
Problem Statement
In this problem, a robot is placed on an infinite 2D grid at the origin faced north. It is programmed to follow a sequence of instructions provided as an array called commands
. Each instruction directs the robot to either turn or move. Specifically:
-2
instructs the robot to turn left by 90 degrees,-1
means turn right by 90 degrees,- Any positive integer between 1 and 9 inclusive tells the robot to move forward by that number of units.
The grid may contain obstacles at specified locations given in the obstacles
array, where each element is the coordinate (xi, yi)
of an obstacle. Encountering an obstacle stops the robot at its current position, ceasing its movement in that direction, but allowing subsequent commands to be processed.
The primary goal is to find the square of the maximum Euclidean distance from the origin the robot achieves at any point during its movement on this grid.
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 3^2 + 4^2 = 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 1^2 + 8^2 = 65 units away.
Example 3
Input:
commands = [6,-1,-1,6], obstacles = [[0,0]]
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 5 units and get blocked by the obstacle at (0,0), robot is at (0, 1). The furthest point the robot ever gets from the origin is (0, 6), which squared is 6^2 = 36 units away.
Constraints
1 <= commands.length <= 10^4
commands[i]
is either-2
,-1
, or an integer in the range[1, 9]
.0 <= obstacles.length <= 10^4
-3 * 10^4 <= xi, yi <= 3 * 10^4
- The answer is guaranteed to be less than
2^31
.
Approach and Intuition
Understanding the movement and turning mechanics is key to solving this problem effectively. Based on the constraints and the input examples given, we can define a clear strategy.
Begin at the origin
(0, 0)
, initially facing north.Use a direction array to handle turning operations. Each entry in this direction array represents a compass point (
N, E, S, W
).For each command:
- If the command is
-2
or-1
, update the robot’s facing direction using modular arithmetic. - If the command is a positive integer, attempt to move the robot forward, step-by-step, checking for obstacles at each step.
- If the command is
Whenever a movement command is processed, calculate the squared Euclidean distance of the robot's position from the origin and track the maximum distance observed.
Handle obstacles by marking the robot’s current position (pre-obstacle) and skipping the remainder of the movement command.
Example Exploration:
The examples help illustrate this approach:
- In the first example, the robot straightforwardly moves north and east without encountering obstacles, resulting in a maximum distance calculation.
- The second example introduces an obstacle which modifies the planned path of the robot. Even though obstructed, the robot's maximum distance calculation is based on the furthest point reached before turning.
- The third example features an initial obstacle at the origin, challenging the robot's return path and demonstrating the mechanism to skip over an initially occupied origin.
Through these steps and explanations, the robot's path computation and maximum distance calculation can be implemented efficiently within the provided constraints.
Solutions
- C++
- Java
- Python
class Solution {
private:
static const long long HASH_FACTOR = 60013; // Hash factor coefficient
// Encodes (x, y) tuple into a unique long long hash
long long hashPosition(long long x, long long y) {
return x + HASH_FACTOR * y;
}
public:
int robotSim(vector<int>& commands, vector<vector<int>>& obstacles) {
unordered_set<long long> obsSet; // set to hold the obstacles hashes
for (const auto& obs : obstacles) {
obsSet.insert(hashPosition(obs[0], obs[1]));
}
vector<vector<int>> dirs = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}}; // directional vectors
vector<int> pos = {0, 0}; // start at origin
int maxDist2 = 0; // initialize max distance squared
int dirIndex = 0; // starting direction index
for (int cmd : commands) {
if (cmd == -1) {
dirIndex = (dirIndex + 1) % 4; // turn right
continue;
}
if (cmd == -2) {
dirIndex = (dirIndex + 3) % 4; // turn left
continue;
}
vector<int> dir = dirs[dirIndex]; // get current direction
for (int steps = 0; steps < cmd; steps++) {
int newX = pos[0] + dir[0];
int newY = pos[1] + dir[1];
if (obsSet.find(hashPosition(newX, newY)) != obsSet.end()) {
break; // obstacle encountered
}
pos[0] = newX;
pos[1] = newY; // update position
}
maxDist2 = max(maxDist2, pos[0] * pos[0] + pos[1] * pos[1]); // update max distance squared
}
return maxDist2;
}
};
The provided C++ solution for the "Walking Robot Simulation" problem implements a robot's movement on a grid where obstacles can block its path. The goal is to calculate the robot's maximum Euclidean distance squared from the origin (0, 0), considering commands for movements and turns.
The solution involves these key components:
Hashing Positions:
- A private method
hashPosition
creates a unique hash for each (x, y) position using a predetermined hash factor. This hash is used to efficiently check for obstacles during simulation.
- A private method
Tracking Obstacles:
- Obstacles are given as a list of positions and are stored in a
unordered_set
calledobsSet
after hashing withhashPosition
. This allows quick lookup to check if a new position has an obstacle.
- Obstacles are given as a list of positions and are stored in a
Processing Commands:
- The robot can turn left, turn right, or move forward. The simulation handles these commands as follows:
-1
turns the robot right by shifting the direction clockwise.-2
turns the robot left by shifting the direction counter-clockwise.- Positive numbers move the robot forward in the current direction for the number of units specified by the command.
- The robot can turn left, turn right, or move forward. The simulation handles these commands as follows:
Calculating Movement:
- Movements are determined by directional vectors stored in
dirs
. For each move command, the robot attempts to move step by step in the current direction. If it encounters an obstacle, the move is halted prematurely.
- Movements are determined by directional vectors stored in
Tracking Maximum Distance:
- After each command, the robot's distance squared from the origin is recalculated. The program retains the maximum value encountered during the entire sequence of commands.
The algorithm concludes by returning the maximum distance squared that the robot has been from the starting point.
This approach allows for efficient simulation of the robot's movements using directional vectors and hashed positions of obstacles, optimizing the check for obstacles and updating the position and direction of the robot effectively.
class Solution {
private static final long HASH_BASE = 60013;
public int robotSimulation(int[] movements, int[][] blocks) {
Set<Long> blockSet = new HashSet<>();
for (int[] block : blocks) {
blockSet.add(encodePosition(block[0], block[1]));
}
int[][] navigation = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
int[] pos = {0, 0};
int greatestDistanceSquared = 0;
int dir = 0;
for (int move : movements) {
if (move == -1) {
dir = (dir + 1) % 4;
continue;
}
if (move == -2) {
dir = (dir + 3) % 4;
continue;
}
int[] vector = navigation[dir];
for (int step = 0; step < move; step++) {
int nextX = pos[0] + vector[0];
int nextY = pos[1] + vector[1];
if (blockSet.contains(encodePosition(nextX, nextY))) {
break;
}
pos[0] = nextX;
pos[1] = nextY;
}
greatestDistanceSquared = Math.max(
greatestDistanceSquared,
pos[0] * pos[0] + pos[1] * pos[1]
);
}
return greatestDistanceSquared;
}
private long encodePosition(long x, long y) {
return x + HASH_BASE * y;
}
}
The provided Java solution involves simulating the movement of a robot on a grid where certain positions are blocked, and computes the highest distance squared that the robot reaches from the origin. The movements are represented by an array of integers, and the blocks are given as coordinates in a 2D array. Specifically:
- Initialize a HashSet to keep track of blocked positions, using a custom hashing function to encode 2D coordinates into a single long integer.
- Define directions for movement based on compass bearings: north, east, south, and west.
- Navigate based on the robot's movements. Direction changes are handled with modulo operations to cycle between the four possible directions.
- Check for blocks before moving. If the next position is blocked, the robot does not move to that position.
- Calculate the squared distance from the origin after each valid move. The solution outputs the maximum of these squared distances.
Use the algorithm efficiently apply direction changes, handle blocked paths, and track the furthest distance reached, yielding a solution to the problem of simulating a walking robot's movements on a potentially obstructed grid.
class Solution:
def __init__(self):
self.PRIME_MULTIPLIER = 60013 # A prime larger than 2x maximum coordinate
def simulateRobot(self, instructions: List[int], blocks: List[List[int]]) -> int:
# Convert obstacles to a set for quick access
obstacle_set = {self._hash_position(i, j) for i, j in blocks}
# Direction vectors corresponding to North, East, South, and West
direction_vectors = [(0, 1), (1, 0), (0, -1), (-1, 0)]
x0, y0 = 0, 0
furthest_distance_squared = 0
facing_direction = 0
for inst in instructions:
if inst == -1: # Turn 90 deg right
facing_direction = (facing_direction + 1) % 4
continue
if inst == -2: # Turn 90 deg left
facing_direction = (facing_direction + 3) % 4
continue
# Execute forward motion
step_x, step_y = direction_vectors[facing_direction]
for _ in range(inst):
new_x, new_y = x0 + step_x, y0 + step_y
if self._hash_position(new_x, new_y) in obstacle_set:
break
x0, y0 = new_x, new_y
furthest_distance_squared = max(furthest_distance_squared, x0**2 + y0**2)
return furthest_distance_squared
# Utility to hash (x, y) as a unique number
def _hash_position(self, ix: int, iy: int) -> int:
return ix + self.PRIME_MULTIPLIER * iy
This Python solution involves simulating a walking robot’s movement given a set of instructions and obstacles. A Solution
class is defined, containing methods to process these instructions and handle the simulation mechanics. Here’s an overview of how this solution works:
Initialization:
PRIME_MULTIPLIER
is set to a large prime number which is used in hashing obstacle coordinates for efficient look-up.
Simulate Robot Process:
- Obstacles are converted into a set using a hashing function for quick access.
- Direction vectors are established for North, East, South, and West to guide the robot's movement.
- The robot starts at the coordinate (0, 0).
furthest_distance_squared
keeps track of the furthest distance the robot reaches from the start (squared, to avoid unnecessary computation of square root). - It iterates over each instruction:
-1
rotates the facing direction to the right.-2
rotates the facing direction to the left.- Positive numbers move the robot forward in the current facing direction, checking for obstacles. If an obstacle is encountered, the movement in that direction stops.
- After moving, the robot calculates the squared distance from the origin and updates
furthest_distance_squared
if this is the maximum distance encountered.
Auxiliary Function:
_hash_position
hashes coordinates for efficient tracking and checking of obstacle locations.
Output:
- The function
simulateRobot
returns the square of the maximum distance from the starting position, solely based on provided valid movements and detected obstacles.
- The function
This implementation uses several efficient programming techniques like hashing and modular arithmetic for rotation handling, making it highly suitable for scenarios with complexities such as a grid full of obstacles and a long list of mixed movement and rotation instructions. The solution ensures efficient computation and clarity, implementing direct operations based on the current state without needing to backtrack or recompute previous states.
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