## 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 `, or if `dir` is not `.

``````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;
}
}
``````

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