Problem
A robot on an infinite XY-plane starts at point (0, 0)
facing north. The robot can receive a sequence of these three possible types of commands
:
-2
: Turn left90
degrees.-1
: Turn right90
degrees.1 <= k <= 9
: Move forwardk
units, one unit at a time.
Some of the grid squares are obstacles
. The ith
obstacle is at grid point obstacles[i] = (xi, yi)
. If the robot runs into an obstacle, then it will instead stay in its current location and move on to the next command.
Return the maximum Euclidean distance that the robot ever gets from the origin squared (i.e. if the distance is 5
, return 25
).
Note:
- North means +Y direction.
- East means +X direction.
- South means -Y direction.
- West means -X direction.
- There can be obstacle in
[0,0]
.
Examples
Example 1:
Input: commands = [4,-1,3], obstacles = []
Output: 25
Explanation: The robot starts at (0, 0):
1. Move north 4 units to (0, 4).
2. Turn right.
3. Move east 3 units to (3, 4).
The furthest point the robot ever gets from the origin is (3, 4), which squared is 32 + 42 = 25 units away.
Example 2:
Input: commands = [4,-1,4,-2,4], obstacles =[[2,4]]
Output: 65
Explanation: The robot starts at (0, 0):
1. Move north 4 units to (0, 4).
2. Turn right.
3. Move east 1 unit and get blocked by the obstacle at (2, 4), robot is at (1, 4).
4. Turn left.
5. Move north 4 units to (1, 8).
The furthest point the robot ever gets from the origin is (1, 8), which squared is 12 + 82 = 65 units away.
Example 3:
Input: commands = [6,-1,-1,6], obstacles = []
Output: 36
Explanation: The robot starts at (0, 0):
1. Move north 6 units to (0, 6).
2. Turn right.
3. Turn right.
4. Move south 6 units to (0, 0).
The furthest point the robot ever gets from the origin is (0, 6), which squared is 62 = 36 units away.
Solution
Method 1 - Run the simulation
Here are the steps we can take:
- Track Directions using an array (North, East, South, West).
- Handle Commands to adjust the direction or move the robot.
- Use a Set to store obstacles for fast lookup.
- Calculate and Track the maximum Euclidean distance squared.
Video explanation
Here is the video explanation:
Code
Java
public class Solution {
public int robotSim(int[] commands, int[][] obstacles) {
// Convert obstacles list to a set of strings for faster lookup
Set<String> obstacleSet = new HashSet<>();
for (int[] obstacle : obstacles) {
obstacleSet.add(obstacle[0] + "," + obstacle[1]);
}
// Directions are in the order of North, East, South, West
int[][] directions = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
int directionIdx = 0; // Start facing north
int x = 0, y = 0; // Robot's starting position
int maxDistanceSq = 0;
for (int command : commands) {
if (command == -1) { // Turn right 90 degrees
directionIdx = (directionIdx + 1) % 4;
} else if (command == -2) { // Turn left 90 degrees
directionIdx = (directionIdx + 3) % 4;
} else { // Move forward k units
int dx = directions[directionIdx][0];
int dy = directions[directionIdx][1];
int steps = 0;
while (steps < command) {
int nextX = x + dx;
int nextY = y + dy;
// Check for obstacle
if (obstacleSet.contains(nextX + "," + nextY)) {
break; // Hit an obstacle, stop moving forward
}
x = nextX;
y = nextY;
// Calculate the current distance squared and update
// maxDistanceSq
maxDistanceSq = Math.max(maxDistanceSq, x * x + y * y);
steps++;
}
}
}
return maxDistanceSq;
}
}
Python
class Solution:
def robotSim(self, commands: List[int], obstacles: List[List[int]]) -> int:
# Convert obstacles list to a set of tuples for faster lookup
obstacle_set = set(map(tuple, obstacles))
# Directions are in the order of North, East, South, West
directions = [(0, 1), (1, 0), (0, -1), (-1, 0)]
direction_idx = 0 # Start facing north
x, y = 0, 0 # Robot's starting position
max_distance_sq = 0
for command in commands:
if command == -1: # Turn right 90 degrees
direction_idx = (direction_idx + 1) % 4
elif command == -2: # Turn left 90 degrees
direction_idx = (direction_idx + 3) % 4
else: # Move forward k units
dx, dy = directions[direction_idx]
steps = 0
while steps < command:
next_x = x + dx
next_y = y + dy
# Check for obstacle
if (next_x, next_y) in obstacle_set:
break # Hit an obstacle, stop moving forward
x, y = next_x, next_y
# Calculate the current distance squared and update max_distance_sq
max_distance_sq = max(max_distance_sq, x * x + y * y)
steps += 1
return max_distance_sq
Complexity
- ⏰ Time complexity:
O(o + c)
whereo
is number of obstacles andc
is number of commands.- Creating obstacle set takes
O(o)
time - Processing all commands takes
O(c)
time
- Creating obstacle set takes
- 🧺 Space complexity:
O(o)
for storing obstacles.