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Formatted question description: https://leetcode.ca/all/1041.html

# 1041. Robot Bounded In Circle (Medium)

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'}

Related Topics: Math

## Solution 1.

• class Solution {
public boolean isRobotBounded(String instructions) {
if (instructions == null || instructions.length() == 0)
return true;
int[][] directions = { {0, 1}, {1, 0}, {0, -1}, {-1, 0} };
int directionIndex = 0;
int xPosition = 0, yPosition = 0;
int length = instructions.length();
for (int i = 0; i < length; i++) {
char instruction = instructions.charAt(i);
if (instruction == 'G') {
int[] direction = directions[directionIndex];
xPosition += direction[0];
yPosition += direction[1];
} else if (instruction == 'L')
directionIndex = (directionIndex + 3) % 4;
else if (instruction == 'R')
directionIndex = (directionIndex + 1) % 4;
}
return xPosition == 0 && yPosition == 0 || directionIndex != 0;
}
}

############

class Solution {
public boolean isRobotBounded(String instructions) {
int[] direction = new int[4];
int cur = 0;
for (char c : instructions.toCharArray()) {
if (c == 'L') {
cur = (cur + 1) % 4;
} else if (c == 'R') {
cur = (cur + 3) % 4;
} else {
++direction[cur];
}
}
return cur != 0 || (direction[0] == direction[2] && direction[1] == direction[3]);
}
}

• // OJ: https://leetcode.com/problems/robot-bounded-in-circle/
// Time: O(N)
// Space: O(1)
class Solution {
public:
bool isRobotBounded(string instructions) {
int x = 0, y = 0, d = 0, dir[4][2] = { {0,1},{-1,0},{0,-1},{1,0} };
for (int i = 0; i < 4; ++i) {
for (char c : instructions) {
if (c == 'G') x += dir[d][0], y += dir[d][1];
else if (c == 'L') d = (d + 1) % 4;
else d = (d + 3) % 4;
}
if (x == 0 && y == 0) return true;
}
return false;
}
};

• class Solution:
def isRobotBounded(self, instructions: str) -> bool:
cur, direction = 0, [0] * 4
for ins in instructions:
if ins == 'L':
cur = (cur + 1) % 4
elif ins == 'R':
cur = (cur + 3) % 4
else:
direction[cur] += 1
return cur != 0 or (
direction[0] == direction[2] and direction[1] == direction[3]
)

############

class Solution(object):
def isRobotBounded(self, instructions):
"""
:type instructions: str
:rtype: bool
"""
dirs = [(0, 1), (-1, 0), (0, -1), (1, 0)]
x, y = 0, 0
curd = 0
for i in instructions:
if i == "G":
x += dirs[curd][0]
y += dirs[curd % 4][1]
elif i == "L":
curd = (curd + 1) % 4
elif i == "R":
curd = (curd - 1) % 4
return (x == 0 and y == 0) or curd != 0

• func isRobotBounded(instructions string) bool {
direction := make([]int, 4)
cur := 0
for _, ins := range instructions {
if ins == 'L' {
cur = (cur + 1) % 4
} else if ins == 'R' {
cur = (cur + 3) % 4
} else {
direction[cur]++
}
}
return cur != 0 || (direction[0] == direction[2] && direction[1] == direction[3])
}

• function isRobotBounded(instructions: string): boolean {
const dist: number[] = new Array(4).fill(0);
let k = 0;
for (const c of instructions) {
if (c === 'L') {
k = (k + 1) % 4;
} else if (c === 'R') {
k = (k + 3) % 4;
} else {
++dist[k];
}
}
return (dist[0] === dist[2] && dist[1] === dist[3]) || k !== 0;
}