Skip to content

1041. Robot Bounded in Circle

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 degrees 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: instructions = "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: instructions = "GG"
Output: false
Explanation: The robot moves north indefinitely.

Example 3:

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

Constraints:

  • 1 <= instructions.length <= 100
  • instructions[i] is 'G', 'L' or, 'R'.

Analysis

  1. if after mimicing the instructions ends up with x = 0, y = 0, that means it's already a circle, so we return true.
  2. if after mimicing we ends up with a direction other than going up (if it's up, then that is the only direcion and we cannot go back), we can still make it a circle

Code

class Solution {
public:
    bool isRobotBounded(string instructions) {
        int sx = 0, sy = 0;
        int dir = 0; // 0: up, 1: left, 2: down, 3: right
        for (char c : instructions) {
            if (c == 'G') {
                if (dir == 0) sx ++;
                else if (dir == 1) sy --;
                else if (dir == 3) sy ++;
                else sx --;
            } else if (c == 'L')  dir = (dir + 1) % 4;
            else dir = (dir + 3) % 4;
        }
        return sx == 0 && sy == 0 || dir > 0;
    }
};
Last update: March 31, 2022