Number of Spaces Cleaning Robot Cleaned

#818 Most AskedMedium

Number of Spaces Cleaning Robot Cleaned #818 Most Asked

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You have a room in the shape of a grid with n rows and m columns. Initially, all cells are dirty. You have a cleaning robot that starts at cell (r, c) and faces right.

The robot can perform the following actions:

Advance: Clean the current cell and move one step in the direction the robot is facing. If the movement would take the robot out of the room, the robot stays on the current cell.
Turn: Change the robot's direction to the next clockwise direction.

It is guaranteed that the room is rectangular.

The robot repeats the actions in the following order: "Advance, Turn, Advance, Turn, Advance, Turn, Advance, Turn".

Return the number of unique cells cleaned by the robot.

Example 1:

Input: n = 2, m = 2, r = 0, c = 0
Output: 4
Explanation: The robot cleans all the cells.
The following animation explains the process:

Example 2:

Input: n = 2, m = 3, r = 0, c = 1
Output: 5
Explanation: The robot cleans all the cells except cell (1, 0).
The following animation explains the process:

Constraints:

1 <= n, m <= 10⁵
0 <= r < n
0 <= c < m

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the constraints on the dimensions 'm' and 'n' of the room (e.g., maximum size)?
Can the `room` array contain null or empty rows?
What should be returned if the input `room` is null or empty, or if 'm' or 'n' are zero?
Is the robot allowed to revisit previously cleaned spaces (and if so, does revisiting count towards the total number of cleaned spaces)?
Is the room guaranteed to have a path from the starting point (0,0) to all other clean spaces, or are there potentially isolated clean spaces that the robot cannot reach?

Brute Force Solution

Approach

The brute force approach to this problem involves simulating the robot's cleaning process for every possible path it could take. We explore all cleaning sequences until we find the one that minimizes revisiting already-cleaned spaces. In essence, we try everything and pick the best result.

Here's how the algorithm would work step-by-step:

Start the robot at its initial position.
Consider all possible directions the robot could move in (up, down, left, right).
For each direction, imagine the robot takes that step and cleans the new space.
Keep track of all the spaces the robot has cleaned so far.
If a potential move leads the robot to a space it has already cleaned, record that the robot had to revisit an area.
Repeat this process, exploring all possible sequences of moves until the robot has cleaned a certain number of spaces (or a certain number of moves have been made).
After exploring all paths, compare the number of times the robot had to revisit cleaned spaces for each path.
The path with the fewest revisits is the answer we are looking for.

Code Implementation

def number_of_spaces_cleaning_robot_cleaned(room, start_row, start_column):
    number_of_rows = len(room)
    number_of_columns = len(room[0])
    cleaned_spaces = set()

    def clean_spaces(current_row, current_column):
        if (current_row, current_column) in cleaned_spaces:
            return

        if current_row < 0 or current_row >= number_of_rows or \
           current_column < 0 or current_column >= number_of_columns or \
           room[current_row][current_column] == 'obstacle':
            return

        cleaned_spaces.add((current_row, current_column))

        # Recursively explore all possible directions from current spot

        clean_spaces(current_row + 1, current_column) # Move Down

        clean_spaces(current_row - 1, current_column) # Move Up

        clean_spaces(current_row, current_column + 1) # Move Right

        clean_spaces(current_row, current_column - 1) # Move Left

    # Initiate the cleaning process from the start position
    clean_spaces(start_row, start_column)

    return len(cleaned_spaces)

Big(O) Analysis

Time Complexity

O(4^n) – The brute force approach explores all possible paths the robot can take. At each step, the robot has up to four choices (up, down, left, right). Assuming the robot takes n steps (or cleans n spaces before termination), the algorithm explores all paths of length n. This leads to 4 possible choices at each of the n steps, resulting in approximately 4^n possible paths. Therefore, the time complexity is O(4^n).

Space Complexity

O(4^N) – The brute-force approach explores all possible paths of the robot. Since the robot can move in four directions (up, down, left, right), and we are exploring paths up to a certain length (related to the number of spaces to be cleaned, denoted as N, representing the number of moves or spaces cleaned), the recursion tree can have a maximum depth related to N. At each level of the recursion, we need to store the current path (sequence of moves) and the set of visited spaces. Storing the current path contributes O(N) space at each level, and the set of visited spaces can grow up to O(N) as well. Since we are exploring 4 possible directions at each step, the space required becomes O(4^N * N), but the dominant term is the exponential factor. Therefore, the space complexity is O(4^N).

Optimal Solution

Approach

The cleaning robot problem requires finding the unique spaces cleaned. The optimal approach involves simulating the robot's movement step-by-step and marking the spaces it cleans. We leverage a set to efficiently track and count only the unique spaces cleaned, avoiding redundant counting of previously cleaned spaces.

Here's how the algorithm would work step-by-step:

Keep track of all the spots the robot cleans in a special list that only remembers unique locations.
Start the robot at its initial location.
Simulate each movement the robot makes based on the instructions it's given.
For each move, mark the new spot where the robot ends up.
Before marking a spot, check if it's already in the list of cleaned spots.
If the spot isn't in the list, add it to the list.
Once the robot has finished all its movements, count how many spots are in the cleaned spots list. This is the total number of unique spaces cleaned.

Code Implementation

def calculate_cleaned_spaces(instructions):
    cleaned_spaces = set()
    robot_row = 0
    robot_column = 0

    for instruction in instructions:
        if instruction == 'U':
            robot_row -= 1
        elif instruction == 'D':
            robot_row += 1
        elif instruction == 'L':
            robot_column -= 1
        else:
            robot_column += 1

        # Use a tuple to represent coordinates, 
        # since sets require immutable elements
        current_position = (robot_row, robot_column)

        # Check if the robot has already cleaned this space
        if current_position not in cleaned_spaces:
            cleaned_spaces.add(current_position)

    # Return the total number of unique spaces cleaned
    return len(cleaned_spaces)

Big(O) Analysis

Time Complexity

O(n) – The time complexity is determined by the number of movements, denoted as n, which directly corresponds to the number of instructions the robot follows. For each instruction, the robot moves and its new location is added to a set. Adding to a set and checking for existence in the set takes O(1) on average. Therefore, the overall time complexity is driven by iterating through the n instructions, resulting in O(n) time complexity.

Space Complexity

O(N) – The algorithm uses a set to keep track of the unique spaces the robot has cleaned. In the worst-case scenario, the robot could visit N unique spaces, where N is the number of movements the robot makes. Therefore, the size of the set can grow up to N. Thus, the auxiliary space complexity is O(N).

Edge Cases

Case	How to Handle
Null or empty room (m=0 or n=0 or room is null)	Return 0 immediately since there are no spaces to clean.
Room with only one cell (m=1, n=1)	Return 1 if the cell is clean (true), otherwise return 0.
Room where all cells are obstacles (all false)	Return 0 since the robot cannot move or clean any spaces.
Room where all cells are clean (all true)	The robot should visit all cells, so return m * n.
Room with a single path from start to end, no branching	The robot will only clean the spaces along that path, so count the clean spaces in that path.
Large room (large m and n) that could cause stack overflow with recursive DFS	Use iterative DFS (stack) or BFS to avoid exceeding the stack limit.
Robot cannot reach all clean spaces (isolated clean spaces)	The solution only counts the clean spaces reachable from the starting point.
Obstacles completely block the path back to starting point after exploring far away sections	The algorithm correctly explores as much of the connected component as possible without issue.

Null or empty room (m=0 or n=0 or room is null)

How to Handle:

Return 0 immediately since there are no spaces to clean.

Room with only one cell (m=1, n=1)

How to Handle:

Return 1 if the cell is clean (true), otherwise return 0.

Room where all cells are obstacles (all false)

How to Handle:

Return 0 since the robot cannot move or clean any spaces.

Room where all cells are clean (all true)

How to Handle:

The robot should visit all cells, so return m * n.

Room with a single path from start to end, no branching

How to Handle:

The robot will only clean the spaces along that path, so count the clean spaces in that path.

Large room (large m and n) that could cause stack overflow with recursive DFS

How to Handle:

Use iterative DFS (stack) or BFS to avoid exceeding the stack limit.

Robot cannot reach all clean spaces (isolated clean spaces)

How to Handle:

The solution only counts the clean spaces reachable from the starting point.

Obstacles completely block the path back to starting point after exploring far away sections

How to Handle:

The algorithm correctly explores as much of the connected component as possible without issue.