Dynamic Programming Techniques for LeetCode Contest 496

Ever look at a LeetCode problem and think, "Oh, this is just simple math," only to see a constraint of $10^{15}$ staring back at you? That was my experience with the "Count Good Numbers" problem today. The core logic was actually pretty straightforward: even indices get even digits (5 options), and odd indices get prime digits (4 options). But trying to calculate 5^even * 4^odd for massive numbers? Immediate Time Limit Exceeded (TLE). I had to scrap the standard approach and write a custom pow() function using Binary Exponentiation. It’s wild how applying modulo 10^9 + 7 at every single multiplication step within the recursive call is the exact difference between a failing solution and a clean O(log n) pass without integer overflow. Definitely a great reminder that knowing the brute-force math is only half the battle—optimizing it is where the real engineering happens. Back to the grind! #LeetCode #DSA #CPP #CompetitiveProgramming #SoftwareEngineering #CodingJourney

🚀 Day 28 / 128 – Coding Challenge Progress Today I worked on the classic "Square Root (x)" problem from LeetCode. 🔍 Problem Summary: Given a non-negative integer x, compute and return its square root rounded down to the nearest integer — without using built-in exponent functions. 💡 Approach I Used: I implemented a Binary Search solution to efficiently find the integer square root. Instead of checking every number, I narrowed down the answer by: Picking a middle value Comparing mid * mid with x Adjusting the search range accordingly ⚡ Why Binary Search? It reduces time complexity from O(n) to O(log n) — making the solution much faster and scalable. 🧠 Key Learning: This problem reinforced how powerful binary search is, even beyond sorted arrays — especially for mathematical computations. 📌 Progress: Day 28 complete, staying consistent and learning something new every day! #128DaysOfCode #CodingChallenge #LeetCode #JavaScript #ProblemSolving #BinarySearch #DeveloperJourney

⚙️ Spent 3 hours fixing asset names in UE5 today. Never again. Inherited a messy UE project? Forgotten to prefix assets as you go? 💡The Solution: I built a Python tool that handles the entire renaming process in one click. 🎯 What makes it different:  • Rules live in a Data Table (not hardcoded): anyone on the team can update prefixes without touching code  • Two-pass collision check: no silent overwrites or mid-rename crashes  • Dry run mode: preview every change before committing  • Skip folders, cancel safely anytime Built for solo devs, students, technical artists, and pipeline TDs who want BP_, SM_, T_ conventions enforced automatically. 🎨 Preview the tool and get the details here: https://lnkd.in/edyCPz3J #UnrealEngine5 #GameDevelopment #PythonScripting #TechnicalArt #GameTools

Here's the thing most developers miss: Claude Code has full terminal access. That's the entire point. It can shell out to wc for accurate token counts (LLMs are terrible at counting). It can use grep, sed, awk for text processing. It can hit APIs, execute Python scripts, deploy to production. It can also rm -rf your entire system if you're not careful with permissions. So the setup isn't optional. It's the foundation. We talked about Claude Code setup and barely opened Claude Code itself yesterday. Because that's how software development works. You build the infrastructure first. #ClaudeCode #DeveloperProductivity #AITools #SoftwareDevelopment

Day 87 on LeetCode — Flatten a Multilevel Doubly Linked List 🔗🧠🔥 Now this was a next-level linked list problem — combining recursion with pointer manipulation 💯 🔹 Flatten a Multilevel Doubly Linked List The goal was to convert a multilevel list (with child pointers) into a single-level doubly linked list. 🔹 Approach Used in My Solution (DFS + Recursion) Key idea: • Traverse the list node by node • If a node has a child:   – Recursively flatten the child list   – Insert the flattened child between current node and next node   – Connect prev and next pointers properly   – Set child = nullptr • Maintain a last pointer to track the tail of the flattened part This is essentially a DFS traversal on a linked list structure. ⚡ Complexity: • Time Complexity: O(n) • Space Complexity: O(n) (due to recursion stack) 💡 Key Takeaways: • Combined recursion + pointer manipulation effectively • Learned how to flatten hierarchical structures using DFS thinking • Careful handling of next, prev, and child pointers is critical 🔥 This is where linked lists start feeling like real problem-solving, not just traversal. #LeetCode #DSA #Algorithms #DataStructures #LinkedList #DoublyLinkedList #Recursion #DFS #ProblemSolving #Coding #Programming #Cpp #STL #SoftwareEngineering #ComputerScience #CodingPractice #DeveloperLife #TechJourney #CodingDaily #Consistency #100DaysOfCode #BuildInPublic #AlgorithmPractice #CodingSkills #Developers #TechCommunity #SoftwareDeveloper #EngineeringJourney

𝗬𝗼𝘂𝗿 𝗣𝘆𝘁𝗵𝗼𝗻 𝗽𝗿𝗼𝗷𝗲𝗰𝘁 𝗵𝗮𝘀 𝗳𝗹𝗮𝗸𝗲𝟴, 𝗶𝘀𝗼𝗿𝘁, 𝗯𝗹𝗮𝗰𝗸, 𝗽𝘆𝗹𝗶𝗻𝘁, 𝗮𝗻𝗱 𝗯𝗮𝗻𝗱𝗶𝘁. Five tools. Five configs. Five places for CI to silently drift. Ruff replaces all of them. Written in Rust, it runs in milliseconds on a full codebase. One config in 𝗽𝘆𝗽𝗿𝗼𝗷𝗲𝗰𝘁.𝘁𝗼𝗺𝗹. One command in CI. 🔹 Replaces: flake8, isort, black, pyupgrade, pydocstyle, and 700+ rules from pylint 🔹 10–100x faster than the tools it replaces — no more slow CI lint steps 🔹 𝗿𝘂𝗳𝗳 𝗰𝗵𝗲𝗰𝗸 --𝗳𝗶𝘅 auto-fixes most violations — not just reports them 🔹 One binary, zero inter-tool conflicts, one source of truth This "BIFES UP" ruleset is my way of beefing up code quality while keeping the stakes high on security. It covers 90% of what matters from security to logic. Want 100%? Just extend it with 𝗪, 𝗔𝗦𝗬𝗡𝗖, and 𝗡 — especially essential for ensuring concurrency safety in FastAPI environments. What other rules do you often include in your projects? #Python #DevSecOps #SoftwareEngineering #BackendDevelopment #SoftwareArchitecture

Day 18: Why the "How" matters more than the "What" 🏗️ Today’s #30DaysOfCode challenge was a classic: The Palindrome Checker. While most would reach for a simple .reverse() method, I took a deeper dive into Linear Data Structures using JavaScript to implement a dual-stream logic: Stacks (LIFO): To handle the reverse flow of data. Queues (FIFO): To maintain the original sequence. The Engineering Takeaway: By comparing the "pop" of a stack against the "dequeue" of a queue, you aren't just checking a string—you’re managing data flow efficiency. As I scale my work in Data Engineering, these fundamentals are the difference between a script that just "works" and a system that is built to scale. Shoutout to my AI collaborator #Gemini for the technical sounding board as I bridge these concepts across my Java and JS projects. #DataEngineering #JavaScript #Algorithms #DataStructures #30DaysOfCode #TechCareer

Day 36 of My DSA Journey Today I solved LeetCode 150 – Evaluate Reverse Polish Notation (RPN) on LeetCode. 📌 Problem Evaluate the value of an arithmetic expression in Reverse Polish Notation. Valid operators are: +, -, *, / Each operand may be an integer. Example: Input: ["2","1","+","3","*"] Output: 9 Explanation: (2 + 1) * 3 = 9 🧠 Approach – Stack This problem is a perfect application of the Stack data structure. Steps I followed: • Traverse the tokens array. • If the element is a number → push it onto the stack. • If it is an operator: Pop the top two elements (a and b) Perform the operation (a operator b) Push the result back into the stack • At the end, the stack contains the final result. ⏱ Time Complexity: O(n) — We process each token once 📦 Space Complexity: O(n) — Stack to store operands 💡 Key Learnings ✔ Understanding Reverse Polish Notation (Postfix Expression) ✔ Applying stack for expression evaluation ✔ Handling operator precedence implicitly using stack This problem connects DSA with compiler/interpreter concepts, which makes it even more interesting 🚀 Consistency continues — improving every day 💪 #100DaysOfCode #DSA #Stack #LeetCode #Java #ProblemSolving #CodingJourney #DeveloperJourney #Programming #SoftwareEngineering

🚀 Day 27/50 – LeetCode Challenge 🧩 Problem: Count and Say Today’s problem focused on generating a sequence based on reading and describing the previous term, which is a great exercise in string manipulation and pattern building. 📌 Problem Summary: The “Count and Say” sequence is defined as: Start with "1" Each next term is formed by reading the previous term aloud Example: 1 → "one 1" → 11 11 → "two 1s" → 21 21 → "one 2, one 1" → 1211 1211 → "one 1, one 2, two 1s" → 111221 🔍 Approach Used ✔ Started from the base case "1" ✔ Iterated to build each next term ✔ Counted consecutive digits ✔ Formed the new string using: count + digit ✔ Repeated until reaching the required n ⏱ Time Complexity: O(n × m) 📦 Space Complexity: O(m) 💡 Key Learning ✔ Understanding pattern generation problems ✔ Working with string grouping and counting ✔ Building results step by step ✔ Improving attention to detail in iteration This problem shows how simple rules can create complex patterns. Consistency is the key to mastery 🚀 🔗 Problem Link: https://lnkd.in/g_cMVP5a #50DaysOfLeetCode #LeetCode #DSA #Strings #ProblemSolving #CodingJourney #FutureAIEngineer #Consistency

LeetCode Day 13 : Problem 42 (Trapping Rain Water) Just solved another LeetCode problem. It was "Trapping Rain Water", sounds like a physics problem, right? But here's what I actually learned: The core insight, for any position, the water it can hold is limited by the shorter of the two tallest walls surrounding it. min(maxLeft, maxRight) - height[i]. Once that clicked, the solution wrote itself. My first approach used two extra arrays, one for the tallest bar to the left of each position, one for the right. Three passes total. O(n) time, O(n) space. Clean and readable. Then I learned the O(1) space version using two pointers. Instead of pre-building arrays, move from both ends toward the middle. Whichever side is shorter determines the water level, process that side and move inward. Same result, no extra arrays. The pattern I keep seeing, when you need both left and right context, either do two passes and store results, or use two pointers and process on the fly. Same idea, different tradeoff between readability and space. Thirteen problems in. The two pass and two pointer patterns keep showing up in different forms. The more problems you solve, the more you recognise the shape of the solution before you write a single line. The real lesson? Understand why the brute force works first. The optimal solution is just the brute force with the extra space removed. #DSA #LeetCode #JavaScript #CodingJourney #Programming