🚀 LeetCode Progress Update | Dynamic Programming Just solved LeetCode – House Robber using Dynamic Programming 🧠✅ This problem is a great example of how breaking a problem into optimal subproblems leads to an efficient solution. 🔍 Key Learning Points: Defined a clear DP state: dp[i] = maximum money that can be robbed from the first i houses At each step, chose between: Stealing from the current house + dp[i-2] Skipping the current house → dp[i-1] Final transition: dp[i] = max(dp[i-1], nums[i-1] + dp[i-2]) ⚙️ Complexity: Time: O(n) Space: O(n) (can be optimized to O(1)) 💡 This problem strengthened my understanding of: DP base cases State transitions Translating problem constraints into DP logic Consistent practice + clarity in fundamentals is the key 🔑 Onward to the next challenge 💪 #LeetCode #DynamicProgramming #DSA #ProblemSolving #Java #CodingJourney #SoftwareEngineering #LearningByDoing
LeetCode House Robber Solved with Dynamic Programming
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One of the best habits I’m currently building is writing pseudocode before writing real code. It forces me to slow down, think clearly, and design the logic first instead of jumping straight into syntax and getting stuck. If you’re new to programming, pseudocode is one of the simplest tools to reduce overwhelm and build real problem-solving skills. Clear thinking first. Clean code later. #OctoPrep #PythonDeveloper #DataScience #DataAnalysis
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🚀 Day 36/100 of My LeetCode Challenge: Mastering Greedy & Dynamic Programming! Just solved LeetCode 3507: Minimum Pair Removal to Sort Array I – an interesting problem that beautifully blends greedy operations with dynamic programming thinking! 🧠 🔍 The Challenge: Given an array, repeatedly replace the adjacent pair with the minimum sum until the array becomes non-decreasing. Return the minimum number of such operations required. 💡 Key Insights: This isn't just about blindly following the operation description (selecting minimum sum pairs) The optimal solution requires recognizing this as a dynamic programming problem We need to find the minimum operations to partition the array into segments that can be merged while maintaining the non-decreasing property Each merge operation happens only when the left segment's sum exceeds the right segment's sum 🛠️ My Approach: Implemented a memoized DP solution that explores all possible partition points For each possible split position, recursively solve left and right subarrays Add a merge cost when the left segment's sum is greater than the right segment's sum Use memoization to avoid redundant computations for O(n²) time complexity 📊 Performance: Runtime: 66.06% Memory: 44.62 MB Beats: 32.51% of Java submissions 🎯 Why This Matters: This problem is a great example of how: Problem understanding matters more than just following instructions Dynamic programming can elegantly solve what seems like a greedy problem Memoization dramatically improves performance for recursive solutions Real-world scenarios often require transforming problem statements into solvable patterns ✨ The Journey Continues: Every day of this 100-day challenge brings new learning opportunities. Today reinforced that sometimes the direct approach isn't optimal, and we need to think one level deeper about the underlying structure of the problem. #LeetCode #CodingChallenge #100DaysOfCode #ProblemSolving #DynamicProgramming #GreedyAlgorithms #Java #SoftwareEngineering #TechCareer #DeveloperJourney #Algorithm #DataStructures #CodingInterview #Programming
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👇 🚀 Day 52 of #100DaysOfCode 💻 Consistency check: still going strong 💪 Today’s problem was a clean and practical string manipulation question — 📝 Length of Last Word (LeetCode) 📌 Problem Summary Given a string consisting of words and spaces, find the length of the last word in the string. Example: "Hello World" → 5 Sounds simple, but edge cases matter: Trailing spaces Multiple spaces between words 🧠 My Approach Instead of splitting the string (which uses extra space), I traversed the string from the end: 1️⃣ Skip all trailing spaces 2️⃣ Count characters until a space is found This keeps the solution efficient and clean. ⚙️ Complexity Analysis ⏱ Time: O(n) 💾 Space: O(1) Optimized and interview-friendly 🚀 🔥 Key Learning Not every problem needs complex logic Backward traversal is a powerful trick for string problems Handling edge cases is what makes solutions robust ✅ Accepted with 0 ms runtime Another step forward in my coding journey ✔️ On to Day 53! 🚀💻 #100DaysOfCode #LeetCode #Java #Strings #ProblemSolving #DSA #CodingJourney #Consistency
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When websites rely on dynamic rendering, Selenium becomes essential. Vilius Dumcius, Product Owner at IPRoyal, explains how to use Ruby Programming Language and Selenium together to extract data reliably and build more resilient scraping workflows for modern web pages: https://lnkd.in/dXfHKqKA
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One thing I always check in code reviews: Exception handling. I've seen too many try-catch blocks that just swallow errors silently. No logs. No alerts. Just empty catch blocks pretending nothing happened. Then something breaks in production and nobody knows why. Good error handling means: Log the error with context. Fail fast when recovery isn't possible. Give the user a clear message, not a generic "something went wrong." The 5 minutes it takes to write proper exception handling saves hours of debugging later. What's something you always look for in code reviews? #Java #SpringBoot #CodeReview #Programming #SoftwareDevelopment
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🚀 Day 41 of #100DaysOfCode Today’s challenge was a classic and powerful Dynamic Programming problem — Matrix Chain Multiplication 🧮 (A textbook example of optimization DP) 📌 Problem Summary Given a sequence of matrices, the task is to find the minimum number of scalar multiplications needed to multiply them together. Important catch ⚠️ Matrix multiplication is associative But different parenthesizations lead to very different costs So the goal isn’t to multiply them — it’s to multiply them optimally. 🧠 My Approach: Dynamic Programming (Interval DP) This problem follows an interval-based DP pattern: Define dp[i][j] as the minimum cost to multiply matrices from i to j Base case: A single matrix → cost = 0 Transition: Try every possible split point k between i and j Choose the split that minimizes total cost This approach systematically explores all valid parenthesizations while avoiding recomputation 💡 ⚙️ Complexity Analysis ⏱ Time: O(n³) 💾 Space: O(n²) 🔥 Key Learning Not all DP problems are linear — some require thinking in ranges/intervals Breaking a big problem into left + right subproblems is a recurring DP theme Optimization problems often look complex, but DP brings structure and clarity ✅ All test cases passed successfully, reinforcing my understanding of advanced DP patterns 💪 Dynamic Programming is no longer intimidating — it’s becoming logical and enjoyable 🚀 Onward to Day 42! #100DaysOfCode #DynamicProgramming #MatrixChainMultiplication #Java #DSA #ProblemSolving #CodingJourney #GeeksForGeeks
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Encapsulation is one of the core pillars of Object-Oriented Programming. By hiding internal data and exposing only what is necessary, we build secure, maintainable, and flexible code. 💡 Protect your data — control access with encapsulation. #Encapsulation #OOPs #Java #ObjectOrientedProgramming #SoftwareDevelopment #CodingConcepts #LearnJava #TechBasics #Programming
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🧠 LeetCode 62: Unique Paths (The Way I Understand It) 🚀 For this problem, I used a 2D Dynamic Programming approach that I can actually visualize 👀. Instead of starting from the beginning, I worked backwards from the destination. ❌ Instead of asking: “How do I get to the end?” ✔️ I asked: “How many ways lead to this cell?” 👉 Each cell answers one question: “How many unique paths lead from here to the end?” 🔍 Key Idea 💲 The robot can only move Right or Down 💲 So if we reverse our thinking, we only care about Up and Left 💲 Every cell’s value represents: The number of unique paths that lead from that cell to the destination 🔑 Logic ✅ The bottom-right cell has exactly 1 path (you’re already there) ✅ Cells in the last row can only move left → 1 path ✅ Cells in the last column can only move up → 1 path ✅ Every other cell: ```paths = paths from right + paths from down``` 🎯 Why I like this ✅ Very visual ✅ Easy to reason about ✅ Easy to debug it turns out this is also one of best optimized solution beating 100% runtime and 81% memory. Nailed it 👍 #LeetCode #DynamicProgramming #Java #ProblemSolving #LearningInPublic 💻🔥
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🚀 Optimizing the Classic FizzBuzz Challenge in Java What started as a simple coding exercise turned into an interesting exploration of performance optimization! I recently implemented FizzBuzz using a counter-based approach that eliminates modulo operations entirely. Here’s what I learned: 💡 Key Insights: • Traditional approach uses modulo (%) for divisibility checks - which involves division operations • Counter-based method uses simple integer comparisons instead • Combined with StringBuilder pre-allocation, this handles millions of iterations efficiently • Result: 5-15% performance improvement on large datasets 🔧 Technical Approach: Instead of checking “if (i % 3 == 0)”, I use counters that increment and reset: - fizzCounter counts 1→2→3→reset - buzzCounter counts 1→2→3→4→5→reset - When counters hit their targets, print the corresponding word This replaces expensive division with lightweight comparisons. 📊 Why It Matters: While FizzBuzz seems simple, optimizing it teaches valuable lessons about: ✓ Understanding computational costs of operations ✓ Choosing the right data structures (StringBuilder vs repeated I/O) ✓ Balancing readability with performance ✓ Measuring and benchmarking improvements The code is clean, scalable, and demonstrates that even classic problems have room for optimization. 🔗 Check out the full implementation on my GitHub: https://lnkd.in/dn8_573E What optimization techniques have you discovered in seemingly simple problems? #Java #Programming #Optimization #SoftwareEngineering #CodingChallenge #CleanCode #PerformanceTuning #TechLearning
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Day 10 | LeetCode Daily Challenge - 712. Minimum ASCII Delete Sum for Two Strings Problem in short: You're given two strings and allowed to delete characters from either one. Each deletion costs the ASCII value of the character. The goal is to make both strings equal with the minimum total deletion cost. Key insight I learned: Instead of directly thinking in terms of deletions, it helps to reframe the problem as keeping the right common subsequence and deleting everything else. Using Dynamic Programming, we compare prefixes of both strings and decide: • when characters match, keep them without any cost • when they don’t, delete from the string that results in a smaller ASCII cost This approach systematically explores all possibilities and guarantees the minimum total cost. Another good example of how reframing the problem makes the DP solution much clearer. On to the next challenge 🚀 #LeetCodeDaily #Day10 #DSA #DynamicProgramming #ProblemSolving #Java #Consistency #LearningByDoing
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