#874

Medium

auto_awesomeArray scanning plus hash lookup

LeetCode Problem Workspace

Walking Robot Simulation

The Walking Robot Simulation problem involves moving a robot in an infinite grid and calculating its furthest distance from the origin, avoiding obstacles.

array hash table simulation

Problem Statement

A robot starts at the origin (0, 0) on an infinite XY-plane, facing north. The robot follows a sequence of commands, where each command is an integer representing movement in one of four directions. The robot can face north (up), east (right), south (down), or west (left). Commands are either numbers indicating steps or negative values that instruct the robot to turn 90 degrees clockwise. Some grid points contain obstacles, and if the robot encounters an obstacle, it will stop moving in that direction and proceed with the next command.

The problem asks you to return the maximum squared Euclidean distance from the origin the robot reaches during its journey. If the distance from the origin is 5, the result should be 25. Your solution must handle large grids and multiple obstacles efficiently, ensuring that the time complexity is manageable even for larger inputs.

Examples

Example 1

Input: commands = [4,-1,3], obstacles = []

Output: 25

The robot starts at (0, 0) : 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

The robot starts at (0, 0) : 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

The robot starts at (0, 0) : 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 <= 104
commands[i] is either -2, -1, or an integer in the range [1, 9].
0 <= obstacles.length <= 104
-3 * 104 <= xi, yi <= 3 * 104
The answer is guaranteed to be less than 231.

Solution Approach

Array Scanning with Directional Movement

To solve this problem, you simulate the robot's movement by processing each command in sequence. The robot's position is updated based on the current facing direction, adjusting for turns and obstacles. The key idea is to keep track of the robot’s position and calculate the squared Euclidean distance after each movement. You can efficiently update the robot's position by considering the direction it faces and modifying the coordinates accordingly.

Hash Table for Obstacle Lookup

To efficiently check for obstacles, store obstacle coordinates in a hash table. This allows constant-time lookups for obstacles during each step of the robot's movement. By checking whether the next position is blocked before proceeding, you can prevent the robot from running into obstacles while still processing the commands in sequence.

Maximizing Squared Distance

After each move, calculate the squared Euclidean distance from the origin. This is done by simply squaring the x and y coordinates of the robot's position and adding them together. Track the maximum squared distance as you process each move, ensuring that you return the greatest distance the robot achieves.

Complexity Analysis

Metric	Value
Time	O(m + n)
Space	O(n)

The time complexity of this solution is O(m + n), where m is the number of commands and n is the number of obstacles. The space complexity is O(n) due to the hash table storing the obstacles. This is efficient enough for the input limits provided, allowing the solution to handle the maximum input sizes within the constraints.

What Interviewers Usually Probe

Test the candidate's ability to break down movement into discrete steps and efficiently track position.
Look for knowledge of hash tables and how they can optimize obstacle detection.
Assess the candidate’s understanding of Euclidean distance and its application to this problem.

Common Pitfalls or Variants

Common pitfalls

Failing to correctly handle turning directions or obstacles can cause incorrect movement simulation.
Not using a hash table for obstacles may lead to inefficient checks and higher time complexity.
Misunderstanding the problem’s requirement to return the squared Euclidean distance rather than the distance itself.

Follow-up variants

Consider different grid sizes or obstacles positioned in challenging ways.
Experiment with larger command sequences and test the performance of your solution.
Allow the robot to face multiple directions with different step sizes to introduce additional complexity.

FAQ

What is the key pattern in the Walking Robot Simulation problem?

The problem mainly relies on array scanning combined with hash table lookups to track robot movement and detect obstacles.

How do I efficiently simulate robot movement in this problem?

You simulate movement by updating the robot’s position according to commands while checking for obstacles using a hash table for efficient lookups.

Why do we return the squared Euclidean distance in this problem?

The squared Euclidean distance is requested to avoid the unnecessary computational cost of taking square roots, as it provides the same relative comparison.

What should I do if the robot runs into an obstacle?

If the robot encounters an obstacle, it should stop moving in that direction and proceed with the next command.

How do I handle multiple obstacles in the grid?

Store all obstacle positions in a hash table, allowing for efficient constant-time checks to see if the robot’s next position is blocked.

terminal

Solution

Solution 1: Hash Table + Simulation

We define a direction array $dirs = [0, 1, 0, -1, 0]$ of length $5$, where each pair of adjacent elements represents a direction. That is, $(dirs[0], dirs[1])$ represents north, $(dirs[1], dirs[2])$ represents east, and so on.

class Solution:
    def robotSim(self, commands: List[int], obstacles: List[List[int]]) -> int:
        dirs = (0, 1, 0, -1, 0)
        s = {(x, y) for x, y in obstacles}
        ans = k = 0
        x = y = 0
        for c in commands:
            if c == -2:
                k = (k + 3) % 4
            elif c == -1:
                k = (k + 1) % 4
            else:
                for _ in range(c):
                    nx, ny = x + dirs[k], y + dirs[k + 1]
                    if (nx, ny) in s:
                        break
                    x, y = nx, ny
                    ans = max(ans, x * x + y * y)
        return ans

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