LeetCode #2596 — MEDIUM

Check Knight Tour Configuration

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

There is a knight on an n x n chessboard. In a valid configuration, the knight starts at the top-left cell of the board and visits every cell on the board exactly once.

You are given an n x n integer matrix grid consisting of distinct integers from the range [0, n * n - 1] where grid[row][col] indicates that the cell (row, col) is the grid[row][col]^th cell that the knight visited. The moves are 0-indexed.

Return true if grid represents a valid configuration of the knight's movements or false otherwise.

Note that a valid knight move consists of moving two squares vertically and one square horizontally, or two squares horizontally and one square vertically. The figure below illustrates all the possible eight moves of a knight from some cell.

Example 1:

Input: grid = [[0,11,16,5,20],[17,4,19,10,15],[12,1,8,21,6],[3,18,23,14,9],[24,13,2,7,22]]
Output: true
Explanation: The above diagram represents the grid. It can be shown that it is a valid configuration.

Example 2:

Input: grid = [[0,3,6],[5,8,1],[2,7,4]]
Output: false
Explanation: The above diagram represents the grid. The 8^th move of the knight is not valid considering its position after the 7^th move.

Constraints:

n == grid.length == grid[i].length
3 <= n <= 7
0 <= grid[row][col] < n * n
All integers in grid are unique.

Step 01

Brute Force Baseline

Problem summary: There is a knight on an n x n chessboard. In a valid configuration, the knight starts at the top-left cell of the board and visits every cell on the board exactly once. You are given an n x n integer matrix grid consisting of distinct integers from the range [0, n * n - 1] where grid[row][col] indicates that the cell (row, col) is the grid[row][col]th cell that the knight visited. The moves are 0-indexed. Return true if grid represents a valid configuration of the knight's movements or false otherwise. Note that a valid knight move consists of moving two squares vertically and one square horizontally, or two squares horizontally and one square vertically. The figure below illustrates all the possible eight moves of a knight from some cell.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

[[0,11,16,5,20],[17,4,19,10,15],[12,1,8,21,6],[3,18,23,14,9],[24,13,2,7,22]]

Example 2

[[0,3,6],[5,8,1],[2,7,4]]

Core Insight

What unlocks the optimal approach

It is enough to check if each move of the knight is valid.
Try all cases of the knight's movements to check if a move is valid.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2596: Check Knight Tour Configuration
class Solution {
    public boolean checkValidGrid(int[][] grid) {
        if (grid[0][0] != 0) {
            return false;
        }
        int n = grid.length;
        int[][] pos = new int[n * n][2];
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                pos[grid[i][j]] = new int[] {i, j};
            }
        }
        for (int i = 1; i < n * n; ++i) {
            int[] p1 = pos[i - 1];
            int[] p2 = pos[i];
            int dx = Math.abs(p1[0] - p2[0]);
            int dy = Math.abs(p1[1] - p2[1]);
            boolean ok = (dx == 1 && dy == 2) || (dx == 2 && dy == 1);
            if (!ok) {
                return false;
            }
        }
        return true;
    }
}

// Accepted solution for LeetCode #2596: Check Knight Tour Configuration
func checkValidGrid(grid [][]int) bool {
	if grid[0][0] != 0 {
		return false
	}
	n := len(grid)
	type pair struct{ x, y int }
	pos := make([]pair, n*n)
	for i, row := range grid {
		for j, x := range row {
			pos[x] = pair{i, j}
		}
	}
	for i := 1; i < n*n; i++ {
		p1, p2 := pos[i-1], pos[i]
		dx := abs(p1.x - p2.x)
		dy := abs(p1.y - p2.y)
		ok := (dx == 2 && dy == 1) || (dx == 1 && dy == 2)
		if !ok {
			return false
		}
	}
	return true
}

func abs(x int) int {
	if x < 0 {
		return -x
	}
	return x
}

# Accepted solution for LeetCode #2596: Check Knight Tour Configuration
class Solution:
    def checkValidGrid(self, grid: List[List[int]]) -> bool:
        if grid[0][0]:
            return False
        n = len(grid)
        pos = [None] * (n * n)
        for i in range(n):
            for j in range(n):
                pos[grid[i][j]] = (i, j)
        for (x1, y1), (x2, y2) in pairwise(pos):
            dx, dy = abs(x1 - x2), abs(y1 - y2)
            ok = (dx == 1 and dy == 2) or (dx == 2 and dy == 1)
            if not ok:
                return False
        return True

// Accepted solution for LeetCode #2596: Check Knight Tour Configuration
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2596: Check Knight Tour Configuration
// class Solution {
//     public boolean checkValidGrid(int[][] grid) {
//         if (grid[0][0] != 0) {
//             return false;
//         }
//         int n = grid.length;
//         int[][] pos = new int[n * n][2];
//         for (int i = 0; i < n; ++i) {
//             for (int j = 0; j < n; ++j) {
//                 pos[grid[i][j]] = new int[] {i, j};
//             }
//         }
//         for (int i = 1; i < n * n; ++i) {
//             int[] p1 = pos[i - 1];
//             int[] p2 = pos[i];
//             int dx = Math.abs(p1[0] - p2[0]);
//             int dy = Math.abs(p1[1] - p2[1]);
//             boolean ok = (dx == 1 && dy == 2) || (dx == 2 && dy == 1);
//             if (!ok) {
//                 return false;
//             }
//         }
//         return true;
//     }
// }

// Accepted solution for LeetCode #2596: Check Knight Tour Configuration
function checkValidGrid(grid: number[][]): boolean {
    if (grid[0][0] !== 0) {
        return false;
    }
    const n = grid.length;
    const pos = Array.from(new Array(n * n), () => new Array(2).fill(0));
    for (let i = 0; i < n; ++i) {
        for (let j = 0; j < n; ++j) {
            pos[grid[i][j]] = [i, j];
        }
    }
    for (let i = 1; i < n * n; ++i) {
        const p1 = pos[i - 1];
        const p2 = pos[i];
        const dx = Math.abs(p1[0] - p2[0]);
        const dy = Math.abs(p1[1] - p2[1]);
        const ok = (dx === 1 && dy === 2) || (dx === 2 && dy === 1);
        if (!ok) {
            return false;
        }
    }
    return true;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n^2)

Space

O(n^2)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.