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Minimum Moves to Reach Target with Rotations

Difficulty: Hard

Problem Description

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, down, or rotate clockwise/counterclockwise under certain conditions. Return the minimum number of moves to reach the target. If there is no way to reach the target, return -1.

Key Insights

  • The problem can be modeled as a pathfinding problem in a grid with specific movement rules.
  • The snake can be in two orientations: horizontal and vertical, which affects movement and rotation.
  • Breadth-First Search (BFS) is suitable for finding the shortest path in an unweighted grid.
  • States can be represented by the position of the snake's head and its orientation.

Space and Time Complexity

Time Complexity: O(n^2) - Each cell can be visited multiple times, but the maximum number of unique states is limited. Space Complexity: O(n^2) - The BFS queue and visited set can store states proportional to the grid size.

Solution

To solve this problem, we will use a Breadth-First Search (BFS) algorithm. The main steps include:

  1. State Representation: Each state in the BFS will be represented by the position of the head of the snake (x, y) and its orientation (0 for horizontal, 1 for vertical).
  2. Movement and Rotation Logic: Define the rules for moving and rotating the snake:
    • Move right if the next cell is empty.
    • Move down if the cell below is empty.
    • Rotate clockwise if the snake is horizontal and the cells below are empty.
    • Rotate counterclockwise if the snake is vertical and the cells to the right are empty.
  3. BFS Execution: Start from the initial position, enqueue valid states, and track the number of moves until the target state is reached or all possibilities are exhausted.

Code Solutions

from collections import deque

def minMoves(grid):
    n = len(grid)
    directions = [(0, 1), (1, 0)]  # right, down
    queue = deque([(0, 0, 0, 0)])  # x, y, orientation, moves
    visited = set((0, 0, 0))

    while queue:
        x, y, orientation, moves = queue.popleft()

        # Check if we've reached the target
        if (x, y, orientation) == (n - 1, n - 2, 0) or (x, y, orientation) == (n - 1, n - 1, 1):
            return moves

        # Move right
        if orientation == 0 and y + 1 < n and grid[x][y + 1] == 0:
            if (x, y + 1, 0) not in visited:
                visited.add((x, y + 1, 0))
                queue.append((x, y + 1, 0, moves + 1))

        # Move down
        if orientation == 0 and x + 1 < n and grid[x + 1][y] == 0:
            if (x + 1, y, 0) not in visited:
                visited.add((x + 1, y, 0))
                queue.append((x + 1, y, 0, moves + 1))

        # Rotate clockwise
        if orientation == 0 and x + 1 < n and grid[x + 1][y] == 0 and grid[x + 1][y + 1] == 0:
            if (x, y, 1) not in visited:
                visited.add((x, y, 1))
                queue.append((x, y, 1, moves + 1))

        # Move left
        if orientation == 1 and y + 1 < n and grid[x][y + 1] == 0:
            if (x, y + 1, 1) not in visited:
                visited.add((x, y + 1, 1))
                queue.append((x, y + 1, 1, moves + 1))

        # Rotate counterclockwise
        if orientation == 1 and y + 1 < n and grid[x][y + 1] == 0 and grid[x + 1][y + 1] == 0:
            if (x, y, 0) not in visited:
                visited.add((x, y, 0))
                queue.append((x, y, 0, moves + 1))

    return -1
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