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RobotRoomCleaner.java
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RobotRoomCleaner.java
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package depth_first_search;
import java.util.HashSet;
import java.util.Objects;
import java.util.Set;
/**
* Created by gouthamvidyapradhan on 08/03/2019
Given a robot cleaner in a room modeled as a grid.
Each cell in the grid can be empty or blocked.
The robot cleaner with 4 given APIs can move forward, turn left or turn right. Each turn it made is 90 degrees.
When it tries to move into a blocked cell, its bumper sensor detects the obstacle and it stays on the current cell.
Design an algorithm to clean the entire room using only the 4 given APIs shown below.
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();
}
Example:
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
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.
Notes:
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 mentioned 4 APIs, without knowing the room
layout and the initial robot's position.
The robot's initial position will always be in an accessible cell.
The initial direction of the robot will be facing up.
All accessible cells are connected, which means the all cells marked as 1 will be accessible by the robot.
Assume all four edges of the grid are all surrounded by wall.
*
*
* Solution: O(N x M)
* Maintain a direction and position of robot in each cell and perform a dfs to clean all rooms.
* Important to note here is that during call back in the dfs recursion the robot has to return back to its original
* cell and orientation
*/
public class RobotRoomCleaner {
//direction
//UP 0, LEFT = 1, DOWN = 2, RIGHT = 3
private final int[] R = {-1, 0, 1, 0};
private final int[] C = {0, -1, 0, 1};
interface Robot {
// 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.
public boolean move();
// Robot will stay in the same cell after calling turnLeft/turnRight.
// Each turn will be 90 degrees.
public void turnLeft();
public void turnRight();
// Clean the current cell.
public void clean();
}
static class Position{
int r, c;
Position(int r, int c){
this.r = r;
this.c = c;
}
@Override
public boolean equals(Object obj) {
int r = ((Position)obj).r;
int c = ((Position)obj).c;
return (this.r == r && this.c == c);
}
@Override
public int hashCode() {
return Objects.hash(r, c);
}
}
private static Set<Position> done;
/**
* Main method
* @param args
*/
public static void main(String[] args) {
}
public void cleanRoom(Robot robot) {
done = new HashSet<>();
dfs(1, 3, done, robot, 0);
}
private void dfs(int r, int c, Set<Position> done, Robot robot, int direction){
done.add(new Position(r, c));
robot.clean();
for(int i = 0; i < 4; i ++){
int newR = r + R[direction];
int newC = c + C[direction];
if(!done.contains(new Position(newR, newC))){
boolean possible = robot.move();
if(possible){
dfs(newR, newC, done, robot, direction);
}
}
robot.turnLeft();
direction = (direction + 1) % 4;
}
robot.turnLeft();
robot.turnLeft();
robot.move();
robot.turnLeft();
robot.turnLeft();
}
}