LeetCode 657 Robot Return to Origin Easy Solution

LeetCode Daily | Day 70 🔥 LeetCode POTD – 874. Walking Robot Simulation (Medium) ✨ 📌 Problem Insight Given a robot on an infinite grid: ✔ Starts at (0, 0), facing North ✔ Commands include: • -2 → turn left • -1 → turn right • k → move forward k steps ✔ Obstacles block movement ✔ Find max distance² from origin reached 🔍 Initial Thinking – Simulation + Directions ✔ 💡 Idea: ✔ Maintain current position (x, y) ✔ Track direction using index (N, E, S, W) ✔ Use dx, dy arrays for movement ✔ Store obstacles in a set for fast lookup ✔ Move step-by-step (important!) • Before each step → check obstacle • If blocked → stop current command ⏱ Complexity: O(n + steps), Space: O(n) ✔ Efficient due to hashing ✔ Each move handled once 💡 Key Optimization – Hashing Obstacles 🔥 👉 Use set / unordered_set ✔ O(1) obstacle check ✔ Encode (x, y) → faster lookup possible 🧠 Key Learning ✔ Simulation problems need careful handling of edge cases ✔ Direction control using modulo is very useful ✔ Step-by-step movement avoids logical bugs ✔ Hashing turns brute-force into efficient solution A great mix of simulation + hashing — classic interview pattern ⚡ #LeetCode #DSA #Algorithms #CPlusPlus #ProblemSolving #CodingJourney #DataStructures

What if Stochastic Gradient Descent didn't actually require gradients? There are many problems in the world that involve functions that don't have nice derivatives. If I want a robot to follow a curve, for example, and using a basic motion control system like a PID loop to control it, I have to optimize that by hand. We can define a function to tell us how good or bad it does: think average speed - average distance from the line, but the intermediate function that relates these values back to kP, kI, and kD is nearly impossible to accurately find. These kinds of problems are why I created Backstaff, a FOSS SGD-like optimizer that doesn't require gradients. Check it out, and feel free to reach out if you have any questions! *Note: The algorithm was written entirely by hand, but I used AI for the UI and some documentation. https://lnkd.in/gQ3DD6bK

Is your A* just Dijkstra in disguise? 🧠 Pathfinding powers everything from games to logistics. While doing LeetCode, we stop at Dijkstra / Floyd–Warshall, but real-world applications are not even close to these. In reality, systems use A* algorithm for path finding. A* isn’t just an algorithm. It’s an engineering framework. The difference comes down to the heuristic (h(n)). At its core: f(n) = g(n) + h(n) • g(n): actual cost • h(n): estimated cost 👉 Weak h(n)-> Dijkstra (slow, blind search) 👉 Aggressive h(n) → breaks optimality Real-world heuristics aren’t simple 🚚 Logistics: traffic, constraints, routing rules 🤖 Robotics: terrain, battery, uncertainty ⚡ Chip design: congestion, delay, power Don’t just apply A*. Engineer the Heuristics. The closer it is to the true cost, the faster your system scales. Can you think of methods to train heuristics, just like regression? #Algorithms #AStar #SystemsDesign #AI #Engineering

#day63 🚀 19/04/26 — Precision Tracking: Robot Return to Origin Today, I solved Robot Return to Origin (LeetCode 657), applying coordinate-based tracking to determine if a sequence of moves returns a robot to its starting position. This problem is a great exercise in Simulation and state management using simple array counters. 🤖 The Origin Logic The objective is to determine if a robot, starting at (0,0) on a 2D plane, returns to (0,0) after executing a string of moves: 'U' (up), 'D' (down), 'L' (left), and 'R' (right). The Logic: Coordinate Simulation State Representation: I used a small integer array arr of size 2 to represent the (x, y) coordinates, initialized at {0, 0}. arr[0] tracks horizontal movement (x-axis). arr[1] tracks vertical movement (y-axis). Linear Execution: I iterated through the moves string exactly once. Vertical: Increment arr[1] for 'U' and decrement for 'D'. Horizontal: Increment arr[0] for 'R' and decrement for 'L'. Verification: The robot returns to the origin if and only if both final coordinates are zero (arr[0] == 0 && arr[1] == 0). Complexity: Time: O(n) where n is the length of the moves string. Space: O(1) as we only use a fixed-size array regardless of the number of moves. 📈 Consistency Report: Foundations of Simulation Today's solution emphasizes that complex tracking problems can often be reduced to basic arithmetic operations on a fixed state. This is a more streamlined version of the Two-Pointer and Sliding Window techniques I've been mastering—while those patterns focus on sub-segments of data, this simulation tracks the global state of an object over time. Huge thanks to the roadmap for building this momentum! Moving from array partitioning to 2D state simulation shows how my problem-solving intuition is expanding. My O(n) coordinate tracking implementation is attached below! 📄👇 #DSA #Java #LeetCode #Simulation #Optimization #Complexity #63DayStreak #LearningInPublic

🚀 LeetCode POTD – Maximum Walls Destroyed by Robots (Hard) Continuing my April challenge 💪 — solving every POTD consistently Today’s problem, Maximum Walls Destroyed by Robots, initially felt like a greedy / simple counting problem… but it turned out to be much deeper than expected 👀 🧠 My Thought Process: First instinct: Each robot has a range → just count walls in that range Then I realized a critical issue: ❌ Same wall can get counted multiple times Tried greedy → failed Tried simple DP → still wrong 👉 Clearly, something was missing in the state 🔥 Key Insight: This is actually a DP + state tracking problem Each robot has 2 choices → left or right But choice affects future robots So we must track: 👉 previous robot direction (left/right) ⚙️ Implementation Strategy: Sort robots based on position Precompute valid range for each robot Use binary search to count walls efficiently Apply Recursion + Memoization (DP) State: dp[i][prevDir] i → current robot prevDir → previous robot went left/right 💡 What I Learned: DP is not just index-based → state matters a lot If decisions affect future → include it in DP state Binary search can turn brute force into efficient solution Always check: “Am I double counting?” 💥 Honestly, I was stuck on this problem for quite some time Concept clear hi nahi ho raha tha Then I watched CodeWithMik’s video — and everything finally clicked 🙌 👉 Big thanks to him — explanation was super clear and intuitive 📈 Slowly improving at: Identifying correct DP states Avoiding overcounting mistakes Converting brute force → optimized solution 📌 April Challenge: Solve every LeetCode POTD consistently Let’s see how far consistency takes me 🚀 #LeetCode #DSA #DynamicProgramming #ProblemSolving #CodingJourney #Consistency #BinarySearch

I’ve been bouncing back and forth between Isaac Lab and Isaac Sim a lot lately. Train in Lab, deploy in Sim, test with ROS, repeat. Training is fine. Deployment… not so much. Not because it’s conceptually hard, it’s just fragile. A long chain of manual steps that are easy to mess up: ➡️ Rebuilding observation tensors by hand ➡️ Matching joint ordering between training configs and USD articulations ➡️ Wiring action scaling and PD gains ➡️ Writing the Omniverse extension boilerplate from scratch Get one index wrong and your robot’s hip commands go to the knee. Ask me how I know. So I built something to fix it. A Lab to Sim tool that automates the whole handoff. You point it at your training outputs (env.yaml, agent.yaml, policy.pt) and it generates a complete, runnable Isaac Sim extension. Config, policy controller, observation builder, scenario runner, keyboard teleop, UI, the lot. Under the hood it: ➡️ Parses Isaac Lab IO Descriptor YAMLs and Python task configs ➡️ Maps training joint names to USD articulation DOF order (exact + fuzzy matching) ➡️ Heuristically infers robot type (manipulator, quadruped, humanoid) based on kinematic structure ➡️ Validates all tensor dimensions before generation so you catch issues early, not at runtime ➡️ Spits out a self‑contained extension you can drop straight into Isaac Sim. It includes a browser‑based wizard with a visual joint‑mapping diagram. So you can see exactly where things would go wrong before they do. #Robotics #IsaacSim #IsaacLab #NVIDIA #ReinforcementLearning #Omniverse

🚀 LeetCode Daily Challenge 🔗 Problem: https://lnkd.in/gzjgiw8d 💡 My thought process: First, we sort the robots by their positions. Instead of changing the original arrays, we store the indices and sort them using the positions array. This way, we process the robots from left to right. This order is necessary because collisions depend on it. We use a vector as a stack to track the robots that are still active. The stack stores indices, allowing us to access their direction and health. For each robot in sorted order, we check if a collision might happen. A collision occurs only when the current robot is moving left and the robot at the top of the stack is moving right. In all other situations, we simply add the current robot to the stack. If a collision is possible, we enter a loop. The current robot may collide with multiple robots in the stack. In each step, we compare the health of the current robot with the robot at the top of the stack: - If the robot in the stack has more health, it survives, its health decreases by one, and the current robot dies. - If the current robot has more health, it survives, its health decreases by one, and we continue checking with the next robot in the stack. - If both have the same health, both are removed. We repeat this until the current robot dies or no more collisions can occur. If the current robot is still alive after the loop, we add it to the stack. In the end, all robots with health greater than zero are survivors. We go through the health array and collect those values. 👉 My Solution: https://lnkd.in/gjSQCeTP If you found this breakdown helpful, feel free to ⭐ the repo or connect with me on LinkedIn 🙂🚀 #️⃣ #leetcode #cpp #dsa #coding #problemsolving #engineering #BDRM #BackendDevWithRahulMaheswari

🚀 #LeetCode Day Problem Solving 🚀 Day-21 📌 Problem: A robot starts at position (0, 0) on an infinite 2D plane, initially facing North. You are given: An array commands representing movement instructions A list of obstacles on the grid 👉 The robot follows these rules: ✔ -2 → Turn left 90° ✔ -1 → Turn right 90° ✔ 1 to 9 → Move forward k steps (one step at a time) ⚠ If the robot encounters an obstacle, it stops before hitting it and continues with the next command. 👉 Your task is to return the maximum squared Euclidean distance from the origin reached at any point during the movement. 🧠 Example 1: Input: commands = [4, -1, 3], obstacles = [] ✅ Output: 25 📖 Explanation: Move to (0,4) Turn right → face East Move to (3,4) 👉 Max distance = (3² + 4² = 25) 🧠 Example 2: Input: commands = [4,-1,4,-2,4], obstacles = [[2,4]] ✅ Output: 65 📖 Explanation: Obstacle blocks movement at (2,4) Final farthest point = (1,8) 👉 Distance = (1² + 8² = 65) 🧠 Example 3: Input: commands = [6,-1,-1,6], obstacles = [[0,0]] ✅ Output: 36 📖 Explanation: Robot cannot return to origin due to obstacle Farthest point = (0,6) → (6² = 36) 💡 Key Insight: ✔ Use direction vectors: North → (0,1) East → (1,0) South → (0,-1) West → (-1,0) ✔ Store obstacles in a set for O(1) lookup ✔ Move step-by-step to detect obstacles ✔ Track max distance using: x² + y² 📊 Complexity Analysis: ⏱ Time Complexity: O(n + k) (n = commands, k = steps) 📦 Space Complexity: O(obstacles) 🧠 What I Learned: ✔ Simulation with direction handling ✔ Efficient obstacle checking using hashing ✔ Tracking maximum distance dynamically ✅ Day 21 Completed 🚀 Leveling up in Simulation & Grid Problems 💪 #Leetcode #DSA #ProblemSolving #BitManipulation #CodingJourney #InterviewPreparation #Consistency

🚀 LeetCode Day Problem Solving 🚀 Day-39 📌 Problem: A robot starts at position (0, 0) on a 2D plane. You are given a string moves representing its movement sequence. Each character in the string denotes a move: ➡ 'R' → Right ⬅ 'L' → Left ⬆ 'U' → Up ⬇ 'D' → Down 👉 Your task is to determine whether the robot returns to the origin (0, 0) after completing all moves. Return true if it returns to origin, otherwise false. 🧠 Example 1: Input: moves = "UD" ✅ Output: true 📖 Explanation: Move Up → (0, 1) Move Down → (0, 0) ✔ Robot returns to origin 🧠 Example 2: Input: moves = "LL" ❌ Output: false 📖 Explanation: Move Left → (-1, 0) Move Left → (-2, 0) ❌ Robot does not return to origin 💡 Key Insight: ✔ Count movements: Number of 'U' should equal 'D' Number of 'L' should equal 'R' ✔ OR simulate coordinates: Track (x, y) position 📊 Complexity Analysis: ⏱ Time Complexity: O(n) 📦 Space Complexity: O(1) 🧠 What I Learned: ✔ Simple simulation problems can be solved efficiently ✔ Importance of balancing movements ✔ Good practice for coordinate tracking ✅ Day 39 Completed 🚀 Keep going strong with DSA & consistency 💪 #Leetcode #DSA #ProblemSolving #BitManipulation #CodingJourney #InterviewPreparation #Consistency #MilanSahoo 🚀

🚀 #LeetCode Day Problem Solving 🚀 Day-20 📌 Problem: A robot starts at position (0, 0) on a 2D plane. You are given a string moves representing its movement sequence. Each character in the string denotes a move: ➡ 'R' → Right ⬅ 'L' → Left ⬆ 'U' → Up ⬇ 'D' → Down 👉 Your task is to determine whether the robot returns to the origin (0, 0) after completing all moves. Return true if it returns to origin, otherwise false. 🧠 Example 1: Input: moves = "UD" ✅ Output: true 📖 Explanation: Move Up → (0, 1) Move Down → (0, 0) ✔ Robot returns to origin 🧠 Example 2: Input: moves = "LL" ❌ Output: false 📖 Explanation: Move Left → (-1, 0) Move Left → (-2, 0) ❌ Robot does not return to origin 💡 Key Insight: ✔ Count movements: Number of 'U' should equal 'D' Number of 'L' should equal 'R' ✔ OR simulate coordinates: Track (x, y) position 📊 Complexity Analysis: ⏱ Time Complexity: O(n) 📦 Space Complexity: O(1) 🧠 What I Learned: ✔ Simple simulation problems can be solved efficiently ✔ Importance of balancing movements ✔ Good practice for coordinate tracking ✅ Day 20 Completed 🚀 Keep going strong with DSA & consistency 💪 #Leetcode #DSA #ProblemSolving #BitManipulation #CodingJourney #InterviewPreparation #Consistency