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: trueExplanation: 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: falseExplanation: The robot moves north indefinitely.`

Example 3:

`Input: instructions = "GL"Output: trueExplanation: 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'`.
`var isRobotBounded = function (instructions) {    const origin = [0,0]    let dir = 0;        const moving = () => {        for(let str of instructions){            if(str === "L"){                dir -= 1                if(dir < 0) dir += 4;            } else if (str === 'R') {                dir += 1                if(dir > 3) dir = dir % 4;            } else {                if(dir === 0){                    origin[1] += 1                } else if(dir === 1) {                    origin[0] -= 1                } else if(dir === 2) {                    origin[1] -= 1                } else {                    origin[0] += 1                }            }        }    }        let times = 0;    while(times < 4){        moving();        if(origin[0] === 0 && origin[1] === 0) return true        times += 1    }    return false};`

Approach 1: One Pass

Intuition

This solution is based on two facts about the limit cycle trajectory.

• After at most 4 cycles, the limit cycle trajectory returns to the initial point `x = 0, y = 0`. That is related to the fact that 4 directions (north, east, south, west) define the repeated cycles' plane symmetry.

Algorithm

• Let’s use numbers from 0 to 3 to mark the directions: `north = 0`, `east = 1`, `south = 2`, `west = 3`. In the array `directions` we could store corresponding coordinates changes, i.e. `directions[0]` is to go north, `directions[1]` is to go east, `directions[2]` is to go south, and `directions[3]` is to go west.
• The initial robot position is in the center `x = y = 0`, facing north `idx = 0`.
• Now everything is ready to iterate over the instructions.
• If the current instruction is `R`, i.e. to turn on the right, the next direction is `idx = (idx + 1) % 4`. Modulo here is needed to deal with the situation - facing west, `idx = 3`, turn to the right to face north, `idx = 0`.
• If the current instruction is `L`, i.e. to turn on the left, the next direction could written in a symmetric way `idx = (idx - 1) % 4`. That means we have to deal with negative indices. A more simple way is to notice that 1 turn to the left = 3 turns to the right: `idx = (idx + 3) % 4`.
• If the current instruction is to move, we simply update the coordinates: `x += directions[idx][0]`, `y += directions[idx][1]`.
• After one cycle we have everything to decide. It’s a limit cycle trajectory if the robot is back to the center: `x = y = 0` or if the robot doesn't face north: `idx != 0`.

Complexity Analysis

• Time complexity: O(N), where N is a number of instructions to parse.
• Space complexity: O(1) because the array `directions` contains only 4 elements.