Problem Description

You have an n x n grid where a snake occupies exactly 2 cells. The snake starts at the top-left corner, occupying cells (0, 0) and (0, 1). The grid contains:

• Empty cells (represented by 0) that the snake can move through
• Blocked cells (represented by 1) that the snake cannot pass

The snake needs to reach the target position at the bottom-right corner, occupying cells (n-1, n-2) and (n-1, n-1).

The snake can perform four types of moves:

1. Move Right: The snake moves one cell to the right while maintaining its orientation (horizontal or vertical). Both cells of the snake must move into empty cells.

2. Move Down: The snake moves one cell down while maintaining its orientation. Both cells of the snake must move into empty cells.

3. Rotate Clockwise (only when horizontal): If the snake is in a horizontal position at (r, c) and (r, c+1), and both cells directly below ((r+1, c) and (r+1, c+1)) are empty, the snake can rotate clockwise. After rotation, the snake occupies (r, c) and (r+1, c), changing from horizontal to vertical orientation.

4. Rotate Counterclockwise (only when vertical): If the snake is in a vertical position at (r, c) and (r+1, c), and both cells to the right ((r, c+1) and (r+1, c+1)) are empty, the snake can rotate counterclockwise. After rotation, the snake occupies (r, c) and (r, c+1), changing from vertical to horizontal orientation.

Each action counts as one move. You need to find the minimum number of moves required for the snake to reach the target position. If it's impossible to reach the target, return -1.

Flowchart Walkthrough

First, let's pin down the algorithm using the Flowchart. Here's a step-by-step walkthrough:

Is it a graph?

• Yes: This problem can be modeled as a graph where each state (position and orientation of the snake) is a node, and valid moves between states are edges.

Is it a tree?

• No: The state space is not a tree structure. Multiple paths can lead to the same state (the snake can reach the same position through different sequences of moves).

Is the problem related to directed acyclic graphs?

• No: The state graph can have cycles (though we avoid revisiting states). The problem isn't about DAG-specific properties.

Is the problem related to shortest paths?

• Yes: We need to find the minimum number of moves (shortest path) from the starting position to the target position.

Is the graph weighted?

• No: Each move has the same cost (1 move), making this an unweighted graph problem.

BFS

• Yes: For unweighted shortest path problems, BFS is the optimal choice.

Conclusion: The flowchart suggests using BFS (Breadth-First Search) for finding the minimum number of moves. This aligns perfectly with the solution approach, as BFS explores states level by level, guaranteeing that when we first reach the target state, we've found it via the shortest path (minimum number of moves).

Intuition

The key insight is to think of this problem as finding the shortest path in a state space. Each state represents not just where the snake is, but also its orientation (horizontal or vertical). This is crucial because the same position with different orientations represents different states that may require different moves to reach the target.

Why do we need to track orientation? Consider that a snake at position (r, c) and (r, c+1) (horizontal) has different movement options than a snake at (r, c) and (r+1, c) (vertical). The horizontal snake can rotate clockwise if there's space below, while the vertical snake can rotate counterclockwise if there's space to the right.

Since each move has equal cost (exactly 1 move), we're dealing with an unweighted graph problem. This immediately suggests BFS as the optimal approach because BFS explores states level by level - all states reachable in 1 move, then 2 moves, then 3 moves, and so on. The first time we encounter the target state, we're guaranteed to have found it via the minimum number of moves.

To represent the snake's position efficiently, we can use the indices of its two cells. Since the grid is n x n, we can flatten the 2D coordinates into 1D indices: a cell at (i, j) becomes index i * n + j. This makes it easier to store and compare positions.

The visited set needs to store both position and orientation to avoid revisiting the same state. We represent this as (tail_position, status) where status is 0 for horizontal and 1 for vertical. This prevents us from exploring the same configuration multiple times, which would lead to inefficiency or infinite loops.

The four possible moves (right, down, clockwise rotation, counterclockwise rotation) are systematically checked for each state. The rotation moves have additional constraints - we need to check that the cells the snake will rotate through are empty, not just the final destination cells.

Learn more about Breadth-First Search patterns.

Solution Approach

The solution implements BFS using a queue to explore all possible states of the snake. Let's walk through the implementation details:

State Representation:

• Each snake position is represented by two indices (a, b) where a is the tail position and b is the head position in the flattened 1D array
• For a cell at (i, j) in the 2D grid, its 1D index is i * n + j
• The snake's orientation is tracked separately: status = 0 for horizontal, status = 1 for vertical

Data Structures:

• Queue q: Stores current positions as tuples (a, b). Initially contains (0, 1) representing the starting position
• Set vis: Tracks visited states as (tail_position, status) to avoid revisiting the same configuration. Initially contains (0, 0) for the starting horizontal state
• Target: Fixed at (n*n - 2, n*n - 1), representing the last two cells of the grid

The move Helper Function: This function validates and adds new positions to the queue:

1. Checks if both cells (i1, j1) and (i2, j2) are within bounds
2. Converts 2D coordinates to 1D indices: a = i1 * n + j1, b = i2 * n + j2
3. Determines orientation: status = 0 if i1 == i2 (horizontal), else status = 1 (vertical)
4. If the state (a, status) hasn't been visited and both cells are empty (grid[i][j] == 0), adds the position to queue and marks as visited

BFS Exploration: For each position (a, b) dequeued:

1. Check Target: If current position equals target (n*n - 2, n*n - 1), return the current move count ans

2. Convert to 2D: Extract coordinates from 1D indices:

   • i1, j1 = a // n, a % n (tail)
   • i2, j2 = b // n, b % n (head)

3. Try All Moves:

   • Move Right: move(i1, j1 + 1, i2, j2 + 1) - both cells shift right
   • Move Down: move(i1 + 1, j1, i2 + 1, j2) - both cells shift down
   • Clockwise Rotation (if horizontal): When i1 == i2, check if grid[i1 + 1][j2] == 0 (cell below head is empty), then move(i1, j1, i1 + 1, j1) rotates tail down
   • Counterclockwise Rotation (if vertical): When j1 == j2, check if grid[i2][j1 + 1] == 0 (cell right of head is empty), then move(i1, j1, i1, j1 + 1) rotates tail right

4. Level Tracking: After processing all positions at current distance, increment ans by 1

The algorithm continues until either the target is reached (return ans) or the queue is empty (return -1 indicating no path exists).

The time complexity is O(n²) as we can visit at most 2n² states (each cell with two orientations), and space complexity is also O(n²) for the visited set and queue.

Example Walkthrough

Let's walk through a small example with a 3x3 grid:

Grid:
[0, 0, 0]
[0, 0, 0]  
[0, 0, 0]

The snake starts at cells (0,0) and (0,1) horizontally, and needs to reach cells (2,1) and (2,2).

Initial State:

• Queue: [(0, 1)] (using 1D indices where 0 = cell(0,0), 1 = cell(0,1))
• Visited: {(0, 0)} (tail at index 0, status 0 for horizontal)
• Moves: 0

Move 1 - Process initial position (0, 1):

• Convert to 2D: tail at (0,0), head at (0,1)
• Try all possible moves:
  • Right: Move to (0,1) and (0,2) → Add (1, 2) to queue
  • Down: Move to (1,0) and (1,1) → Add (3, 4) to queue
  • Clockwise rotation: Rotate to (0,0) and (1,0) → Add (0, 3) to queue
  • Counterclockwise: Not applicable (snake is horizontal)
• Queue after: [(1, 2), (3, 4), (0, 3)]
• Visited adds: {(1, 0), (3, 0), (0, 1)}
• Increment moves to 1

Move 2 - Process position (1, 2):

• Snake at (0,1) and (0,2) horizontally
• Try moves:
  • Right: Would go out of bounds
  • Down: Move to (1,1) and (1,2) → Add (4, 5) to queue
  • Clockwise rotation: Rotate to (0,1) and (1,1) → Add (1, 4) to queue

Move 2 - Process position (3, 4):

• Snake at (1,0) and (1,1) horizontally
• Try moves:
  • Right: Move to (1,1) and (1,2) → Add (4, 5) (already added)
  • Down: Move to (2,0) and (2,1) → Add (6, 7) to queue
  • Clockwise rotation: Rotate to (1,0) and (2,0) → Add (3, 6) to queue

Move 2 - Process position (0, 3):

• Snake at (0,0) and (1,0) vertically
• Try moves:
  • Right: Move to (0,1) and (1,1) → Add (1, 4) (already added)
  • Down: Move to (1,0) and (2,0) → Add (3, 6) (already added)
  • Counterclockwise: Rotate to (0,0) and (0,1) → Already visited

After processing all positions at distance 1, increment moves to 2.

Move 3 - Continue processing: The BFS continues exploring positions at distance 2. Eventually, we'll reach position (7, 8) which represents the snake at cells (2,1) and (2,2) - our target!

Key Observations:

1. The snake explores multiple paths simultaneously through BFS
2. Each state tracks both position AND orientation (horizontal vs vertical)
3. Rotations are only possible when there's enough empty space
4. The first time we reach the target is guaranteed to be the shortest path

In this example, the minimum number of moves would be 3: starting horizontally at top-left, the snake can move down, then down again, then rotate and move to reach the target position at bottom-right.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n^2)

The algorithm uses BFS to explore all possible states of the snake. The snake is represented by two cells, and each cell can be in one of n^2 positions. However, the snake has only two orientations (horizontal or vertical), and the valid positions are constrained. The total number of unique states is bounded by O(n^2) because:

• We track states using (head_position, orientation) where head_position can be at most n^2 locations
• Each state has 2 possible orientations (horizontal=0, vertical=1)
• The visited set vis ensures each state is processed only once

For each state, we perform at most 4 moves (right, down, and potentially 2 rotations), each taking O(1) time. Therefore, the overall time complexity is O(n^2).

Space Complexity: O(n^2)

The space is dominated by:

• The BFS queue q which can contain at most O(n^2) states in the worst case
• The visited set vis which stores at most O(n^2) unique states (each state is a tuple of position and orientation)
• The input grid takes O(n^2) space

Therefore, the total space complexity is O(n^2).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrect Snake Position Representation

A critical pitfall is confusing which cell represents the "head" and which represents the "tail" of the snake. In the provided code, there's actually a naming inconsistency - the variables are named head_index and tail_index, but based on the problem description and initial state (0, 1), the snake starts at cells (0, 0) and (0, 1). The code treats the first value as position at (0, 0) and second as (0, 1), but names suggest otherwise.

Fix: Use consistent and clear naming:

# Better naming convention
queue = deque([(0, 1)])  # (first_cell, second_cell) or (tail, head)
# where first_cell is at (0,0) and second_cell is at (0,1)

2. Rotation Boundary Check Missing

When performing rotations, the code checks if the pivot cell's adjacent cell is empty, but doesn't verify if the rotation would place the snake out of bounds. While the current implementation includes boundary checks in try_move, a common mistake is to forget checking the intermediate cell needed for rotation.

Issue Example: For clockwise rotation from horizontal position, both (r+1, c) and (r+1, c+1) must be empty AND within bounds for the rotation to be valid.

Fix: Add explicit validation before rotation:

# For clockwise rotation (horizontal to vertical)
if head_row == tail_row:  # Currently horizontal
    # Need to check BOTH cells below are valid and empty
    if (head_row + 1 < grid_size and 
        grid[head_row + 1][head_col] == 0 and 
        grid[head_row + 1][tail_col] == 0):
        try_move(head_row, head_col, head_row + 1, head_col)

3. Visited State Tracking Error

The visited set stores (position, orientation) but uses only one position index instead of both. This could theoretically cause issues if the snake can reach the same cell pair in different orders.

Current approach: visited.add((head_index, orientation))

Potential issue: Two different snake configurations with the same head position and orientation but different tail positions would be considered the same state.

Fix: Store complete state information:

# More precise state tracking
visited.add((min(head_index, tail_index), max(head_index, tail_index), orientation))

4. Target Position Assumption

The code assumes the target is always at (n*n - 2, n*n - 1) in horizontal orientation. However, the problem states the snake should occupy (n-1, n-2) and (n-1, n-1), which translates to indices (n*n - n - 1, n*n - 1) not (n*n - 2, n*n - 1).

Fix: Correct target calculation:

# Correct target position
# (n-1, n-2) -> index = (n-1) * n + (n-2) = n*n - 2
# (n-1, n-1) -> index = (n-1) * n + (n-1) = n*n - 1
target_position = (grid_size * grid_size - 2, grid_size * grid_size - 1)  # This is actually correct

5. Orientation Detection Logic

The orientation is determined by comparing rows (head_row == tail_row), but this assumes a specific ordering of the snake cells. If the cell indices are swapped, the orientation detection could fail.

Fix: Use absolute position comparison:

# More robust orientation detection
def get_orientation(pos1_row, pos1_col, pos2_row, pos2_col):
    if pos1_row == pos2_row:
        return 0  # Horizontal
    elif pos1_col == pos2_col:
        return 1  # Vertical
    else:
        raise ValueError("Invalid snake position")