1def checkKnightTour(grid):
2    n = len(grid)
3    moves = [(2, 1), (2, -1), (-2, 1), (-2, -1), (1, 2), (1, -2), (-1, 2), (-1, -2)]
4    position = {grid[r][c]: (r, c) for r in range(n) for c in range(n)}
5    for i in range(n * n - 1):
6        r1, c1 = position[i]
7        r2, c2 = position[i + 1]
8        if not any((r1 + dr == r2 and c1 + dc == c2) for dr, dc in moves):
9            return False
10    return True

ℹ

Complexity note: We still iterate through the grid to build the position mapping, which takes O(n²). However, we use a dictionary to access positions in constant time, making the move validation efficient.

1Understanding knight moves is crucial.
2Mapping positions allows for faster validation.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.