Dynamic Programming Solution for Maximum Earning Robot

🚀 Day 32 of #LeetCode Journey ✅ Problem: Robot Return to Origin (LeetCode 657) Today’s problem was simple yet a great way to strengthen basic logic and coordinate tracking skills. 🔍 Problem Statement: Given a string of moves (U, D, L, R), determine if the robot returns to the origin after completing all moves. 💡 Approach: * Start from position (0,0) * Track horizontal (x) and vertical (y) movements * Update position based on each move * Finally, check if we are back at (0,0) 🧠 Key Insight: Equal number of opposite moves cancel each other out: * U cancels D * L cancels R ⏱️ Complexity: Time: O(n) Space: O(1) 📌 Takeaway: Even simple problems help build strong fundamentals in problem-solving and coding logic. #Java #LeetCode #CodingJourney #100DaysOfCode #Programming #Developer

🚀 Day 5 of Daily LeetCode Challenge (06 April 2026) Staying consistent and pushing forward 💪 Today’s problem: “Robot Return to Origin” (Easy) 📌 Problem Insight: A robot moves on a 2D plane using steps (U, D, L, R). The goal is to determine whether it returns back to the origin after completing all moves. 🧠 My Approach: Used (x, y) coordinates to track movement Increment/decrement based on direction Final check: (x == 0 && y == 0) → back to origin ✅ ✨ Key Takeaway: Even easy problems strengthen logic and help build consistency — the real game is discipline, not difficulty. 💻 Time Complexity: O(n) 💾 Space Complexity: O(1) 🔥 5 days in — building momentum, one problem at a time! #LeetCode #Day5 #DSA #CodingJourney #Java #ProblemSolving #Consistency #100DaysOfCode

🚀 Day 553 of #750DaysOfCode 🚀 🤖 LeetCode 657: Robot Return to Origin Today’s problem was simple yet a great reminder of how powerful basic logic can be. 📌 Problem Summary: A robot starts at (0,0) and follows a sequence of moves: 'U' → Up 'D' → Down 'L' → Left 'R' → Right 👉 Goal: Check whether the robot returns back to the origin after all moves. 💡 Approach: Instead of overthinking, I focused on tracking movement on axes: Vertical → U cancels D Horizontal → L cancels R If both balances are zero → back to origin ✅ 🧠 Key Insight: This problem is not about simulation, but about balance. ⚡ Complexity: Time → O(n) Space → O(1) 🎯 Takeaway: Even easy problems strengthen your fundamentals. Consistency > Complexity 💯 🔥 553 days down, 197 to go! Let’s keep building 🚀 #LeetCode #Java #DSA #CodingJourney #Consistency #ProblemSolving #Developers #100DaysOfCode #Tech

#Day358 of #1001DaysOfCode 📘 LeetCode Daily Challenge Problem: Robot Return to Origin (LeetCode 657) 💡 Approach: Simulated the robot’s movement on a 2D plane by tracking vertical and horizontal positions. Each move updates the position, and if the robot returns to (0,0), it means all movements are balanced. ⏱ Time Complexity: O(n) 🧠 Space Complexity: O(1) Consistency is key — one problem every day 🚀 #DSA #Java #LeetCode #ProblemSolving #Coding

Solved a basic but interesting problem: Robot Return to Origin 🤖 In this problem, a robot moves based on given directions (Up, Down, Left, Right). The goal is to check whether the robot comes back to the starting point (0,0) after completing all moves. 👉 Approach: I tracked the movement using x and y coordinates. Left/Right affects x Up/Down affects y If both x and y become 0 at the end, it means the robot returned to origin. Simple logic, but a good reminder that strong basics are very important 💡 #LeetCode #DSA #ProblemSolving #CodingJourney #Java #Stack #Algorithms #Consistency #Learning #Tech

#Day359 of #1001DaysOfCode 📘 LeetCode Daily Challenge Problem: Walking Robot Simulation (LeetCode 874) 💡 Approach: Simulated the robot’s movement step-by-step while tracking direction changes. Used a HashSet to store obstacle positions for O(1) lookup, ensuring the robot stops correctly when encountering obstacles. Updated the maximum distance from origin after each move. ⏱ Time Complexity: O(n + k) 🧠 Space Complexity: O(m) Staying consistent and improving problem-solving skills every day 🚀 #DSA #Java #LeetCode #ProblemSolving #Coding

I used to write a lot of "if let" when dealing with "Option" in Rust. It works. It’s clear. No problem with it. But after a while, it started to feel a bit... repetitive. Then I came across a small pattern that changed how I write this kind of code: Instead of this: let mut items = vec![1, 2, 3]; if let Some(x) = maybe_value { items.push(x); } I started doing this: let mut items = vec![1, 2, 3]; items.extend(maybe_value); At first, it feels almost too simple. But that’s kind of the point. No branching. No extra lines. Just: “if there’s something, add it.” And it scales really nicely: items.extend(maybe_value); items.extend(other_optional); items.extend(optional_vec); This small shift helped me think differently about "Option<T>" — not as a special case, but as a tiny collection (zero or one element). Since then, my code feels a bit more uniform… and honestly, easier to read. Nothing groundbreaking here. Just one of those small improvements that quietly make your code better over time. #Rust #CleanCode #SoftwareEngineering #RustLang #Programming #CodeQuality #DevTips #BackendDevelopment

Clean code is not about being clever. It’s about surviving 6 months later. That “smart” one-liner you wrote today? Even you won’t understand it after a few sprints. Good code: → is boring → is readable → is predictable Because code is read 10x more than it’s written. Optimize for the next developer. (It might be you.) #DeveloperTips #CleanCode #SoftwareEngineering #Coding

Day 45 of solving LeetCode problems. Solved: Defuse the Bomb This problem looks simple, but it tests a key concept — handling circular arrays with a sliding window. Instead of recalculating sums repeatedly (O(n*k)), I used a rolling window to bring it down to O(n). The tricky part was managing indices correctly when moving forward vs backward (k > 0 vs k < 0). Key takeaway: Efficient thinking is about avoiding redundant work, not just getting the correct answer. Consistency is good, but improving how you think matters more. #leetcode #dsa #problemSolving #coding #100DaysOfCode

Day 56 of solving LeetCode. Today’s problem: Count Primes Result: Accepted ✔️ (66/66 test cases) Performance: • Runtime: 96 ms (beats 75.25%) • Memory: 51.26 MB Key takeaway: Brute force thinking doesn’t scale. The real shift is recognizing patterns like the Sieve of Eratosthenes — precompute smartly instead of checking repeatedly. Most people stay stuck trying to optimize loops. The smarter move is changing the approach entirely. Consistency > motivation. 56 days in. No breaks. No excuses. ∆ #LeetCode #CodingJourney #DSA #Programming #ProblemSolving #100DaysOfCode #DeveloperLife #CodingDaily #Algorithms #Consistency #GrowthMindset