LeetCode #1210 — HARD

Minimum Moves to Reach Target with Rotations

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

In an n*n grid, there is a snake that spans 2 cells and starts moving from the top left corner at (0, 0) and (0, 1). The grid has empty cells represented by zeros and blocked cells represented by ones. The snake wants to reach the lower right corner at (n-1, n-2) and (n-1, n-1).

In one move the snake can:

Move one cell to the right if there are no blocked cells there. This move keeps the horizontal/vertical position of the snake as it is.
Move down one cell if there are no blocked cells there. This move keeps the horizontal/vertical position of the snake as it is.
Rotate clockwise if it's in a horizontal position and the two cells under it are both empty. In that case the snake moves from (r, c) and (r, c+1) to (r, c) and (r+1, c).
Rotate counterclockwise if it's in a vertical position and the two cells to its right are both empty. In that case the snake moves from (r, c) and (r+1, c) to (r, c) and (r, c+1).

Return the minimum number of moves to reach the target.

If there is no way to reach the target, return -1.

Example 1:

Input: grid = [[0,0,0,0,0,1],
               [1,1,0,0,1,0],
               [0,0,0,0,1,1],
               [0,0,1,0,1,0],
               [0,1,1,0,0,0],
               [0,1,1,0,0,0]]
Output: 11
Explanation:
One possible solution is [right, right, rotate clockwise, right, down, down, down, down, rotate counterclockwise, right, down].

Example 2:

Input: grid = [[0,0,1,1,1,1],
               [0,0,0,0,1,1],
               [1,1,0,0,0,1],
               [1,1,1,0,0,1],
               [1,1,1,0,0,1],
               [1,1,1,0,0,0]]
Output: 9

Constraints:

2 <= n <= 100
0 <= grid[i][j] <= 1
It is guaranteed that the snake starts at empty cells.

Step 01

Brute Force Baseline

Problem summary: In an n*n grid, there is a snake that spans 2 cells and starts moving from the top left corner at (0, 0) and (0, 1). The grid has empty cells represented by zeros and blocked cells represented by ones. The snake wants to reach the lower right corner at (n-1, n-2) and (n-1, n-1). In one move the snake can: Move one cell to the right if there are no blocked cells there. This move keeps the horizontal/vertical position of the snake as it is. Move down one cell if there are no blocked cells there. This move keeps the horizontal/vertical position of the snake as it is. Rotate clockwise if it's in a horizontal position and the two cells under it are both empty. In that case the snake moves from (r, c) and (r, c+1) to (r, c) and (r+1, c). Rotate counterclockwise if it's in a vertical position and the two cells to its right are both empty. In that case the snake moves from (r, c) and (r+1, c)

Baseline thinking

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

Pattern signal: Array

Example 1

[[0,0,0,0,0,1],[1,1,0,0,1,0],[0,0,0,0,1,1],[0,0,1,0,1,0],[0,1,1,0,0,0],[0,1,1,0,0,0]]

Example 2

[[0,0,1,1,1,1],[0,0,0,0,1,1],[1,1,0,0,0,1],[1,1,1,0,0,1],[1,1,1,0,0,1],[1,1,1,0,0,0]]

Step 02

Core Insight

What unlocks the optimal approach

Use BFS to find the answer.
The state of the BFS is the position (x, y) along with a binary value that specifies if the position is horizontal or vertical.

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

Largest constraint values

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 #1210: Minimum Moves to Reach Target with Rotations
class Solution {
    private int n;
    private int[][] grid;
    private boolean[][] vis;
    private Deque<int[]> q = new ArrayDeque<>();

    public int minimumMoves(int[][] grid) {
        this.grid = grid;
        n = grid.length;
        vis = new boolean[n * n][2];
        int[] target = {n * n - 2, n * n - 1};
        q.offer(new int[] {0, 1});
        vis[0][0] = true;
        int ans = 0;
        while (!q.isEmpty()) {
            for (int k = q.size(); k > 0; --k) {
                var p = q.poll();
                if (p[0] == target[0] && p[1] == target[1]) {
                    return ans;
                }
                int i1 = p[0] / n, j1 = p[0] % n;
                int i2 = p[1] / n, j2 = p[1] % n;
                // 尝试向右平移（保持身体水平/垂直状态）
                move(i1, j1 + 1, i2, j2 + 1);
                // 尝试向下平移（保持身体水平/垂直状态）
                move(i1 + 1, j1, i2 + 1, j2);
                // 当前处于水平状态，且 grid[i1 + 1][j2] 无障碍，尝试顺时针旋转90°
                if (i1 == i2 && i1 + 1 < n && grid[i1 + 1][j2] == 0) {
                    move(i1, j1, i1 + 1, j1);
                }
                // 当前处于垂直状态，且 grid[i2][j1 + 1] 无障碍，尝试逆时针旋转90°
                if (j1 == j2 && j1 + 1 < n && grid[i2][j1 + 1] == 0) {
                    move(i1, j1, i1, j1 + 1);
                }
            }
            ++ans;
        }
        return -1;
    }

    private void move(int i1, int j1, int i2, int j2) {
        if (i1 >= 0 && i1 < n && j1 >= 0 && j1 < n && i2 >= 0 && i2 < n && j2 >= 0 && j2 < n) {
            int a = i1 * n + j1, b = i2 * n + j2;
            int status = i1 == i2 ? 0 : 1;
            if (!vis[a][status] && grid[i1][j1] == 0 && grid[i2][j2] == 0) {
                q.offer(new int[] {a, b});
                vis[a][status] = true;
            }
        }
    }
}

// Accepted solution for LeetCode #1210: Minimum Moves to Reach Target with Rotations
func minimumMoves(grid [][]int) int {
	n := len(grid)
	type pair struct{ a, b int }
	target := pair{n*n - 2, n*n - 1}
	q := []pair{pair{0, 1}}
	vis := make([][2]bool, n*n)
	vis[0][0] = true

	move := func(i1, j1, i2, j2 int) {
		if i1 >= 0 && i1 < n && j1 >= 0 && j1 < n && i2 >= 0 && i2 < n && j2 >= 0 && j2 < n {
			a, b := i1*n+j1, i2*n+j2
			status := 1
			if i1 == i2 {
				status = 0
			}
			if !vis[a][status] && grid[i1][j1] == 0 && grid[i2][j2] == 0 {
				q = append(q, pair{a, b})
				vis[a][status] = true
			}
		}
	}

	ans := 0
	for len(q) > 0 {
		for k := len(q); k > 0; k-- {
			p := q[0]
			q = q[1:]
			if p == target {
				return ans
			}
			a, b := p.a, p.b
			i1, j1 := a/n, a%n
			i2, j2 := b/n, b%n
			// 尝试向右平移（保持身体水平/垂直状态）
			move(i1, j1+1, i2, j2+1)
			// 尝试向下平移（保持身体水平/垂直状态）
			move(i1+1, j1, i2+1, j2)
			// 当前处于水平状态，且 grid[i1 + 1][j2] 无障碍，尝试顺时针旋转90°
			if i1 == i2 && i1+1 < n && grid[i1+1][j2] == 0 {
				move(i1, j1, i1+1, j1)
			}
			// 当前处于垂直状态，且 grid[i2][j1 + 1] 无障碍，尝试逆时针旋转90°
			if j1 == j2 && j1+1 < n && grid[i2][j1+1] == 0 {
				move(i1, j1, i1, j1+1)
			}
		}
		ans++
	}
	return -1
}

# Accepted solution for LeetCode #1210: Minimum Moves to Reach Target with Rotations
class Solution:
    def minimumMoves(self, grid: List[List[int]]) -> int:
        def move(i1, j1, i2, j2):
            if 0 <= i1 < n and 0 <= j1 < n and 0 <= i2 < n and 0 <= j2 < n:
                a, b = i1 * n + j1, i2 * n + j2
                status = 0 if i1 == i2 else 1
                if (a, status) not in vis and grid[i1][j1] == 0 and grid[i2][j2] == 0:
                    q.append((a, b))
                    vis.add((a, status))

        n = len(grid)
        target = (n * n - 2, n * n - 1)
        q = deque([(0, 1)])
        vis = {(0, 0)}
        ans = 0
        while q:
            for _ in range(len(q)):
                a, b = q.popleft()
                if (a, b) == target:
                    return ans
                i1, j1 = a // n, a % n
                i2, j2 = b // n, b % n
                # 尝试向右平移（保持身体水平/垂直状态）
                move(i1, j1 + 1, i2, j2 + 1)
                # 尝试向下平移（保持身体水平/垂直状态）
                move(i1 + 1, j1, i2 + 1, j2)
                # 当前处于水平状态，且 grid[i1 + 1][j2] 无障碍，尝试顺时针旋转90°
                if i1 == i2 and i1 + 1 < n and grid[i1 + 1][j2] == 0:
                    move(i1, j1, i1 + 1, j1)
                # 当前处于垂直状态，且 grid[i2][j1 + 1] 无障碍，尝试逆时针旋转90°
                if j1 == j2 and j1 + 1 < n and grid[i2][j1 + 1] == 0:
                    move(i1, j1, i1, j1 + 1)
            ans += 1
        return -1

// Accepted solution for LeetCode #1210: Minimum Moves to Reach Target with Rotations
struct Solution;

use std::collections::HashSet;
use std::collections::VecDeque;

type State = (usize, usize, bool);

impl Solution {
    fn minimum_moves(grid: Vec<Vec<i32>>) -> i32 {
        let n = grid.len();
        let mut visited: HashSet<State> = HashSet::new();
        let start = (0, 0, false);
        let finish = (n - 1, n - 2, false);
        visited.insert(start);
        let mut queue: VecDeque<State> = VecDeque::new();
        queue.push_back(start);

        let mut res = 0;
        while !queue.is_empty() {
            let m = queue.len();
            for _ in 0..m {
                let state = queue.pop_front().unwrap();
                if state == finish {
                    return res;
                }
                let i = state.0;
                let j = state.1;
                let d = state.2;
                let right = (i, j + 1, d);
                let down = (i + 1, j, d);
                let rotate = (i, j, !d);
                if d {
                    if j + 1 < n && grid[i][j + 1] == 0 && grid[i + 1][j + 1] == 0 {
                        if visited.insert(right) {
                            queue.push_back(right);
                        }
                        if visited.insert(rotate) {
                            queue.push_back(rotate);
                        }
                    }
                    if i + 2 < n && grid[i + 2][j] == 0 && visited.insert(down) {
                        queue.push_back(down);
                    }
                } else {
                    if i + 1 < n && grid[i + 1][j] == 0 && grid[i + 1][j + 1] == 0 {
                        if visited.insert(down) {
                            queue.push_back(down);
                        }
                        if visited.insert(rotate) {
                            queue.push_back(rotate);
                        }
                    }
                    if j + 2 < n && grid[i][j + 2] == 0 && visited.insert(right) {
                        queue.push_back(right);
                    }
                }
            }
            res += 1;
        }
        -1
    }
}

#[test]
fn test() {
    let grid = vec_vec_i32![
        [0, 0, 0, 0, 0, 1],
        [1, 1, 0, 0, 1, 0],
        [0, 0, 0, 0, 1, 1],
        [0, 0, 1, 0, 1, 0],
        [0, 1, 1, 0, 0, 0],
        [0, 1, 1, 0, 0, 0]
    ];
    let res = 11;
    assert_eq!(Solution::minimum_moves(grid), res);
    let grid = vec_vec_i32![
        [0, 0, 1, 1, 1, 1],
        [0, 0, 0, 0, 1, 1],
        [1, 1, 0, 0, 0, 1],
        [1, 1, 1, 0, 0, 1],
        [1, 1, 1, 0, 0, 1],
        [1, 1, 1, 0, 0, 0]
    ];
    let res = 9;
    assert_eq!(Solution::minimum_moves(grid), res);
}

// Accepted solution for LeetCode #1210: Minimum Moves to Reach Target with Rotations
function minimumMoves(grid: number[][]): number {
    const n = grid.length;
    const target: number[] = [n * n - 2, n * n - 1];
    const q: number[][] = [[0, 1]];
    const vis = Array.from({ length: n * n }, () => Array(2).fill(false));
    vis[0][0] = true;

    const move = (i1: number, j1: number, i2: number, j2: number) => {
        if (i1 >= 0 && i1 < n && j1 >= 0 && j1 < n && i2 >= 0 && i2 < n && j2 >= 0 && j2 < n) {
            const a = i1 * n + j1;
            const b = i2 * n + j2;
            const status = i1 === i2 ? 0 : 1;
            if (!vis[a][status] && grid[i1][j1] == 0 && grid[i2][j2] == 0) {
                q.push([a, b]);
                vis[a][status] = true;
            }
        }
    };

    let ans = 0;
    while (q.length) {
        for (let k = q.length; k; --k) {
            const p: number[] = q.shift();
            if (p[0] === target[0] && p[1] === target[1]) {
                return ans;
            }
            const [i1, j1] = [~~(p[0] / n), p[0] % n];
            const [i2, j2] = [~~(p[1] / n), p[1] % n];
            // 尝试向右平移（保持身体水平/垂直状态）
            move(i1, j1 + 1, i2, j2 + 1);
            // 尝试向下平移（保持身体水平/垂直状态）
            move(i1 + 1, j1, i2 + 1, j2);
            // 当前处于水平状态，且 grid[i1 + 1][j2] 无障碍，尝试顺时针旋转90°
            if (i1 == i2 && i1 + 1 < n && grid[i1 + 1][j2] == 0) {
                move(i1, j1, i1 + 1, j1);
            }
            // 当前处于垂直状态，且 grid[i2][j1 + 1] 无障碍，尝试逆时针旋转90°
            if (j1 == j2 && j1 + 1 < n && grid[i2][j1 + 1] == 0) {
                move(i1, j1, i1, j1 + 1);
            }
        }
        ++ans;
    }
    return -1;
}

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.