Robot Bounded In Circle
3 min readAug 2, 2021
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 arraydirections
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, anddirections[3]
is to go west. - The initial robot position is in the center
x = y = 0
, facing northidx = 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 isidx = (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 wayidx = (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.