LeetCode 657 | Robot Return to Origin | Java | Easy | String & Simulation | CodeByTushu
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In today’s video, we solve LeetCode 657 — Robot Return to Origin using String and Simulation in Java.
In this problem, a robot begins at position (0, 0) on a 2D Cartesian plane and executes a series of directional moves specified by a string of characters: U for up, D for down, L for left, and R for right. Each move carries an identical step magnitude regardless of which direction the robot is facing. Our objective is to determine whether the robot returns precisely to its starting position (0, 0) after completing the entire sequence of moves.
To solve this efficiently, we maintain two integer variables, x and y, to track horizontal and vertical displacement respectively. As we iterate through each move command in the string, we increment or decrement the coordinates accordingly: U increments y, D decrements y, R increments x, and L decrements x. After processing all commands, the robot has successfully returned to the origin if and only if both x and y evaluate to zero. This delivers a single-pass linear time complexity of O(N) and optimal O(1) auxiliary space.
🔥 Problem Highlights
✅ 2D coordinate displacement simulation using horizontal and vertical offsets
✅ Direct state updates without memory overhead for grid allocation
✅ Up/Down and Left/Right parity cancellation principle
✅ Clean handling of move strings of any length up to 20,000 characters
✅ Invariant check: origin return requires both x == 0 and y == 0 simultaneously
✅ Optimal O(N) time and strict O(1) auxiliary space complexity
⏱️ TIMESTAMPS
00:00 Intro & Daily Challenge Kickoff (Streak 98 Days)
01:23 Problem Understanding & 2D Coordinate Geometry
03:03 Key Insights: Axis Independence & Displacement Invariant
04:46 Optimal Linear Simulation Algorithm
06:10 Clean Java Solution Walkthrough Line by Line
07:15 Test Cases Dry Run & Benchmark Verification (3 ms, Beats 98.7%)
08:36 Time Complexity Analysis (O(N) Time)
09:25 Space Complexity Analysis (O(1) Auxiliary Space)
10:05 Day 5 Wrap-up & Channel Resources
📂 Problem Folder & Detailed Documentation:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/tree/main/solutions/April/05-April/657-Robot-Return-to-Origin
☕ Java Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/April/05-April/657-Robot-Return-to-Origin/Solution.java
🐍 Python Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/April/05-April/657-Robot-Return-to-Origin/solution.py
⚡ C++ Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/April/05-April/657-Robot-Return-to-Origin/solution.cpp
🌐 JavaScript Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/April/05-April/657-Robot-Return-to-Origin/solution.js
🔗 LeetCode Source Code
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🔥 Perfect For
LeetCode Easy Problems
Java DSA Preparation
2D Plane Simulation
String Traversal Patterns
Beginner Friendly DSA
Time Complexity Optimization
Coding Interview Preparation
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💬 Question of the Day
What is the necessary and sufficient condition for a sequence of moves to return the robot to the origin?
1️⃣ count('U') == count('D') and count('L') == count('R')
2️⃣ The total length of moves must be an odd number
3️⃣ Every 'U' must be followed immediately by a 'D'
4️⃣ count('U') + count('D') must equal count('L') + count('R')
#LeetCode657 #LeetCode #RobotReturnToOrigin #Java #JavaDSA #String #Simulation #DSA #CodingInterview #ProblemSolving #LeetCodeEasy #DailyLeetCode #JavaProgramming #CodeByTushu
👇 Comment the answer below! 🔥 Close description
In today’s video, we solve LeetCode 657 — Robot Return to Origin using String and Simulation in Java.
In this problem, a robot begins at position (0, 0) on a 2D Cartesian plane and executes a series of directional moves specified by a string of characters: U for up, D for down, L for left, and R for right. Each move carries an identical step magnitude regardless of which direction the robot is facing. Our objective is to determine whether the robot returns precisely to its starting position (0, 0) after completing the entire sequence of moves.
To solve this efficiently, we maintain two integer variables, x and y, to track horizontal and vertical displacement respectively. As we iterate through each move command in the string, we increment or decrement the coordinates accordingly: U increments y, D decrements y, R increments x, and L decrements x. After processing all commands, the robot has successfully returned to the origin if and only if both x and y evaluate to zero. This delivers a single-pass linear time complexity of O(N) and optimal O(1) auxiliary space.
🔥 Problem Highlights
✅ 2D coordinate displacement simulation using horizontal and vertical offsets
✅ Direct state updates without memory overhead for grid allocation
✅ Up/Down and Left/Right parity cancellation principle
✅ Clean handling of move strings of any length up to 20,000 characters
✅ Invariant check: origin return requires both x == 0 and y == 0 simultaneously
✅ Optimal O(N) time and strict O(1) auxiliary space complexity
⏱️ TIMESTAMPS
00:00 Intro & Daily Challenge Kickoff (Streak 98 Days)
01:23 Problem Understanding & 2D Coordinate Geometry
03:03 Key Insights: Axis Independence & Displacement Invariant
04:46 Optimal Linear Simulation Algorithm
06:10 Clean Java Solution Walkthrough Line by Line
07:15 Test Cases Dry Run & Benchmark Verification (3 ms, Beats 98.7%)
08:36 Time Complexity Analysis (O(N) Time)
09:25 Space Complexity Analysis (O(1) Auxiliary Space)
10:05 Day 5 Wrap-up & Channel Resources
📂 Problem Folder & Detailed Documentation:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/tree/main/solutions/April/05-April/657-Robot-Return-to-Origin
☕ Java Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/April/05-April/657-Robot-Return-to-Origin/Solution.java
🐍 Python Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/April/05-April/657-Robot-Return-to-Origin/solution.py
⚡ C++ Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/April/05-April/657-Robot-Return-to-Origin/solution.cpp
🌐 JavaScript Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/April/05-April/657-Robot-Return-to-Origin/solution.js
🔗 LeetCode Source Code
📌 Admin type the link
🌐 All LeetCode Solutions
📌 https://codebytushu.com/Leetcode/
📱 Join CodeByTushu on WhatsApp
Get coding updates, LeetCode solutions, daily questions and more.
https://whatsapp.com/channel/0029Vb8k5mCDeON2Xwlkxe40
❤️ Donate
📌 https://codebytushu.com/Leetcode/donate.php
🔥 Perfect For
LeetCode Easy Problems
Java DSA Preparation
2D Plane Simulation
String Traversal Patterns
Beginner Friendly DSA
Time Complexity Optimization
Coding Interview Preparation
Problem Solving
🌐 Follow CodeByTushu
▶️ YouTube: https://www.youtube.com/@codebytushu
📸 Instagram: https://instagram.com/codebytushu/
💼 LinkedIn: https://www.linkedin.com/company/codebytushu/
🌍 Website: https://codebytushu.com
🐙 GitHub: https://github.com/CodeByTushu
❤️ Don’t Forget To
👍 Like | 💬 Comment | 🔔 Subscribe
💬 Question of the Day
What is the necessary and sufficient condition for a sequence of moves to return the robot to the origin?
1️⃣ count('U') == count('D') and count('L') == count('R')
2️⃣ The total length of moves must be an odd number
3️⃣ Every 'U' must be followed immediately by a 'D'
4️⃣ count('U') + count('D') must equal count('L') + count('R')
#LeetCode657 #LeetCode #RobotReturnToOrigin #Java #JavaDSA #String #Simulation #DSA #CodingInterview #ProblemSolving #LeetCodeEasy #DailyLeetCode #JavaProgramming #CodeByTushu
👇 Comment the answer below! 🔥 Close description
