Maximizing Walls Destroyed by Robots with Binary Search and DP

🔥 LeetCode Hard Problem Breakdown — Maximum Walls Destroyed by Robots Recently, I worked on a really challenging problem that completely changed how I think about greedy vs dynamic programming. 🧩 Problem Summary (Simple Words) You are given: 🤖 Robots placed on a number line 🧱 Walls placed on the same line Each robot has a shooting distance Each robot can: Shoot LEFT ⬅️ or RIGHT ➡️ (only once) Destroy all walls in that range BUT ❌ bullet stops if another robot comes in between 👉 Goal: Destroy maximum UNIQUE walls 🚨 Initial Thought (Where Most Go Wrong) At first, it feels like: “For each robot, choose the direction that destroys more walls.” ❌ This greedy approach fails Because: Robots can destroy overlapping walls Decisions of one robot affect others 💡 Key Insight This is NOT an independent decision problem. 👉 It’s a global optimization problem 👉 Requires Dynamic Programming (DP) 🧠 Approach Breakdown Step 1: Preprocessing Sort robots and walls Maintain mapping of robot → distance Step 2: For each robot, calculate: left[i] → walls destroyed if robot shoots LEFT right[i] → walls destroyed if robot shoots RIGHT num[i] → walls between robot[i-1] and robot[i] (overlap region) Step 3: DP State We maintain: subLeft → max walls till i if current robot shoots LEFT subRight → max walls till i if current robot shoots RIGHT Step 4: Transition currentLeft = max( subLeft + left[i], subRight - right[i-1] + min(left[i] + right[i-1], num[i]) ); currentRight = max( subLeft + right[i], subRight + right[i] ); 👉 The tricky part is handling overlap correctly 🎯 Key Learning 👉 If decisions: overlap affect each other Then: Greedy ❌ DP ✅ 🚀 Final Thoughts: This problem taught me. Don’t trust the greedy blindly. Always think about overlapping effects. DP often hides behind “simple-looking” problems.. 💬 Would love to hear: Have you faced similar problems where greedy failed? #DSA #LeetCode #DynamicProgramming #Java #CodingInterview #SoftwareEngineering

🚀 Day 4 of Problem Solving Journey Today, I tackled LeetCode 3661 – Maximum Walls Destroyed by Robots 🤖💥 🔍 Problem Summary: Given positions of robots and walls on a straight line, each robot can shoot once either left or right within a limited distance. The challenge is to calculate the maximum number of unique walls that can be destroyed without bullets passing through other robots. 🧠 Key Learnings: Sorting helps simplify positional problems Binary search can optimize finding valid ranges Careful handling of edge cases (like robots blocking bullets) Efficient mapping of robot positions to distances ⚡ Approach: Used sorting + binary search to determine reachable walls for each robot and ensured no overlap or blockage affects the final count. ✅ Result: Accepted ✔️ 📌 Consistency is the key — one problem a day keeps improving problem-solving skills! #Day4 #LeetCode #ProblemSolving #Java #CodingJourney #DataStructures #Algorithms #Consistency

#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

Day 95: Back to Square One 🤖📍 Problem 657: Robot Return to Origin Today’s problem was a clean exercise in coordinate tracking. The goal: determine if a sequence of moves (U, D, L, R) brings a robot back to its starting position (0,0). The Logic: • The "Odd" Shortcut: Realized that if the total number of moves is odd, it’s mathematically impossible to return to the origin. Every move needs an opposite to cancel it out. • Coordinate Mapping: Used two variables, x and y, to represent the robot's position. ∘ Horizontal: 'R' increments x, 'L' decrements x. ∘ Vertical: 'U' increments y, 'D' decrements y. • The Verdict: If both x and y are zero after the final move, the robot successfully completed the circle. Sometimes the most effective solutions are the ones that stick to the basics. No complex data structures needed—just simple arithmetic and a bit of parity logic. 🚀 #LeetCode #Java #Algorithms #ProblemSolving #DailyCode

Excited to finally open-source NervLynx — a modular robotics runtime framework I’ve been building. In my experience working on robotics projects (and the ones I’ve built), most start with quick Python scripts for prototypes. That works great for demos… until you need to move to production. Suddenly you’re fighting non-deterministic behavior, impossible-to-debug data flows, missing observability, and fragile lifecycle management. NervLynx is my attempt to close that exact gap. It’s not a full autonomy stack or a ROS replacement. It’s the runtime foundation that sits underneath your nodes — giving you: Deterministic + async execution with priority scheduling Traceable dataflow (every message carries topic, source, sequence, timestamp, schema, and trace_id) Built-in safety primitives (watchdogs, backpressure detection, startup dependency supervision) Full observability out of the box (replayable JSONL traces, latency/flow stats, Prometheus-style metrics) Config-driven graph wiring via YAML + a clean plugin SDK First-class support for both Python and C++ runtimes Ready-to-use deployment profiles (Docker + systemd) Everything is MIT licensed, thoroughly tested, and ships with working smoke tests and examples so you can try it immediately. I'm currently working on sharing realtime results from actual sensors using NervLynx — and I'll be posting those results soon. If you’re building mobile robot software (perception → planning → actuation pipelines, surveillance stacks, or any hardware-in-the-loop system) and you’ve ever struggled moving from prototype scripts to something maintainable and observable, I’d genuinely love your feedback. 👉 Repo: https://lnkd.in/gGu29zMW #Robotics #OpenSource #RobotSoftware #AutonomousSystems #Embedded #Python #CPP

Just solved LeetCode 657: Robot Return to Origin and it’s a great reminder that clean, fundamental thinking often beats over-engineering. The Problem: Given a string of moves (U, D, L, R), determine if a robot starting at (0,0) returns to the origin after executing all commands. My Approach: Coordinate Tracking Simulation Instead of reaching for complex data structures, I simply tracked position: • x → horizontal (R = +1, L = -1) • y → vertical (U = +1, D = -1) • Return true if x == 0 && y == 0 Time: O(n) | Space: O(1) Key Insight: Opposite moves cancel each other out. If count(U) == count(D) and count(L) == count(R), the robot must return home. Sometimes a quick count beats step-by-step simulation! This pattern (coordinate tracking + cancellation logic) is foundational for grid problems, path simulations, and even early-stage BFS/DFS intuition. Mastering the basics makes medium problems feel much more approachable. #LeetCode #Algorithms #CodingInterview #SoftwareEngineering #Python #ProblemSolving #TechCareers #DataStructures

Internal tool, not a product. Sharing (link in comments) in case the setup overhead is familiar: Setting up IKFast on a new robot (6R industrial, 7-DOF redundant etc.) requires OpenRAVE, an old Ubuntu image, a ROS container, and a working collada export. We run this on every new robot we onboard, so between projects we wrote a replacement. We call it miniIK: one Python file, no dependencies, drop in any URDF, get verified IK solutions at IKFast precision. In pure Python on stock CPU it handles real-time Cartesian tracking above 1 kHz and full branch-discovery planning in about a second, no numpy, no compilation step. What it does: - Reads URDFs directly — no DH parameter table, no preliminary arm-modeling step. - One solver for every robot: 6R industrial, 7-DOF redundant, SCARA, gantry+arm, humanoid upper body. No topology-specific code paths. - Returns every valid solution, not just one. A 6R arm typically has up to 8 IK branches and you want all of them when planning around obstacles. - Ranks them by joint-space distance to the robot's current configuration. The closest branch comes first. - Verifies every solution against forward kinematics before returning it. Anything outside tolerance is dropped.

#100DaysOfLeetcode journey 🚀 Day 58/100 — Optimizing Perpetual Motion! Today’s Problem: 2069. Walking Robot Simulation II 🔹 The Goal: Design a robot that traverses the perimeter of a grid. It moves forward until it hits a wall, then turns 90 degrees counter-clockwise. We need to support millions of steps and instantly return the robot's $(x, y)$ coordinates and direction. 🔹 The Insight: A naive simulation would take $O(\text{steps})$, which fails for large inputs. The key is to realize the robot's path is a Fixed Cycle. By calculating the perimeter length and using the modulo operator, we can reduce any number of steps into a single coordinate on the boundary. 🔹 The Logic: Modulo Striding: I used pos = (pos + num) % loopSize to handle infinite movement in constant time. Perimeter Segmentation: I mapped the 1D "distance traveled" into 2D coordinates by dividing the perimeter into four segments (Bottom, Right, Top, Left). The "Facing" Trap: The trickiest part is the direction. If the robot lands on $(0,0)$ after moving, it’s not facing "East" anymore—it’s facing "South" because it technically hit the boundary and turned before stopping. ✨ Achievement: Day 58! Successfully transformed a simulation problem into a coordinate geometry and modular arithmetic challenge. Crossing the 58% mark by mastering "cycle-based" optimizations is a great feeling. 🔍 Steps followed: ✔ Cycle Length Derivation: Mathematically defined the unique cells on the grid boundary. ✔ Constant-Time Positioning: Implemented $O(1)$ jumps to any future state. ✔ State-Aware Logic: Handled the edge cases of initial vs. post-movement orientation at the origin. 🔧 Complexity Analysis: Time Complexity: $O(1)$ per operation. Space Complexity: $O(1)$. 58 days down! The logic is getting cleaner, and the "Hard" constraints are becoming much more manageable with the right math. 🛠️ #Java #DSA #LeetCode #CodingJourney #100DaysOfCode #SoftwareEngineer #InternshipBound #Programming #Simulation #Geometry #Optimization #Day58

💡Day 46 of LeetCode Problem Solved! 🔧 🌟657. Robot Return to Origin🌟 🔗 Solution Code: https://lnkd.in/gffUSuzJ Task : • There is a robot starting at the position (0, 0), the origin, on a 2D plane. Given a sequence of its moves, judge if this robot ends up at (0, 0) after it completes its moves. • You are given a string moves that represents the move sequence of the robot where moves[i] represents its ith move. Valid moves are 'R' (right), 'L' (left), 'U' (up), and 'D' (down). • Return true if the robot returns to the origin after it finishes all of its moves, or false otherwise. • Note: The way that the robot is "facing" is irrelevant. 'R' will always make the robot move to the right once, 'L' will always make it move left, etc. Also, assume that the magnitude of the robot's movement is the same for each move. Example 1: Input: moves = "UD" Output: true Explanation: The robot moves up once, and then down once. All moves have the same magnitude, so it ended up at the origin where it started. Therefore, we return true. Example 2: Input: moves = "LL" Output: false Explanation: The robot moves left twice. It ends up two "moves" to the left of the origin. We return false because it is not at the origin at the end of its moves. #LeetCode #Java #DSA #ProblemSolving #Consistency #100DaysOfChallenge #CodingJourney #KeepGrowing

🚀 Day 36 of 100 Days LeetCode Challenge Problem: Robot Return to Origin Day 36 is a simple but important simulation + counting problem 🔥 💡 Key Insight: To return to origin (0,0): Total moves Right = Left Total moves Up = Down 👉 If both conditions satisfy → ✅ robot returns to origin 🔍 Core Approach: 1️⃣ Initialize Counters x = 0, y = 0 2️⃣ Process Each Move 'R' → x++ 'L' → x-- 'U' → y++ 'D' → y-- 3️⃣ Final Check If (x == 0 && y == 0) → true Else → false 💡 Alternative Approach: Count occurrences: count('R') == count('L') count('U') == count('D') 🔥 What I Learned Today: Simple problems test basic logic clarity Simulation problems are about correct tracking Don’t overcomplicate—keep it clean 📈 Challenge Progress: Day 36/100 ✅ Consistency still going strong! LeetCode, Simulation, Strings, Counting, Coordinates, Arrays, DSA Practice, Coding Challenge, Problem Solving #100DaysOfCode #LeetCode #DSA #CodingChallenge #Simulation #Strings #ProblemSolving #TechJourney #ProgrammerLife #SoftwareDeveloper #CodingLife #LearnToCode #Developers #Consistency #GrowthMindset #InterviewPrep