Walking Robot Simulation - Greedy - Easy - LeetCode
💻 coding

Walking Robot Simulation - Greedy - Easy - LeetCode

2 min read 383 words
2 min read
ShareWhatsAppPost on X
  • 1The robot starts at (0, 0) and can turn left, turn right, or move forward based on commands.
  • 2Obstacles prevent the robot from moving onto certain grid squares, causing it to remain in its previous position.
  • 3The solution calculates the maximum Euclidean distance squared from the origin based on the robot's movements and obstacles.

AI-generated summary · May not capture all nuances

Key Insight
AskGif

"The robot starts at (0, 0) and can turn left, turn right, or move forward based on commands."

Walking Robot Simulation - Greedy - Easy - LeetCode

A robot on an infinite grid starts at point (0, 0) and faces north. The robot can receive one of three possible types of commands:

-2: turn left 90 degrees -1: turn right 90 degrees 1 <= x <= 9: move forward x units Some of the grid squares are obstacles.

The i-th obstacle is at grid point (obstacles[i][0], obstacles[i][1])

If the robot would try to move onto them, the robot stays on the previous grid square instead (but still continues following the rest of the route.)

Return the square of the maximum Euclidean distance that the robot will be from the origin.

Example 1:

Input: commands = [4,-1,3], obstacles = [] Output: 25 Explanation: robot will go to (3, 4) Example 2:

Input: commands = [4,-1,4,-2,4], obstacles = [[2,4]] Output: 65 Explanation: robot will be stuck at (1, 4) before turning left and going to (1, 8)

Note:

0 <= commands.length <= 10000 0 <= obstacles.length <= 10000 -30000 <= obstacle[i][0] <= 30000 -30000 <= obstacle[i][1] <= 30000 The answer is guaranteed to be less than 2 ^ 31.

public class Solution {
 char face = 'N';
 public int RobotSim(int[] commands, int[][] obstacles) { 
 int x =0;
 int y = 0;
 int maxDistance = 0;
 var set = new HashSet<string>();
 for(int i=0;i<obstacles.Length;i++){
 set.Add(GetKey(obstacles[i][0],obstacles[i][1]));
 }
 
 for(int i=0;i<commands.Length;i++){
 if(commands[i]==-2){
 TurnLeft();
 }
 else if(commands[i]==-1){
 TurnRight();
 }
 else{
 for(int j=1;j<=commands[i];j++){
 int tx = x + GetValue('x');
 int ty = y + GetValue('y');
 if(set.Contains(GetKey(tx,ty))){
 break;
 }
 x = tx;
 y = ty;
 }
 }
 maxDistance = Math.Max(maxDistance, x*x+y*y);
 }
 
 return maxDistance;
 }
 
 private string GetKey(int x, int y){
 return x+" "+y;
 }
 
 private int GetValue(char ch){
 if(ch=='x'){
 if(face=='E'){
 return 1;
 }
 else if(face == 'W'){
 return -1;
 }
 }
 else if(ch=='y'){
 if(face=='N'){
 return 1;
 }
 else if(face == 'S'){
 return -1;
 }
 }
 return 0;
 }
 
 private void TurnLeft(){
 switch(face){
 case 'E':
 face = 'N';
 break;
 case 'W':
 face = 'S';
 break;
 case 'N':
 face = 'W';
 break;
 case 'S':
 face = 'E';
 break;
 }
 }
 
 private void TurnRight(){
 switch(face){
 case 'E':
 face = 'S';
 break;
 case 'W':
 face = 'N';
 break;
 case 'N':
 face = 'E';
 break;
 case 'S':
 face = 'W';
 break;
 }
 }
}

Time Complexity: O(n*m)

Space Complexity: O(n)

Where n is the number of inputs and m is the length of steps

Enjoyed this article?

Share it with someone who'd find it useful.

ShareWhatsAppPost on X

sumitc91

Published on 5 October 2020 · 2 min read · 383 words

Part of AskGif Blog · coding

You might also like