Number of Spaces Cleaning Robot Cleaned

Updated on 14 July, 2025
Number of Spaces Cleaning Robot Cleaned header image

Problem Statement

In this problem, we're presented with a cleaning robot located in a 2D binary matrix called room, representing spaces where 0 denotes an empty space and 1 denotes a space occupied by an object. Commencing at the matrix's top-left corner, the robot faces towards the right. It continues moving in its current direction until it encounters either the room's boundary or an object. Upon such an encounter, it turns 90 degrees clockwise and follows the same rules indefinitely. The task is to calculate how many unique empty spaces the robot manages to clean when operating non-stop under these conditions.

Examples

Example 1

Input:

room = [[0,1,0],[1,0,0],[0,0,0]]

Output:

1

Explanation:

1. The robot cleans the space at (0, 0).
2. The robot hits an object, so it turns 90 degrees clockwise and now faces down.
3. The robot hits an object, so it turns 90 degrees clockwise and now faces left.
4. The robot is at the edge of the room, so it turns 90 degrees clockwise and now faces up.
5. The robot is at the edge of the room, so it turns 90 degrees clockwise and now faces right.
6. The robot is back at its starting position.
7. The robot has cleaned 1 space, so return 1.

Example 2

Input:

room = [[0,0,0],[0,0,0],[0,0,0]]

Output:

8​​​​​​​

Constraints

  • m == room.length
  • n == room[r].length
  • 1 <= m, n <= 300
  • room[r][c] is either 0 or 1.
  • room[0][0] == 0

Approach and Intuition

Understanding the problem through the examples:

  • Example 1:

    room = [[0,1,0],[1,0,0],[0,0,0]]

    • The robot starts at position (0,0), facing right.
    • The robot cleans (0,0). Moving right, it strikes an object at (0,1) and turns down to face the next open direction.
    • The collision with an object directly below at (1,0) forces another clockwise turn to face left.
    • The robot is already at the row's start, prompting another turn to now face upward.
    • As it can't move upward from the top row, it turns right again, ending back at the start.
    • Here the robot has cleaned only one distinct space which is (0,0).
  • Example 3:

    room = [[0,0,0],[0,0,0],[0,0,0]]

    • Here the room lacks obstacles.
    • Starting from (0,0) and cleaning that cell, the robot moves all the way to the right end, turns down, shifts left cleaning all cells rounding over to the left, and then upwards till it reaches the start point again.
    • The robot efficiently cleans all places except (2,2), giving a count of 8 cleaned unique spaces.

Observations and Insights

  1. The robot's path is significantly affected by the arrangement of objects (1) in the room.
  2. By keeping a track of spaces cleaned (perhaps with a set to ensure each space is counted just once) and the robot's positional updates after each move, we can determine the total number of unique spaces cleaned.
  3. The robot's direction changes are pivotal. Upon encountering an object or boundary, turning 90 degrees clockwise allows the robot to attempt new path explorations.
  4. Loops occur if the robot returns to a previous position facing the same direction. Hence, detecting such loops could be crucial to understanding when further space cleaning is infeasible.
  5. The complications to consider are the room's size and object density, which could influence computational challenges, particularly given that both room dimensions can scale up to 300x300, making the matrix computationally large to handle with simplistic algorithms.

The problem can be approached by simulating the robot's movements in the room, altering directions upon hitting boundaries or objects, and maintaining a record of cleaned spots to ensure uniqueness.

Solutions

  • C++
cpp
class Solution {
public:
    int countCleanedRooms(vector<vector<int>>& grid) {
        int m = grid.size();
        int n = grid[0].size();
        int total_clean = 0;
    
        int x = 0, y = 0;
        int dir = 0;
    
        while (!(grid[x][y] >> (dir + 1) & 1)) {
            if (grid[x][y] == 0) {
                total_clean++;
            }
    
            grid[x][y] |= 1 << (dir + 1);
    
            int newX = x + MOVEMENTS[dir];
            int newY = y + MOVEMENTS[dir + 1];
            if (0 <= newX && newX < m && 0 <= newY && newY < n && grid[newX][newY] != 1) {
                x = newX;
                y = newY;
            } else {
                dir = (dir + 1) % 4;
            }
        }
        return total_clean;
    }
    
private:
    const vector<int> MOVEMENTS = {0, 1, 0, -1, 0};
};

The provided C++ code defines a function countCleanedRooms that simulates the movement of a cleaning robot within a 2D grid, represented by the grid parameter, where each cell indicates whether a room has been cleaned (using specific bits in integers for tracking direction-specific visits).

Here is an overview of how the function operates:

  • Variables m and n are initialized to represent the dimensions of the grid (rows and columns, respectively).
  • The robot starts at the top-left corner of the grid (x = 0, y = 0).
  • The direction dir is initially set to 0, which will correlate to east based on the MOVEMENTS array indexing.
  • The primary loop continues as long as the current cell (grid[x][y]) hasn't been visited from the current direction dir.
  • Marking the room as cleaned involves setting a specific bit for the direction the robot enters the cell. Cleaner status is checked and updated using bitwise operations.
  • The robot moves according to the encoded MOVEMENTS array which offers directional movement modulations.
  • If moving in the current direction leads to a wall or an off-grid position, the robot changes direction clockwise.
  • The loop breaks once the robot has attempted to visit a cell from all four possible directions, which is tracked using bit marking per direction.
  • The function eventually returns the total count of cleaned unique rooms.

The robot uses a bit manipulation strategy to prevent cleaning the same room from the same direction multiple times, thereby optimizing its path. This technique ensures that once all directions for a room are visited, the robot does not re-enter the room from the same direction, effectively handling the challenge of revisiting rooms.

Overall, the solution efficiently tracks both the cleaned rooms and the travel directions using compact storage (bit fields within integers), which is particularly useful for robots operating in constrained environments with limited memory.

  • Java
java
class Solution {
    
    private static final int[] STEPS = { 0, 1, 0, -1, 0 };
    
    public int countCleanedRooms(int[][] room) {
        int numberOfRows = room.length;
        int numberOfColumns = room[0].length;
        int countOfCleaned = 0;
    
        int currentRow = 0, currentColumn = 0;
        int currentDirection = 0;
    
        while (((room[currentRow][currentColumn] >> (currentDirection + 1)) & 1) == 0) {
            if (room[currentRow][currentColumn] == 0) {
                countOfCleaned += 1;
            }
    
            room[currentRow][currentColumn] |= 1 << (currentDirection + 1);
    
            int newRow = currentRow + STEPS[currentDirection];
            int newColumn = currentColumn + STEPS[currentDirection + 1];
            if (
                0 <= newRow &&
                newRow < numberOfRows &&
                0 <= newColumn &&
                newColumn < numberOfColumns &&
                room[newRow][newColumn] != 1
            ) {
                currentRow = newRow;
                currentColumn = newColumn;
            } else {
                currentDirection = (currentDirection + 1) % 4;
            }
        }
        return countOfCleaned;
    }
}

The Java code provided aims to calculate the number of different rooms a cleaning robot cleans in a grid representation of a room, where the grid elements can be either clean (0) or an obstacle (1). The method countCleanedRooms takes a 2D array room as input, which represents the different rooms.

  • Start by initializing the size variables for the grid (numberOfRows by numberOfColumns).
  • countOfCleaned is used to keep track of the number of spaces cleaned by the robot.
  • The robot's starting position is at the top-left corner of the grid (currentRow = 0, currentColumn = 0).
  • A direction variable currentDirection defines the current direction of the robot, based on the STEPS array that helps in moving the robot up, down, left, or right.
  • A while loop continues the cleaning process as long as there are directions not marked as cleaned in the current room. This is checked using bitwise operations.
  • If the current space is unclean (0), it marks this space as cleaned, thereby increasing the countOfCleaned.
  • Updates the room's status to mark the direction the robot just cleaned.
  • Calculates the next row and column based on the current direction.
  • If the new positions are within bounds and not an obstacle, the robot moves there. Otherwise, the robot changes direction.
  • The loop terminates when all possible cleaning directions in the current room are exhausted, and the function returns the total count of rooms cleaned.

The code effectively employs a simulation approach with bitwise operations to track directions cleaned in each room, preventing the robot from cleaning the same place more than once. Each while iteration updates the robot's status and position within the grid, adapting its path according to obstacles and already cleaned rooms.

  • Python
python
class RobotCleaner:
    def countCleanRooms(self, grid: List[List[int]]) -> int:
        MOVES = (0, 1, 0, -1, 0)
        total_rows, total_cols = len(grid), len(grid[0])
        total_cleaned = 0
    
        current_row, current_col = 0, 0
        current_direction = 0
    
        while not grid[current_row][current_col] >> (current_direction + 1) & 1:
            if grid[current_row][current_col] == 0:
                total_cleaned += 1
    
            grid[current_row][current_col] |= 1 << (current_direction + 1)
    
            next_position_row = current_row + MOVES[current_direction]
            next_position_col = current_col + MOVES[current_direction + 1]
            if (
                0 <= next_position_row < total_rows
                and 0 <= next_position_col < total_cols
                and grid[next_position_row][next_position_col] != 1
            ):
                current_row = next_position_row
                current_col = next_position_col
            else:
                current_direction = (current_direction + 1) % 4
    
        return total_cleaned

The Python code provided outlines a method to simulate the movement of a cleaning robot on a grid-based layout and count the number of unique cells it cleans. The approach uses a class RobotCleaner with a method countCleanRooms, which takes a grid as its parameter.

  • Use a bitwise operation to track the robot's movements and cleaned status on the grid, ensuring that each cell records not just its cleaned status, but also the directions from which it has been approached.
  • The robot can only clean unmarked cells and marks each cell as cleaned by updating its value in the grid using bitwise OR operation.
  • The robot follows a specific direction set by MOVES tuple until an obstacle (represented by '1' in the grid) or a grid boundary is encountered.
  • If the robot can't proceed in the current direction due to obstacles or boundaries, it will rotate clockwise, which is managed by adjusting the current_direction.
  • The process continues in loop until the robot revisits a cell in the same direction, meaning all possible paths have been cleaned or are blocked.
  • The function then returns the total count of unique cleaned cells.

This code efficiently keeps track of both cleaning coverage and movement direction, ensuring compact storage and avoiding redundant cleaning by maintaining each cell's state with respect to the directions from which it has been accessed and cleaned. All operations within the loop rely on boundary and state checking for decision making, ensuring proper navigation through the grid.

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