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:

Example 2:

Example 3:

Constraints:

  • 1 <= instructions.length <= 100
  • instructions[i] is 'G', 'L' or, 'R'.

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.

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