What is the time complexity of Robot Bounded In Circle?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Robot Bounded In Circle?

This problem uses Math, String, Simulation. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Robot Bounded In Circle?

Insight 1: The robot is bounded if it returns to the origin after executing the instructions once. Insight 2: The robot is also bounded if, after executing the instructions once, it is facing a direction other than North. If it's facing any other direction, each repetition of the instructions will rotate the direction vector, eventually leading back towards the origin.

Which companies ask LeetCode 1041. Robot Bounded In Circle in interviews?

This problem is frequently asked at Airbnb, Chewy, Nvidia, ZScaler.

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Robot Bounded In Circle - LeetCode 1041 Solution

Robot Bounded In Circle - Complete Solution Guide

Robot Bounded In Circle is LeetCode problem 1041, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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. The south direction is the negative direction of the y-axis. The east direction is the positive direction of the x-axis. The west direction is the negative direction of the x-axis. The robot can receive one of three instructions: "G" : go straight 1 unit. "L" : turn 90 degrees to the left (i.e., anti-clockwise direction). "R" : turn 90 degrees to the r

Detailed Explanation

The problem asks us to determine whether a robot, starting at (0, 0) and facing North, will remain within a bounded circular area on an infinite plane after repeatedly executing a given sequence of instructions. The robot can move one unit forward ('G'), turn 90 degrees left ('L'), or turn 90 degrees right ('R'). We need to return `true` if the robot stays within a circle, and `false` otherwise. The instructions are repeated infinitely.

Solution Approach

The solution simulates the robot's movement by iterating through the instructions once. It keeps track of the robot's current position (x, y) and direction. After executing the instructions, it checks if either the robot is back at the origin or it is not facing North. If either condition is true, it means the robot is bounded; otherwise, it is not.

Step-by-Step Algorithm

Step 1: Initialize the robot's position to (0, 0) and its direction to North (represented by 0). Use an array to store displacement vectors for each direction (North, East, South, West).
Step 2: Iterate through the instructions string.
Step 3: For each instruction, update the robot's position and direction. If the instruction is 'G', update the x and y coordinates based on the current direction. If the instruction is 'L', turn the robot 90 degrees to the left (decrement direction modulo 4). If the instruction is 'R', turn the robot 90 degrees to the right (increment direction modulo 4).
Step 4: After processing all instructions, check if the robot has returned to the origin (x == 0 and y == 0) or if the robot is not facing North (direction != 0).
Step 5: Return `true` if either condition from Step 4 is met; otherwise, return `false`.

Key Insights

Insight 1: The robot is bounded if it returns to the origin after executing the instructions once.
Insight 2: The robot is also bounded if, after executing the instructions once, it is facing a direction other than North. If it's facing any other direction, each repetition of the instructions will rotate the direction vector, eventually leading back towards the origin.
Insight 3: If the robot ends up at a non-origin point facing North, it will continue to move away from the origin indefinitely, and thus is not bounded.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Math, String, Simulation.

Companies

Asked at: Airbnb, Chewy, Nvidia, ZScaler.

Robot Bounded In Circle | LeetCode 1041

Medium

Math String Simulation

Time: O(n)Space: O(1)

Solution Code

Available in:

class Solution {
    public boolean isRobotBounded(String instructions) {
        int x = 0, y = 0;
        // 0: North, 1: East, 2: South, 3: West
        int direction = 0;
        
        // Displacements for each direction (N, E, S, W)
        int[][] directions = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
        
        for (char instruction : instructions.toCharArray()) {
            if (instruction == 'L') {
                direction = (direction + 3) % 4;
            } else if (instruction == 'R') {
                direction = (direction + 1) % 4;
            } else {  // instruction == 'G'
                int dx = directions[direction][0];
                int dy = directions[direction][1];
                x += dx;
                y += dy;
            }
        }
        
        // The path is bounded if the robot returns to the origin
        // or if it does not face North at the end of the instructions.
        // If direction is not North, the net displacement vector rotates
        // with each cycle, forming a larger circle that eventually returns
        // to the origin.
        // If the robot is at a non-origin point and facing North,
        // it will continue to move away from the origin indefinitely.
        return (x == 0 && y == 0) || (direction != 0);
    }
}

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Problem Info

Description

Problem 1041: Robot Bounded In Circle - Detailed explanation and solution approaches.

Topics

Math String Simulation

Companies

AirbnbChewyNvidiaZScaler

Stats

Difficulty:Medium

Languages:4

Companies:4

Problem #:1041

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

5 steps

1
Step 1: Initialize the robot's position to (0, 0) and its direction to North (represented by 0). Use an array to store displacement vectors for each direction (North, East, South, West).
2
Step 2: Iterate through the instructions string.
3
Step 3: For each instruction, update the robot's position and direction. If the instruction is 'G', update the x and y coordinates based on the current direction. If the instruction is 'L', turn the robot 90 degrees to the left (decrement direction modulo 4). If the instruction is 'R', turn the robot 90 degrees to the right (increment direction modulo 4).
4
Step 4: After processing all instructions, check if the robot has returned to the origin (x == 0 and y == 0) or if the robot is not facing North (direction != 0).
5
Step 5: Return `true` if either condition from Step 4 is met; otherwise, return `false`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is linear because the algorithm processes each instruction in the input string exactly once. The space complexity is constant because the amount of extra space used does not depend on the size of the input.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Brute-force simulation with multiple iterations

Simulate the robot's movement for a large number of instruction repetitions. If the robot's distance from the origin exceeds a certain threshold, it can be assumed unbounded.

Time: O(k*n), where n is the length of the instructions and k is the number of repetitions.Space: O(1)

Trade-offs: This approach is less efficient than the optimal solution and requires selecting an appropriate threshold value to determine boundedness. The optimal solution guarantees correctness in O(n) time, while the brute-force may not reach a conclusive answer if k is insufficient.

✨

Key Insights

1Insight 1: The robot is bounded if it returns to the origin after executing the instructions once.
2Insight 2: The robot is also bounded if, after executing the instructions once, it is facing a direction other than North. If it's facing any other direction, each repetition of the instructions will rotate the direction vector, eventually leading back towards the origin.
3Insight 3: If the robot ends up at a non-origin point facing North, it will continue to move away from the origin indefinitely, and thus is not bounded.

📚

Key Terms Explained

For Students

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: Empty instructions string. The loop doesn't execute and the condition `(x == 0 && y == 0) || (direction != 0)` becomes `(0 == 0 && 0 == 0) || (0 != 0)`, which simplifies to `true || false` which results in `true`. This is correct because an empty instruction set results in no movement, hence the robot is bounded at the origin.
Edge case 2: Instructions that result in no net movement but change direction. For example, "L". The robot stays at the origin but turns left. This results in `(x == 0 && y == 0) || (direction != 0)` becoming `(0 == 0 && 0 == 0) || (0 != 0)`, resulting in true, which is the correct output.
Edge case 3: Instructions that move the robot away from the origin facing North. For example, "G". The robot moves one unit North and faces North. This results in `(x == 0 && y == 0) || (direction != 0)` becoming `(0 == 0 && 1 == 0) || (0 != 0)`, resulting in false, which is the correct output.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly updating the direction after a 'L' or 'R' instruction. Using the wrong modulo operator can lead to incorrect direction calculations.

✗Common mistake 2: Failing to consider both the position and direction after the instructions are executed. Checking only if the robot returns to the origin is not sufficient, as the robot might be facing a different direction, leading to a bounded path.

✗Common mistake 3: Incorrect implementation of direction updates, leading to incorrect coordinate updates, and ultimately, an incorrect conclusion about boundedness.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Path planning for autonomous vehicles or drones. Understanding whether a series of movements will result in a bounded or unbounded trajectory is crucial for safe navigation.

🏦Real-world scenario 2: Modeling the behavior of a robot arm in a manufacturing environment. Ensuring the arm stays within its operational workspace after a series of movements is essential to prevent collisions and maintain efficiency.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: October 20, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 13

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 6

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJun 29

❓

Frequently Asked Questions

What is the sliding window technique for string problems?

Sliding window maintains a "window" of characters with two pointers. Expand the right pointer to include characters, shrink from left when window violates constraints. Useful for substring problems, anagrams, and longest/shortest substring with properties.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Robot Bounded In Circle"?

For this medium-level Math problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether string techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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