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Leetcode #1041: Robot Bounded In Circle

In this guide, we solve Leetcode #1041 Robot Bounded In Circle in Python and focus on the core idea that makes the solution efficient.

You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.

Leetcode

Problem Statement

On an infinite plane, a robot initially stands at (0, 0) and faces north. Note that: The north direction is the positive direction of the y-axis.

Quick Facts

  • Difficulty: Medium
  • Premium: No
  • Tags: Math, String, Simulation

Intuition

There is a mathematical invariant or formula that directly leads to the result.

Using math avoids unnecessary loops and reduces complexity.

Approach

Derive the formula or update rule, then compute the answer directly.

Handle edge cases like overflow or zero carefully.

Steps:

  • Identify the math relationship.
  • Compute the result with a loop or formula.
  • Handle edge cases.

Example

Input: instructions = "GGLLGG" Output: true Explanation: The robot is initially at (0, 0) facing the north direction. "G": move one step. Position: (0, 1). Direction: North. "G": move one step. Position: (0, 2). Direction: North. "L": turn 90 degrees anti-clockwise. Position: (0, 2). Direction: West. "L": turn 90 degrees anti-clockwise. Position: (0, 2). Direction: South. "G": move one step. Position: (0, 1). Direction: South. "G": move one step. Position: (0, 0). Direction: South. Repeating the instructions, the robot goes into the cycle: (0, 0) --> (0, 1) --> (0, 2) --> (0, 1) --> (0, 0). Based on that, we return true.

Python Solution

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

Complexity

The time complexity is O(n)O(n)O(n), and the space complexity is O(1)O(1)O(1), where nnn is the length of the instruction string instructions\textit{instructions}instructions. The space complexity is O(1)O(1)O(1), where nnn is the length of the instruction string instructions\textit{instructions}instructions.

Edge Cases and Pitfalls

Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.

Summary

This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.


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