Leetcode #489: Robot Room Cleaner
In this guide, we solve Leetcode #489 Robot Room Cleaner in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Hard
- Premium: Yes
- Tags: Backtracking, Interactive
Intuition
We must explore combinations of choices, but many branches can be pruned early.
Backtracking enumerates valid candidates while keeping the search space under control.
Approach
Use DFS to build candidates step by step, and backtrack when constraints are violated.
Pruning keeps the exploration practical for typical constraints.
Steps:
- Define the decision tree.
- DFS through choices and backtrack.
- Prune invalid paths early.
Example
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();
}
Python Solution
# """
# This is the robot's control interface.
# You should not implement it, or speculate about its implementation
# """
# class Robot:
# def move(self):
# """
# Returns true if the cell in front is open and robot moves into the cell.
# Returns false if the cell in front is blocked and robot stays in the current cell.
# :rtype bool
# """
#
# def turnLeft(self):
# """
# Robot will stay in the same cell after calling turnLeft/turnRight.
# Each turn will be 90 degrees.
# :rtype void
# """
#
# def turnRight(self):
# """
# Robot will stay in the same cell after calling turnLeft/turnRight.
# Each turn will be 90 degrees.
# :rtype void
# """
#
# def clean(self):
# """
# Clean the current cell.
# :rtype void
# """
class Solution:
def cleanRoom(self, robot):
"""
:type robot: Robot
:rtype: None
"""
def dfs(i, j, d):
vis.add((i, j))
robot.clean()
for k in range(4):
nd = (d + k) % 4
x, y = i + dirs[nd], j + dirs[nd + 1]
if (x, y) not in vis and robot.move():
dfs(x, y, nd)
robot.turnRight()
robot.turnRight()
robot.move()
robot.turnRight()
robot.turnRight()
robot.turnRight()
dirs = (-1, 0, 1, 0, -1)
vis = set()
dfs(0, 0, 0)
Complexity
The time complexity is Exponential (worst case). The space complexity is O(depth).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.