1041. Robot Bounded In Circle

Description

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'.

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Solutions

Solution 1

Intuition

if back to start point or direction is not to up/north after one set of instructions, it means it returns true

Code

class Solution {

    private final int[][] DIRECTIONS = new int[][] {
            { 1, 0 }, // up
            { 0, 1 }, // right
            { -1, 0 }, // down
            { 0, -1 }// left
    };

    public boolean isRobotBounded(String instructions) {
        // start point
        int x = 0, y = 0;
        // default direction index is 0, means go up
        int index = 0;
        for (char instruction : instructions.toCharArray()) {
            switch (instruction) {
                case 'L':
                    // index = 3 means go left
                    index = (index + 3) % 4;
                    break;
                case 'R':
                    // index = 1 means go right
                    index = (index + 1) % 4;
                    break;
                default:
                    // move towards direction
                    x += DIRECTIONS[index][0];
                    y += DIRECTIONS[index][1];
            }
        }
        // back start point or direction is not go up
        return x == 0 && y == 0 || index != 0;
    }
}

Complexity

• Time: O(n)
• Space: O(1)