Introduction

In this blog post, we will explore how to solve the LeetCode problem "1210" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

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.

Explanation

Here's the solution to the snake in a grid problem:

• BFS Approach: Use Breadth-First Search (BFS) to explore possible snake movements. The state of the snake (its position and orientation) will be represented by a tuple.
• State Representation: Each state consists of (row1, col1, orientation), where (row1, col1) is the top-left cell the snake occupies. Orientation 0 represents horizontal, and 1 represents vertical.

• Visited Set: Maintain a set of visited states to avoid cycles.

• Runtime Complexity: O(N³) - where N is the grid dimension.

• Storage Complexity: O(N³)

Code

    from collections import deque

def snake_in_grid(grid):
    n = len(grid)
    target = (n - 1, n - 2, 0)  # Target state (horizontal)

    q = deque([(0, 0, 0, 0)])  # (row, col, orientation, moves)
    visited = set()
    visited.add((0, 0, 0))

    while q:
        r, c, orientation, moves = q.popleft()

        if (r, c, orientation) == target:
            return moves

        # Possible moves: right, down, rotate (clockwise or counterclockwise)

        # Move right (if horizontal)
        if orientation == 0 and c + 2 < n and grid[r][c + 2] == 0:
            new_state = (r, c + 1, 0)
            if new_state not in visited:
                visited.add(new_state)
                q.append((r, c + 1, 0, moves + 1))

        # Move down (if vertical)
        if orientation == 1 and r + 2 < n and grid[r + 2][c] == 0:
            new_state = (r + 1, c, 1)
            if new_state not in visited:
                visited.add(new_state)
                q.append((r + 1, c, 1, moves + 1))

        # Move right (if vertical)
        if orientation == 1 and c + 1 < n and grid[r][c+1] == 0 and grid[r+1][c+1] == 0:
             new_state = (r, c + 1, 1)
             if new_state not in visited:
                visited.add(new_state)
                q.append((r, c + 1, 1, moves + 1))

        #Move down (if horizontal)
        if orientation == 0 and r + 1 < n and grid[r+1][c] == 0 and grid[r+1][c+1] == 0:
             new_state = (r + 1, c, 0)
             if new_state not in visited:
                visited.add(new_state)
                q.append((r + 1, c, 0, moves + 1))


        # Rotate clockwise (if horizontal)
        if orientation == 0 and r + 1 < n and grid[r + 1][c] == 0 and grid[r + 1][c + 1] == 0:
            new_state = (r, c, 1)
            if new_state not in visited:
                visited.add(new_state)
                q.append((r, c, 1, moves + 1))

        # Rotate counterclockwise (if vertical)
        if orientation == 1 and c + 1 < n and grid[r][c + 1] == 0 and grid[r + 1][c + 1] == 0:
            new_state = (r, c, 0)
            if new_state not in visited:
                visited.add(new_state)
                q.append((r, c, 0, moves + 1))

    return -1