How to Solve the Robot Bounded In Circle Challenge in Java

The challenge

On an infinite plane, a robot initially stands at (0, 0) and faces north.  The robot can receive one of three instructions:

  • "G": go straight 1 unit;
  • "L": turn 90 degrees to the left;
  • "R": turn 90 degress to the right.

The robot performs the instructions given in order, and repeats them forever.

Return true if and only if there exists a circle in the plane such that the robot never leaves the circle.

Example 1:

Input: "GGLLGG"
Output: true
Explanation: 
The robot moves from (0,0) to (0,2), turns 180 degrees, and then returns to (0,0).
When repeating these instructions, the robot remains in the circle of radius 2 centered at the origin.

Example 2:

Input: "GG"
Output: false
Explanation: 
The robot moves north indefinitely.

Example 3:

Input: "GL"
Output: true
Explanation: 
The robot moves from (0, 0) -> (0, 1) -> (-1, 1) -> (-1, 0) -> (0, 0) -> ...

Note:

  1. 1 <= instructions.length <= 100
  2. instructions[i] is in {'G', 'L', 'R'}

The solution in Java

First it is important to validate our inputs.

Then we can create some variables to store the x, y (for coordinates) and a further dir to store the direction we are travelling in.

Looping through the list of instructions and checking for which coordinate we need to adjust as well as turning left or right allows for a simple complete block of code.

We can determine a successful robotic circle if x and y are both 0, or if dir is not 0.

class Solution { public boolean isRobotBounded(String instructions) { // perform some initial sanity checks if (instructions.length()>=1 && instructions.length()<=100) { // create `x`, `y` and `dir` variables // for the coordinate system, as well as the direction int x = 0; int y = 0; int dir = 0; // loop through all instructions for (int i=0; i<instructions.length(); i++) { // get the current instruction char ch = instructions.charAt(i); if (ch=='G') { // if Going Forward // _____ // | 0 | // |3 + 1| // |__2__| // // works like a clock: // 0==up (y++) // 1==right (x++) // 2==down (y--) // 3==left (x--) if (dir==0) y++; if (dir==3) x--; if (dir==2) y--; if (dir==1) x++; } else if (ch=='L') { // change direction if left dir = (dir+5)%4; } else if (ch=='R') { // change direction if right dir = (dir+3)%4; } } return (x==0 && y==0) ? true : (dir!=0); } else return false; } }
Code language: Java (java)

This solution is also performed in O(n) time complexity.

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Tee Jay
Tee Jay
5 months ago

dir = (dir+5)%4;

Can you explain the logic behind this?

Anonym
Anonym
4 months ago
Reply to  Tee Jay

No logic because its obvious mistake
They should’ve replace ch=’L’ with ‘R’ and vice versa