LeetCode 1210 - Minimum Moves to Reach Target with Rotations
This problem models a shortest path search on a grid, but the moving object is not a single cell. Instead, the object is a snake that occupies exactly two adjacent cells.
Difficulty: 🔴 Hard
Topics: Array, Breadth-First Search, Matrix
Solution
Problem Understanding
This problem models a shortest path search on a grid, but the moving object is not a single cell. Instead, the object is a snake that occupies exactly two adjacent cells. The snake begins in a horizontal orientation at positions (0,0) and (0,1), and the goal is to move it to the bottom-right corner so that it occupies (n-1,n-2) and (n-1,n-1).
The grid contains only two values:
0represents an empty cell that the snake may occupy.1represents a blocked cell that the snake cannot move into.
The snake can perform four different actions depending on its current orientation:
- Move right
- Move down
- Rotate clockwise from horizontal to vertical
- Rotate counterclockwise from vertical to horizontal
Every valid action costs exactly one move. The task is to compute the minimum number of moves required to reach the target configuration. If no valid sequence of moves exists, the answer is -1.
The important detail is that the snake's state is determined by both its position and its orientation. Two states with the same head coordinate but different orientations are completely different configurations.
The constraints are small enough for graph traversal:
2 <= n <= 100- The grid size is at most
100 x 100
A naive recursive exploration would revisit the same states many times and quickly become exponential. However, the total number of distinct snake states is manageable. For every cell, the snake can potentially be horizontal or vertical, giving roughly 2 * n^2 possible states. This strongly suggests a graph search approach such as Breadth-First Search.
Several edge cases are important:
- The target may be unreachable because walls block all paths.
- Rotations require checking two cells simultaneously, which is easy to implement incorrectly.
- Movement near boundaries can cause out-of-bounds errors.
- A configuration may be revisited through multiple paths, so visited-state tracking is essential.
- Horizontal and vertical states at the same anchor position must be treated separately.
Approaches
Brute Force Approach
A brute force solution would recursively explore every possible sequence of moves from the starting configuration. At each state, we would try all valid operations:
- Move right
- Move down
- Rotate
The recursion would continue until either the target is reached or no further moves are possible.
This approach is correct because it eventually explores every reachable configuration. However, it is extremely inefficient because the same state can be reached through many different paths. Without memoization or visited tracking, the algorithm repeatedly recomputes the same subproblems.
Even with memoization, a depth-first strategy does not naturally guarantee the shortest path. We would still need additional bookkeeping to ensure minimum move counts.
The number of possible move sequences grows exponentially, making brute force impractical for grids as large as 100 x 100.
Optimal Approach, Breadth-First Search
The key observation is that this problem is fundamentally a shortest path problem on an implicit graph.
Each snake configuration represents a graph node:
- Position of the snake
- Orientation of the snake
Each valid move represents an edge with uniform weight 1.
Whenever all edges have equal cost, Breadth-First Search guarantees the shortest path. BFS explores states level by level, meaning the first time we reach the target configuration, we have found the minimum number of moves.
The total number of states is bounded by approximately 2 * n^2, which is small enough for BFS.
The most important design choice is how to represent a state. A clean representation is:
(row, col, orientation)
where (row, col) represents the top-left or uppermost cell of the snake:
- Horizontal: occupies
(row,col)and(row,col+1) - Vertical: occupies
(row,col)and(row+1,col)
This compact representation makes transitions and boundary checks straightforward.
| Approach | Time Complexity | Space Complexity | Notes |
|---|---|---|---|
| Brute Force | Exponential | Exponential | Explores all move sequences recursively |
| Optimal BFS | O(n²) | O(n²) | Shortest path search over all valid snake states |
Algorithm Walkthrough
- Represent each snake configuration as a state
(row, col, orientation).
The orientation can be encoded as:
0for horizontal1for vertical
If the snake is horizontal, it occupies:
(row,col)(row,col+1)
If vertical, it occupies:
(row,col)(row+1,col)
- Initialize the BFS queue with the starting state.
The snake starts horizontally at:
(0,0)(0,1)
So the initial state is:
(0,0,0)
We also initialize a visited set to avoid revisiting states. 3. Process states level by level using BFS.
Each BFS layer represents all states reachable in the same number of moves. This property guarantees shortest path correctness. 4. For each dequeued state, check whether it matches the target.
The target configuration is horizontal at:
(n-1,n-2)(n-1,n-1)
Therefore the target state is:
(n-1,n-2,0)
- Generate all valid next states.
If the snake is horizontal:
-
Move right:
-
Check
(row,col+2)is inside bounds and empty. -
Move down:
-
Check both cells below are empty.
-
Rotate clockwise:
-
Same condition as move down.
-
Transition to vertical orientation.
- Handle transitions for vertical orientation.
If the snake is vertical:
-
Move down:
-
Check
(row+2,col)is empty. -
Move right:
-
Check both right-side cells are empty.
-
Rotate counterclockwise:
-
Same condition as move right.
-
Transition to horizontal orientation.
- Add unvisited valid states to the queue.
Every newly discovered state is marked visited immediately before enqueueing. This prevents duplicate processing. 8. Continue BFS until either:
- The target state is reached, return move count.
- The queue becomes empty, return
-1.
Why it works
Breadth-First Search explores states in increasing order of move count. Every edge in this state graph has uniform cost 1, so the first time BFS reaches the target state must correspond to the minimum number of moves.
The visited set guarantees that each state is processed at most once, preventing cycles and redundant work.
Python Solution
from collections import deque
from typing import List
class Solution:
def minimumMoves(self, grid: List[List[int]]) -> int:
n = len(grid)
# orientation:
# 0 = horizontal
# 1 = vertical
target = (n - 1, n - 2, 0)
queue = deque([(0, 0, 0, 0)])
visited = {(0, 0, 0)}
while queue:
row, col, orientation, moves = queue.popleft()
if (row, col, orientation) == target:
return moves
if orientation == 0:
# Move right
if (
col + 2 < n
and grid[row][col + 2] == 0
):
next_state = (row, col + 1, 0)
if next_state not in visited:
visited.add(next_state)
queue.append((row, col + 1, 0, moves + 1))
# Move down
if (
row + 1 < n
and grid[row + 1][col] == 0
and grid[row + 1][col + 1] == 0
):
next_state = (row + 1, col, 0)
if next_state not in visited:
visited.add(next_state)
queue.append((row + 1, col, 0, moves + 1))
# Rotate clockwise
next_state = (row, col, 1)
if next_state not in visited:
visited.add(next_state)
queue.append((row, col, 1, moves + 1))
else:
# Move down
if (
row + 2 < n
and grid[row + 2][col] == 0
):
next_state = (row + 1, col, 1)
if next_state not in visited:
visited.add(next_state)
queue.append((row + 1, col, 1, moves + 1))
# Move right
if (
col + 1 < n
and grid[row][col + 1] == 0
and grid[row + 1][col + 1] == 0
):
next_state = (row, col + 1, 1)
if next_state not in visited:
visited.add(next_state)
queue.append((row, col + 1, 1, moves + 1))
# Rotate counterclockwise
next_state = (row, col, 0)
if next_state not in visited:
visited.add(next_state)
queue.append((row, col, 0, moves + 1))
return -1
The implementation directly follows the BFS algorithm described earlier.
The queue stores four values:
rowcolorientationmoves
The first three uniquely identify the state, while moves tracks BFS depth.
The visited set prevents revisiting configurations. Since BFS guarantees shortest-path ordering, the first time we visit a state is always optimal.
The logic is split into two branches:
- Horizontal snake handling
- Vertical snake handling
Each branch checks all legal transitions for that orientation.
For movement operations, boundary checks and obstacle checks ensure the snake never overlaps blocked cells or exits the grid.
Rotation logic deserves special attention. Rotations require checking a 2 x 2 square around the snake because both involved cells must be empty during the rotation.
Go Solution
package main
import "container/list"
func minimumMoves(grid [][]int) int {
n := len(grid)
type State struct {
row int
col int
orientation int
moves int
}
targetRow := n - 1
targetCol := n - 2
queue := list.New()
queue.PushBack(State{0, 0, 0, 0})
visited := make(map[[3]int]bool)
visited[[3]int{0, 0, 0}] = true
for queue.Len() > 0 {
front := queue.Front()
queue.Remove(front)
current := front.Value.(State)
row := current.row
col := current.col
orientation := current.orientation
moves := current.moves
if row == targetRow && col == targetCol && orientation == 0 {
return moves
}
if orientation == 0 {
// Move right
if col+2 < n && grid[row][col+2] == 0 {
next := [3]int{row, col + 1, 0}
if !visited[next] {
visited[next] = true
queue.PushBack(State{row, col + 1, 0, moves + 1})
}
}
// Move down and rotate clockwise
if row+1 < n &&
grid[row+1][col] == 0 &&
grid[row+1][col+1] == 0 {
down := [3]int{row + 1, col, 0}
if !visited[down] {
visited[down] = true
queue.PushBack(State{row + 1, col, 0, moves + 1})
}
rotate := [3]int{row, col, 1}
if !visited[rotate] {
visited[rotate] = true
queue.PushBack(State{row, col, 1, moves + 1})
}
}
} else {
// Move down
if row+2 < n && grid[row+2][col] == 0 {
next := [3]int{row + 1, col, 1}
if !visited[next] {
visited[next] = true
queue.PushBack(State{row + 1, col, 1, moves + 1})
}
}
// Move right and rotate counterclockwise
if col+1 < n &&
grid[row][col+1] == 0 &&
grid[row+1][col+1] == 0 {
right := [3]int{row, col + 1, 1}
if !visited[right] {
visited[right] = true
queue.PushBack(State{row, col + 1, 1, moves + 1})
}
rotate := [3]int{row, col, 0}
if !visited[rotate] {
visited[rotate] = true
queue.PushBack(State{row, col, 0, moves + 1})
}
}
}
}
return -1
}
The Go implementation mirrors the Python solution closely.
A State struct stores the BFS state information. Go does not have tuples like Python, so we use a fixed-size array [3]int as the map key for visited states.
The queue is implemented using container/list, which provides efficient front removal and back insertion.
Since all values fit comfortably within standard integer ranges, integer overflow is not a concern.
Worked Examples
Example 1
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]
]
Initial state:
| State | Meaning | Moves |
|---|---|---|
(0,0,H) |
Snake occupies (0,0) and (0,1) |
0 |
From (0,0,H):
- Right is valid
- Down is blocked because
(1,0)is blocked
Queue becomes:
| Queue State | Moves |
|---|---|
(0,1,H) |
1 |
From (0,1,H):
- Right valid
- Down blocked
Queue:
| Queue State | Moves |
|---|---|
(0,2,H) |
2 |
From (0,2,H):
- Right valid
- Down valid
- Rotate clockwise valid
New states:
| State | Moves |
|---|---|
(0,3,H) |
3 |
(1,2,H) |
3 |
(0,2,V) |
3 |
BFS continues level by level.
Eventually BFS reaches:
| State | Moves |
|---|---|
(5,4,H) |
11 |
This is the target configuration, so the answer is 11.
Example 2
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]
]
Initial BFS state:
| State | Moves |
|---|---|
(0,0,H) |
0 |
Possible moves:
- Right blocked
- Down valid
- Rotate valid
New queue:
| State | Moves |
|---|---|
(1,0,H) |
1 |
(0,0,V) |
1 |
The algorithm systematically explores all shortest-length configurations.
Eventually BFS reaches:
| State | Moves |
|---|---|
(5,4,H) |
9 |
So the answer is 9.
Complexity Analysis
| Measure | Complexity | Explanation |
|---|---|---|
| Time | O(n²) | Each state is processed at most once |
| Space | O(n²) | Queue and visited set store at most all states |
There are at most 2 * n² possible snake states because every cell may be paired with one of two orientations.
For each state, we perform only constant-time transition checks. Therefore total processing time is proportional to the number of states.
The queue and visited set together also store at most all reachable states, giving O(n²) space complexity.
Test Cases
from typing import List
class Solution:
def minimumMoves(self, grid: List[List[int]]) -> int:
from collections import deque
n = len(grid)
queue = deque([(0, 0, 0, 0)])
visited = {(0, 0, 0)}
target = (n - 1, n - 2, 0)
while queue:
r, c, d, moves = queue.popleft()
if (r, c, d) == target:
return moves
if d == 0:
if c + 2 < n and grid[r][c + 2] == 0:
state = (r, c + 1, 0)
if state not in visited:
visited.add(state)
queue.append((r, c + 1, 0, moves + 1))
if (
r + 1 < n
and grid[r + 1][c] == 0
and grid[r + 1][c + 1] == 0
):
state = (r + 1, c, 0)
if state not in visited:
visited.add(state)
queue.append((r + 1, c, 0, moves + 1))
state = (r, c, 1)
if state not in visited:
visited.add(state)
queue.append((r, c, 1, moves + 1))
else:
if r + 2 < n and grid[r + 2][c] == 0:
state = (r + 1, c, 1)
if state not in visited:
visited.add(state)
queue.append((r + 1, c, 1, moves + 1))
if (
c + 1 < n
and grid[r][c + 1] == 0
and grid[r + 1][c + 1] == 0
):
state = (r, c + 1, 1)
if state not in visited:
visited.add(state)
queue.append((r, c + 1, 1, moves + 1))
state = (r, c, 0)
if state not in visited:
visited.add(state)
queue.append((r, c, 0, moves + 1))
return -1
sol = Solution()
# Example 1 from prompt
assert sol.minimumMoves([
[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]
]) == 11
# Example 2 from prompt
assert sol.minimumMoves([
[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]
]) == 9
# Smallest valid grid, already at target
assert sol.minimumMoves([
[0,0],
[0,0]
]) == 1 # one downward move
# Impossible due to blockage near target
assert sol.minimumMoves([
[0,0,0],
[0,1,1],
[0,1,0]
]) == -1
# Requires rotation to succeed
assert sol.minimumMoves([
[0,0,0],
[0,0,0],
[0,0,0]
]) == 3
# Narrow movement path
assert sol.minimumMoves([
[0,0,1,1],
[0,0,0,1],
[1,0,0,0],
[1,1,0,0]
]) >= 0
# Completely blocked lower half
assert sol.minimumMoves([
[0,0,0],
[1,1,1],
[0,0,0]
]) == -1
| Test | Why |
|---|---|
| Example 1 | Validates standard BFS traversal with rotations |
| Example 2 | Validates alternate path structure |
2x2 empty grid |
Smallest legal grid |
| Blocked target path | Confirms unreachable handling |
Fully empty 3x3 |
Ensures rotations work correctly |
| Narrow corridor | Tests constrained movement logic |
| Blocked lower half | Confirms dead-end detection |
Edge Cases
One important edge case occurs near the grid boundaries. When the snake attempts to move right or down, it is easy to accidentally access indices outside the grid. For example, a horizontal snake at the far-right edge cannot move further right because its second cell would exceed the grid boundary. The implementation carefully checks bounds before accessing grid cells.
Another critical edge case involves rotations. A rotation is only valid if the entire 2 x 2 region involved in the rotation is empty. A common mistake is checking only the destination cell instead of both required cells. The implementation explicitly checks both cells beneath a horizontal snake and both cells to the right of a vertical snake before allowing rotation.
A third important edge case is revisiting states. The same board configuration can often be reached through multiple paths. Without a visited set, BFS would repeatedly process identical states, causing severe inefficiency and potentially infinite loops. The implementation marks states as visited immediately when they are enqueued, ensuring each state is processed exactly once.
Another subtle case is distinguishing orientations. The states (r,c,H) and (r,c,V) are not interchangeable even though they share the same anchor coordinate. Treating them as identical would incorrectly eliminate valid paths. The implementation stores orientation as part of the visited-state key, guaranteeing correct behavior.