What is the time complexity of Walking Robot Simulation II?

The optimal solution has a time complexity of O(1).

What data structures are used to solve Walking Robot Simulation II?

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

What are the key insights for solving Walking Robot Simulation II?

Insight 1: The robot's movement repeats after a certain number of steps equal to the perimeter of the grid minus 4 (2 * (width + height - 2)). This is because after walking along all the edges, the robot will return to the starting point. Insight 2: Instead of simulating each step individually, we can calculate the final position and direction by using the modulo operator (%) with the perimeter to determine the effective number of steps taken.

Which companies ask LeetCode 2069. Walking Robot Simulation II in interviews?

This problem is frequently asked at Block.

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Walking Robot Simulation II - LeetCode 2069 Solution

Walking Robot Simulation II - Complete Solution Guide

Walking Robot Simulation II is LeetCode problem 2069, 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

A width x height grid is on an XY-plane with the bottom-left cell at (0, 0) and the top-right cell at (width - 1, height - 1) . The grid is aligned with the four cardinal directions ( "North" , "East" , "South" , and "West" ). A robot is initially at cell (0, 0) facing direction "East" . The robot can be instructed to move for a specific number of steps . For each step, it does the following. Attempts to move forward one cell in the direction it is facing. If the cell the robot is moving to is o

Detailed Explanation

The problem simulates a robot walking on a rectangular grid of size `width x height`. The robot starts at `(0, 0)` facing East. The robot takes a given number of steps. If a step would take the robot out of bounds, it turns 90 degrees counterclockwise and retries the step. The goal is to implement a `Robot` class with methods to move the robot, get its current position, and get its current direction.

Solution Approach

The solution avoids simulating each step individually to achieve O(1) time complexity. It calculates the perimeter of the grid and then uses the modulo operator to determine the effective distance the robot travels within one cycle of the perimeter. Based on this effective distance, the code determines the robot's final position and direction by checking which edge it lies on.

Step-by-Step Algorithm

Step 1: Initialize the Robot object with the given width and height, calculating the perimeter as `2 * (width + height - 2)`. Store the width, height, total steps taken, and intermediate points representing the end of each side.
Step 2: In the `step` method, increment the `moved` variable by the number of steps taken.
Step 3: In the `getPos` method, calculate the effective number of steps (`m`) taken by taking the modulo of `moved` with the `perimeter`. Handle the edge case where `moved` is 0 or `m` is 0 (robot is at starting position)
Step 4: Determine the robot's position based on the value of `m`. Check if `m` is within the boundaries of each side of the grid and calculate the coordinates accordingly. Use the intermediate point values(p1,p2,p3) to determine what edge the robot is currently on.
Step 5: In the `getDir` method, similar to `getPos`, determine the robot's direction based on the value of `m`. Check which side of the grid the robot is on and return the corresponding direction string.

Key Insights

Insight 1: The robot's movement repeats after a certain number of steps equal to the perimeter of the grid minus 4 (2 * (width + height - 2)). This is because after walking along all the edges, the robot will return to the starting point.
Insight 2: Instead of simulating each step individually, we can calculate the final position and direction by using the modulo operator (%) with the perimeter to determine the effective number of steps taken.
Insight 3: Carefully handle the edge cases when the robot has not moved at all (moved = 0) and when the number of steps is exactly a multiple of the perimeter.

Complexity Analysis

Time Complexity: O(1)

Space Complexity: O(1)

Topics

This problem involves: Design, Simulation.

Companies

Asked at: Block.

Walking Robot Simulation II | LeetCode 2069

Medium

Design Simulation

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

Solution Code

Available in:

class Robot {

    private int w;
    private int h;
    private long moved;
    private int perimeter;
    private int p1;
    private int p2;
    private int p3;

    public Robot(int width, int height) {
        this.w = width;
        this.h = height;
        this.moved = 0;
        this.perimeter = 2 * (width + height - 2);
        if (this.perimeter > 0) {
            this.p1 = width - 1;
            this.p2 = this.p1 + height - 1;
            this.p3 = this.p2 + width - 1;
        }
    }
    
    public void step(int num) {
        this.moved += num;
    }
    
    public int[] getPos() {
        if (this.moved == 0) {
            return new int[] {0, 0};
        }

        long m = this.moved % this.perimeter;
        if (m == 0) {
            return new int[] {0, 0};
        }
        
        if (m <= this.p1) {
            return new int[] {(int)m, 0};
        } else if (m <= this.p2) {
            return new int[] {this.w - 1, (int)(m - this.p1)};
        } else if (m <= this.p3) {
            return new int[] {this.w - 1 - (int)(m - this.p2), this.h - 1};
        } else {
            return new int[] {0, this.h - 1 - (int)(m - this.p3)};
        }
    }
    
    public String getDir() {
        if (this.moved == 0) {
            return "East";
        }
        
        long m = this.moved % this.perimeter;
        if (m == 0) {
            return "South";
        }

        if (m <= this.p1) {
            return "East";
        } else if (m <= this.p2) {
            return "North";
        } else if (m <= this.p3) {
            return "West";
        } else {
            return "South";
        }
    }
}

/**
 * Your Robot object will be instantiated and called as such:
 * Robot obj = new Robot(width, height);
 * obj.step(num);
 * int[] param_2 = obj.getPos();
 * String param_3 = obj.getDir();
 */

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Problem Info

Description

Problem 2069: Walking Robot Simulation II - Detailed explanation and solution approaches.

Topics

Design Simulation

Companies

Block

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2069

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 object with the given width and height, calculating the perimeter as `2 * (width + height - 2)`. Store the width, height, total steps taken, and intermediate points representing the end of each side.
2
Step 2: In the `step` method, increment the `moved` variable by the number of steps taken.
3
Step 3: In the `getPos` method, calculate the effective number of steps (`m`) taken by taking the modulo of `moved` with the `perimeter`. Handle the edge case where `moved` is 0 or `m` is 0 (robot is at starting position)
4
Step 4: Determine the robot's position based on the value of `m`. Check if `m` is within the boundaries of each side of the grid and calculate the coordinates accordingly. Use the intermediate point values(p1,p2,p3) to determine what edge the robot is currently on.
5
Step 5: In the `getDir` method, similar to `getPos`, determine the robot's direction based on the value of `m`. Check which side of the grid the robot is on and return the corresponding direction string.

💡

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(1)

⚡ Constant time - Super fast! Same speed regardless of input size.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

All operations (step, getPos, and getDir) involve constant time calculations and comparisons. The space used by the Robot class is constant, regardless of the input width, height, or number of steps.

📖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

Simulate each step

Simulate the robot's movement step by step, checking for out-of-bounds conditions and changing direction accordingly.

Time: O(num) where num is the number of stepsSpace: O(1)

Trade-offs: This approach is simpler to understand, but it's much slower for large values of `num`. Use the main solution for better performance, especially with many calls to `step`.

✨

Key Insights

1Insight 1: The robot's movement repeats after a certain number of steps equal to the perimeter of the grid minus 4 (2 * (width + height - 2)). This is because after walking along all the edges, the robot will return to the starting point.
2Insight 2: Instead of simulating each step individually, we can calculate the final position and direction by using the modulo operator (%) with the perimeter to determine the effective number of steps taken.
3Insight 3: Carefully handle the edge cases when the robot has not moved at all (moved = 0) and when the number of steps is exactly a multiple of the perimeter.

📚

Key Terms Explained

For Students

Edge Cases

Edge case 1: The robot hasn't moved yet (moved = 0). In this case, the position is (0, 0) and the direction is East. The code handles this explicitly.
Edge case 2: The number of steps taken is exactly a multiple of the perimeter. In this case, the robot returns to the starting position (0,0) and is at the 'South' direction instead of East if perimeter>0, the code correctly handles this edge case by setting to South.
Edge case 3: width = 2, height = 2. The perimeter is 0, division by zero is avoided using a conditional to make the points only if perimeter >0.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not considering the perimeter and simulating each step individually, leading to a Time Limit Exceeded (TLE) error.

✗Common mistake 2: Incorrectly calculating the effective number of steps by forgetting to use the modulo operator with the perimeter.

✗Common mistake 3: Not handling edge cases like the robot not moving or moving a multiple of the perimeter, leading to incorrect positions or directions.

✅

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 in robotics, where robots need to navigate within defined boundaries.

🏦Real-world scenario 2: Game development, simulating character movement in a confined map with collision detection.

💡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

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ 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: August 29, 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

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 13

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 6

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

@AlgoMasterJun 29

❓

Frequently Asked Questions

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 "Walking Robot Simulation II"?

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

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