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Minimum Moves to Reach Target with Rotations

Number: 1322

Difficulty: Hard

Paid? No

Companies: Kakao

Problem Description

We are given an n x n grid with cells that can be either empty (0) or blocked (1). A snake occupies two consecutive cells and starts at the top left in a horizontal position covering (0,0) and (0,1). The snake can move:

  • Right: if the next cell(s) in the same row are empty.
  • Down: if the next cell(s) in the same column(s) are empty.
  • Rotate clockwise or counterclockwise, depending on its current orientation, if the required cells for rotation are empty. The goal is to reach the target position where the snake occupies the cells (n-1, n-2) and (n-1, n-1). Return the minimum number of moves required. If it is not possible, return -1.

Key Insights

  • Use Breadth-First Search (BFS) to explore states in increasing order of moves.
  • Represent each state by (row, col, orientation) where orientation is horizontal (0) or vertical (1).
  • Transitions depend on the snake’s orientation:
    • Horizontal: move right if cell (r, c+2) is free; move down if both (r+1, c) and (r+1, c+1) are free; rotate clockwise if (r+1, c) and (r+1, c+1) are free.
    • Vertical: move down if cell (r+2, c) is free; move right if both (r, c+1) and (r+1, c+1) are free; rotate counterclockwise if (r, c+1) and (r+1, c+1) are free.
  • Use a visited set to avoid reprocessing states.
  • The grid size n is up to 100; thus, BFS is efficient if states are pruned properly.

Space and Time Complexity

Time Complexity: O(n^2) — Each cell with each orientation is visited at most once. Space Complexity: O(n^2) — For storing the visited states and the BFS queue.

Solution

We solve the problem using a BFS approach where each state is denoted by its top-left cell and orientation (0 for horizontal, 1 for vertical). The snake can perform three types of moves: move right, move down, and perform a rotation (clockwise for horizontal states, counterclockwise for vertical states) subject to the grid constraints. We enqueue valid moves and count the number of steps until the target state is reached. If BFS terminates without reaching the target, we return -1. The use of a visited state structure ensures we do not process the same state multiple times, providing efficiency.

Code Solutions

Below are implementations in Python, JavaScript, C++, and Java with line-by-line comments.

from collections import deque

def minimumMoves(grid):
    n = len(grid)
    # state: (row, col, orientation) where orientation 0 = horizontal, 1 = vertical
    start = (0, 0, 0)
    target = (n-1, n-2, 0)
    
    # directions for BFS: (row, col, orientation)
    queue = deque([(0, 0, 0, 0)])  # (row, col, orientation, moves)
    visited = set([(0, 0, 0)])
    
    while queue:
        r, c, orient, moves = queue.popleft()
        if (r, c, orient) == target:
            return moves
        
        # if snake is horizontal
        if orient == 0:
            # Move right: check that the new head (r, c+2) is within bounds and cell is free
            if c + 2 < n and grid[r][c+2] == 0 and (r, c+1, 0) not in visited:
                visited.add((r, c+1, 0))
                queue.append((r, c+1, 0, moves + 1))
            # Move down: check the cells below both parts (r+1, c) and (r+1, c+1)
            if r + 1 < n and grid[r+1][c] == 0 and grid[r+1][c+1] == 0 and (r+1, c, 0) not in visited:
                visited.add((r+1, c, 0))
                queue.append((r+1, c, 0, moves + 1))
            # Rotate clockwise: only possible if the two cells below are empty
            if r + 1 < n and grid[r+1][c] == 0 and grid[r+1][c+1] == 0 and (r, c, 1) not in visited:
                visited.add((r, c, 1))
                queue.append((r, c, 1, moves + 1))
        
        # if snake is vertical
        else:  # orient == 1
            # Move down: check that the new tail (r+2, c) is within bounds and cell is free
            if r + 2 < n and grid[r+2][c] == 0 and (r+1, c, 1) not in visited:
                visited.add((r+1, c, 1))
                queue.append((r+1, c, 1, moves + 1))
            # Move right: check the cells to the right of both parts (r, c+1) and (r+1, c+1)
            if c + 1 < n and grid[r][c+1] == 0 and grid[r+1][c+1] == 0 and (r, c, 1) not in visited:
                # Note: staying in vertical orientation but column remains same for top cell.
                visited.add((r, c, 1))  # actually this state remains same; handled in rotation below.
                # Instead, to move right in vertical state, we simulate as a rotation possibility.
                # Better: directly use rotation counterclockwise to horizontal.
            # Rotate counterclockwise: only if the two cells to the right are empty
            if c + 1 < n and grid[r][c+1] == 0 and grid[r+1][c+1] == 0 and (r, c, 0) not in visited:
                visited.add((r, c, 0))
                queue.append((r, c, 0, moves + 1))
            # Additionally, for vertical state, let's allow a move to the right without rotation
            # Move right: check the right cells for vertical orientation.
            if c + 1 < n and grid[r][c+1] == 0 and grid[r+1][c+1] == 0 and (r, c+1, 1) not in visited:
                visited.add((r, c+1, 1))
                queue.append((r, c+1, 1, moves + 1))
    return -1

# Example usage:
grid1 = [[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]]
print(minimumMoves(grid1))
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