Problem

You are controlling a robot that is located somewhere in a room. The room is modeled as an m x n binary grid where 0 represents a wall and 1 represents an empty slot.

The robot starts at an unknown location in the room that is guaranteed to be empty, and you do not have access to the grid, but you can move the robot using the given API Robot.

You are tasked to use the robot to clean the entire room (i.e., clean every empty cell in the room). The robot with the four given APIs can move forward, turn left, or turn right. Each turn is 90 degrees.

When the robot tries to move into a wall cell, its bumper sensor detects the obstacle, and it stays on the current cell.

Design an algorithm to clean the entire room using the following APIs:

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interface Robot {
  // returns true if next cell is open and robot moves into the cell.
  // returns false if next cell is obstacle and robot stays on the current cell.
  boolean move();

  // Robot will stay on the same cell after calling turnLeft/turnRight.
  // Each turn will be 90 degrees.
  void turnLeft();
  void turnRight();

  // Clean the current cell.
  void clean();
}

Note that the initial direction of the robot will be facing up. You can assume all four edges of the grid are all surrounded by a wall.

Custom testing:

The input is only given to initialize the room and the robot’s position internally. You must solve this problem “blindfolded”. In other words, you must control the robot using only the four mentioned APIs without knowing the room layout and the initial robot’s position.

Examples

Example 1:

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Input: room = [  
  [1,1,1,1,1,0,1,1],  
  [1,1,1,1,1,0,1,1],  
  [1,0,1,1,1,1,1,1],  
  [0,0,0,1,0,0,0,0],  
  [1,1,1,1,1,1,1,1]  
],  
row = 1,  
col = 3

Output: Robot cleaned all rooms.

Explanation:
All grids in the room are marked by either 0 or 1.
0 means the cell is blocked, while 1 means the cell is accessible.
The robot initially starts at the position of row=1, col=3.
From the top left corner, its position is one row below and three columns right.

Solution

Method 1 - Backtracking

This is a typical problem that can be solved with the backtracking paradigm, as many of you might have figured out from the description of the problem.

The problem requires designing an algorithm for a robot to clean an entire room represented as an unknown binary grid. The robot starts at an unknown location and uses given APIs to move and clean the room. Since the room’s layout and initial position are unknown, the algorithm must ensure the robot can navigate and clean all empty slots by leveraging Depth-First Search (DFS). The robot must also handle obstacles (walls) and avoid re-cleaning cells.

Here is the approach:

  1. DFS Traversal: Use DFS to move the robot to each cell. Mark cleaned cells to avoid re-cleaning.
  2. Backtracking: To backtrack from dead ends or obstacles, the robot must retreat to the previous position and continue exploring.
  3. Direction Management: The robot can move in four possible directions (up, right, down, left). Manage direction using an array and turning functions.
  4. Visited Set: Use a set to keep track of visited cells, ensuring the robot does not revisit cleaned cells.

Code

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interface Robot {
    boolean move();
    void turnLeft();
    void turnRight();
    void clean();
}

public class Solution {
    private static final int[][] DIRECTIONS = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}};
    
    public void cleanRoom(Robot robot) {
        Set<String> visited = new HashSet<>();
        dfs(robot, 0, 0, 0, visited);
    }
    
    private void dfs(Robot robot, int x, int y, int d, Set<String> visited) {
        String pos = x + "," + y;
        if (visited.contains(pos)) {
            return;
        }
        
        robot.clean();
        visited.add(pos);

        for (int i = 0; i < 4; i++) {
            int newDir = (d + i) % 4;
            int newX = x + DIRECTIONS[newDir][0];
            int newY = y + DIRECTIONS[newDir][1];

            if (robot.move()) {
                dfs(robot, newX, newY, newDir, visited);
                goBack(robot);
            }
            robot.turnRight();
        }
    }

    private void goBack(Robot robot) {
        robot.turnLeft();
        robot.turnLeft();
        robot.move();
        robot.turnLeft();
        robot.turnLeft();
    }
}
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class Solution:
    def cleanRoom(self, robot: 'Robot') -> None:
        DIRECTIONS = [(-1, 0), (0, 1), (1, 0), (0, -1)]
        visited: Set[Tuple[int, int]] = set()
        
        def dfs(x: int, y: int, d: int) -> None:
            if (x, y) in visited:
                return
            
            robot.clean()
            visited.add((x, y))

            for i in range(4):
                new_dir = (d + i) % 4
                new_x, new_y = x + DIRECTIONS[new_dir][0], y + DIRECTIONS[new_dir][1]

                if robot.move():
                    dfs(new_x, new_y, new_dir)
                    goBack()
                robot.turnRight()

        def goBack() -> None:
            robot.turnLeft()
            robot.turnLeft()
            robot.move()
            robot.turnLeft()
            robot.turnLeft()

        dfs(0, 0, 0)

Complexity

  • ⏰ Time complexity:  O(m*n) where m is the number of rows and n is the number of columns because the robot visits each cell exactly once.
  • 🧺 Space complexity: O(m*n) due to the recursive stack and the storage of visited cells.