# 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'`

.

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.