Subbareddy karri’s Post

Ever wondered how searching in a sorted array can go from slow to lightning fast? ⚡🔍 Hey everyone! Day 290 of my 365-day coding journey, and today I focused on a fundamental algorithm: LeetCode's "Binary Search." Understanding search algorithms is crucial for efficient data retrieval, and this problem is a perfect way to strengthen that skill. Let’s dive in! 🛠️ The Problem Given a sorted array of integers nums and a target value, the task is to search for the target in nums. If the target exists, return its index; otherwise, return -1. The array is guaranteed to be sorted in ascending order. 🎯 The Approach Solution 1: Linear Search - Iterate through each element from start to end. - Compare each element with the target; return the index if matched. - Simple, but O(n) time complexity makes it inefficient for large datasets. Solution 2: Binary Search - Much more efficient for sorted arrays. - Use two pointers, left and right, to track search boundaries. - Calculate mid, compare nums[mid] with target, and adjust pointers accordingly. - Repeat until target is found or pointers cross. - O(log n) time complexity makes it ideal for large datasets. ⚡ 🧠 Key Takeaways - For sorted data, always consider Binary Search first due to its logarithmic efficiency.   - Understanding midpoint calculation and pointer adjustments is critical for correct implementation.   - Linear search is simple but quickly becomes inefficient as input size grows.  💡 Challenge for you! Can you think of scenarios where linear search might still be better than binary search, even though the latter is faster? Share your thoughts! 💬 📺 Check out my video walkthrough I demonstrate both methods and discuss their nuances in detail: https://lnkd.in/dk89dtnJ 🔥 Join the Conversation If you’re sharpening your algorithms skills or tackling daily coding challenges, let’s connect! Always great to learn and share with fellow developers. 🚀 #CodingJourney #DSA #Algorithms #BinarySearch #LeetCode #JavaScript #ProblemSolving #DataStructures #Programming #DeveloperLife #LearningEveryDay #CodeNewbies

⚙️ 𝗖𝗿𝗮𝗳𝘁𝗶𝗻𝗴 𝗖𝗹𝗮𝗿𝗶𝘁𝘆 𝘄𝗶𝘁𝗵 𝗙𝗹𝗮𝘀𝗸 In software, there’s a quiet power in simplicity. Frameworks come and go — each promising speed, structure, and scale. But at the heart of great engineering lies one timeless question: “𝘿𝙤 𝙄 𝙘𝙤𝙣𝙩𝙧𝙤𝙡 𝙩𝙝𝙚 𝙨𝙮𝙨𝙩𝙚𝙢, 𝙤𝙧 𝙙𝙤𝙚𝙨 𝙩𝙝𝙚 𝙨𝙮𝙨𝙩𝙚𝙢 𝙘𝙤𝙣𝙩𝙧𝙤𝙡 𝙢𝙚?” That question defined my choice. And that choice was Flask. Flask doesn’t try to impress with noise. It gives you space — to architect your logic, to shape your flow, to own your performance. It’s not just a microframework; it’s a philosophy of clarity, precision, and freedom. I’ve built APIs where every route has intent, every response has meaning, and every decision feels deliberate — not dictated. Through Flask, I learned that scalability isn’t born from complexity — it’s born from understanding. Because in the end, true craftsmanship in code is not about frameworks. It’s about the mind behind them. #Flask #Python #APIDesign #Microservices #SoftwareArchitecture #EngineeringPhilosophy #CleanCode #LeadershipInTech

What if generating all subsets wasn’t the hard part — but avoiding duplicates was? ⚡ Hey everyone! Day 306 of my 365-day coding journey took me into an interesting combinatorial challenge — LeetCode 90: Subsets II. This problem is a great test of recursion, backtracking, and handling duplicates efficiently. Let’s break it down 🛠️ The Problem Given an integer array that may contain duplicates, the goal is to return all possible subsets (the power set) — but without any duplicate subsets. Example: Input: [1,2,2] Output: [[], [1], [2], [1,2], [1,2,2]] 🎯 The Approaches Solution 1: Brute Force + Set - Generate all possible subsets using recursion or bitmasking. - Store each subset in a set to automatically remove duplicates. - Simple to implement, but storing all combinations before filtering makes it inefficient in terms of both time and memory. Solution 2: Backtracking (Optimized) - Sort the array first to group duplicates. - During recursion, skip duplicate elements at the same decision level. - This avoids generating duplicate subsets altogether, making it clean and efficient. - It’s a perfect example of pruning unnecessary branches in recursion. 🧠 Key Takeaways 💡 Sorting helps manage duplicates effectively in combination problems.   ⚙️ Using sets removes duplicates but doesn’t scale well for larger inputs.   🚀 Backtracking with skipping logic is both elegant and optimal — generating only unique subsets. 💬 Challenge for You When solving recursion problems, do you prefer brute force first for clarity or jump straight into optimization?   Share your thoughts — I’d love to hear your approach! 💬 📺 Watch My Full Walkthrough I’ve explained both brute force and optimized backtracking solutions step-by-step in my latest video: https://lnkd.in/gAEYMPHu 🔥 Join the Conversation If you’re passionate about DSA and love solving problems that sharpen your logical thinking, let’s connect! Every problem solved is one step closer to mastery. 🚀 #CodingJourney #DSA #LeetCode #Subsets #Backtracking #ProblemSolving #Algorithms #DataStructures #Python #JavaScript #TechLearning #CodeNewbies #DeveloperLife #Programming #365DaysOfCode #LearningEveryDay

I have realized that good documentation and clear naming conventions often matter as much as the code itself. When we’re learning or working on fast-paced projects, it’s easy to skip these — but they quietly shape how teams collaborate and how future developers (including our future selves 😅) understand what’s going on. 💡 Why they matter: ✅ They make handovers smoother and reduce confusion. ✅ They help debug faster when something breaks. ✅ They turn messy projects into scalable, professional ones. Clear names and small comments can save hours of digging later. I’ve started following a simple rule: ✍️ “Code for humans first, machines second.” I’d love to hear from others — how do you handle documentation and naming in your projects? Any tips or conventions that work best for your team? 💬 #DataEngineering #LearningJourney #CareerGrowth #SQL #Python #SoftwareBestPractices

𝐔𝐧𝐝𝐞𝐫𝐬𝐭𝐚𝐧𝐝𝐢𝐧𝐠 𝐃𝐞𝐩𝐭𝐡 𝐅𝐢𝐫𝐬𝐭 𝐒𝐞𝐚𝐫𝐜𝐡 (𝐃𝐅𝐒) — 𝐄𝐱𝐩𝐥𝐨𝐫𝐢𝐧𝐠 𝐭𝐡𝐞 𝐃𝐞𝐩𝐭𝐡𝐬 𝐨𝐟 𝐃𝐚𝐭𝐚 𝐒𝐭𝐫𝐮𝐜𝐭𝐮𝐫𝐞𝐬 🔍⁣ ⁣ Follow PrivoLabs For More Interesting Posts When it comes to traversing trees or graphs, Depth First Search (DFS) is one of the most fundamental and powerful algorithms every developer should master. 💡⁣ ⁣ Here’s a quick breakdown 👇⁣ ⁣ 🔹 𝐖𝐡𝐚𝐭 𝐢𝐬 𝐃𝐅𝐒?⁣ DFS is a graph traversal algorithm that explores as far as possible along each branch before backtracking — just like diving deep into one path before exploring others.⁣ ⁣ 🔹 𝐇𝐨𝐰 𝐢𝐭 𝐖𝐨𝐫𝐤𝐬:⁣ 1️⃣ Start from a source node⁣ 2️⃣ Visit the node and mark it as visited⁣ 3️⃣ Recursively visit all its unvisited neighbors⁣ 4️⃣ Backtrack when no unvisited nodes remain⁣ ⁣ 🔹 𝐃𝐅𝐒 𝐂𝐚𝐧 𝐁𝐞 𝐈𝐦𝐩𝐥𝐞𝐦𝐞𝐧𝐭𝐞𝐝 𝐔𝐬𝐢𝐧𝐠:⁣ ⁣ Recursion (call stack)⁣ ⁣ Explicit Stack (iterative method)⁣ ⁣ 🔹 𝐂𝐨𝐦𝐦𝐨𝐧 𝐀𝐩𝐩𝐥𝐢𝐜𝐚𝐭𝐢𝐨𝐧𝐬:⁣ ⁣ Pathfinding in mazes and games 🧩⁣ ⁣ Detecting cycles in graphs 🔁⁣ ⁣ Topological sorting 📈⁣ ⁣ Solving puzzles (like Sudoku) 🧠⁣ ⁣ 💬 𝐏𝐫𝐨 𝐓𝐢𝐩:⁣ DFS is ideal when you want to explore deeply before widely.⁣ For exploring all neighbors first, go for 𝐁𝐫𝐞𝐚𝐝𝐭𝐡 𝐅𝐢𝐫𝐬𝐭 𝐒𝐞𝐚𝐫𝐜𝐡 (𝐁𝐅𝐒).⁣ ⁣ Follow Shivani Sharma For More Interesting Posts #datastructures #algorithms #programming #coding #graphalgorithm #softwareengineering #computerscience #developerlife #techlearning #webdevelopment #python #java #cpp #frontend #backend #devlife #dfs #bfs #codingjourney #problemSolving

Ever wondered what a binary tree looks like from the right side? 🌳👀 Let’s find out! Hey everyone! Day 300 of my 365-day coding journey brought me to a classic tree traversal challenge: LeetCode's "Binary Tree Right Side View." This problem is a great way to understand level-order traversal and how to extract meaningful patterns from it. Let’s dive in. ⚡ 🛠️ The Problem Given the root of a binary tree, imagine standing on the right side of it — your goal is to return the values of the nodes you can see, ordered from top to bottom. In simple terms, at each level of the tree, you only see the rightmost node. 🌲➡️ 🎯 The Approach I solved this using two efficient variations of Breadth-First Search (BFS), both leveraging a queue for level-order traversal. Solution 1: BFS using Queue (with pop) - Used a queue to process nodes level by level. - Captured the queue size before processing each level. - The last node in each level is the one visible from the right side — added it to the result. Solution 2: BFS using Queue (without pop, using indexing) - Instead of popping nodes, I accessed them by their level range indices. - The last node in each level’s range becomes the right-side view node. - Avoids extra pop operations, keeping traversal clean and efficient. 🧠 Key Takeaways 🌿 BFS is the go-to method for problems that involve levels or hierarchies in trees.   🌿 Tracking level boundaries using queue size is a simple yet powerful trick.   🌿 You can also solve this using DFS by recording the last node visited at each depth — multiple ways, same goal! 💡 Challenge for You! Do you prefer BFS or DFS for tree traversal problems?   If you’re a BFS fan, do you use the queue size trick or a sentinel marker to detect level changes? Drop your thoughts below! 💬 📺 Watch My Full Walkthrough I explain both BFS techniques step-by-step in my latest video: https://lnkd.in/gx3xz9mi 🔥 Join the Conversation If you're exploring data structures and algorithms like me, let’s connect! Always great to share ideas and learn together. 🚀 #CodingJourney #DSA #LeetCode #BinaryTree #TreeTraversal #BFS #ProblemSolving #Algorithms #DataStructures #JavaScript #LearningEveryDay #DeveloperLife #Programming #CodeNewbies #TechLearning #365DaysOfCode

Ever wondered how to find a duplicate number without touching the array or using extra space? Let’s decode that today! 🔍✨ Hey everyone! Day 296 of my 365-day coding journey, and today I tackled a classic problem: LeetCode's "Find the Duplicate Number." This challenge is an excellent test of algorithmic thinking — balancing performance, space, and creativity. 🌟 🛠️ The Problem You’re given an array of n + 1 integers, each between 1 and n. One number is repeated — your task is to find it. The catch? You can’t modify the array and should use only constant extra space. 🎯 The Approach I explored three different methods, each with its own strengths and trade-offs: Solution 1: Sort + Linear Search Sort the array first, then perform a single linear scan to check adjacent elements. Simple and effective, but sorting costs O(n log n) time. Solution 2: Using a Set (HashSet) Iterate through the array and store elements in a set. If a number already exists, that’s your duplicate! O(n) time but O(n) extra space. Solution 3: Floyd’s Two Pointer (Cycle Detection) The optimal and most elegant approach. Treat the array as a linked list where each value points to the next index. Using slow and fast pointers, detect the cycle and find the duplicate. This achieves O(n) time and O(1) space — brilliant use of logic and pattern recognition! 🧠 Key Takeaways • Every problem has layers — start simple, then optimize.   • The time-space trade-off is one of the most important concepts in algorithm design.   • Floyd’s cycle detection is a game changer — a true example of thinking beyond the obvious.  💡 Challenge for You Have you used the fast and slow pointer technique outside of linked lists? Where else have you applied it creatively? Drop your insights below! 💬 📺 Watch My Full Walkthrough I’ve broken down all three methods (especially the two-pointer logic) in my latest video: https://lnkd.in/gYGpdZXN 🔥 Join the Conversation If you love exploring algorithms and understanding the “why” behind them, let’s connect and grow together! 🚀 #CodingJourney #DSA #LeetCode #Algorithms #ProblemSolving #DataStructures #Pointers #CycleDetection #LearningEveryDay #Programming #DeveloperLife #TechCommunity #CodeNewbies #SoftwareEngineering #365DaysOfCode

Ever wondered how to find the longest path in a binary tree without getting lost in recursion? 🌳 Let’s break it down. Hey everyone! Day 298 of my 365-day coding journey, and today’s challenge was a classic tree problem: LeetCode’s “Diameter of Binary Tree.” This one really tests how well you understand recursion and tree traversal logic. Let’s dive in! ⚡ 🛠️ The Problem Given the root of a binary tree, the goal is to find the length of the tree’s diameter — the longest path between any two nodes. The path may or may not pass through the root. 🎯 The Approach I explored two different solutions to understand both logic and optimization: Solution 1: Brute Force My initial solution used a simple O(n²) approach: 1. For every node, calculate the height of its left and right subtrees. 2. Compute the possible diameter as left_height + right_height. 3. Keep track of the maximum diameter across all nodes. It worked, but recalculating heights multiple times made it inefficient. Solution 2: DFS (Optimized) The optimized O(n) solution uses Depth-First Search to calculate the height and diameter simultaneously. Here’s the key idea: 1. A helper function returns the height of a node (1 + max(left_height, right_height)). 2. While returning, it also updates a variable tracking the max diameter by checking left_height + right_height. This way, a single recursive pass gives both results efficiently — no repeated calculations. 🧠 Key Takeaways - Many tree problems can be optimized by combining calculations in a single DFS traversal.   - The brute force approach helps you understand the logic, but recognizing overlapping computations leads to real optimization.   - Recursion becomes powerful when you learn how to return one value and update another along the way.  💡 Challenge for you! When solving tree problems, do you prefer starting with a brute-force approach to understand it or jump straight to an optimized one? Let me know below! 💬 📺 Check out my detailed video walkthrough: https://lnkd.in/g9jd55ZK 🔥 Join the Conversation If you’re learning DSA or practicing tree problems, let’s connect! Always great to share ideas and grow together. 🚀 #CodingJourney #365DaysOfCode #LeetCode #DSA #BinaryTree #Trees #Recursion #ProblemSolving #DepthFirstSearch #CodingCommunity #DeveloperLife #LearningEveryDay #TechLearning #Programming

𝐃𝐞𝐛𝐮𝐠𝐠𝐢𝐧𝐠 𝐬𝐚𝐯𝐞𝐬 𝐡𝐨𝐮𝐫𝐬, 𝐜𝐨𝐦𝐦𝐞𝐧𝐭𝐢𝐧𝐠 𝐬𝐚𝐯𝐞𝐬 𝐲𝐨𝐮𝐫 𝐬𝐚𝐧𝐢𝐭𝐲 One thing that makes a massive difference in the real world is 𝐡𝐨𝐰 𝐰𝐞𝐥𝐥 𝐲𝐨𝐮 𝐜𝐨𝐦𝐦𝐞𝐧𝐭 𝐚𝐧𝐝 𝐝𝐞𝐛𝐮𝐠 𝐲𝐨𝐮𝐫 𝐬𝐜𝐫𝐢𝐩𝐭𝐬. I can’t count how many times I’ve gone back to a Python script weeks later and thought, “Who wrote this?”, only to realise it was me. Adding proper logging, print statements, and comments doesn’t just help others understand your work, it helps you when things inevitably break. 𝐀 𝐟𝐞𝐰 𝐬𝐢𝐦𝐩𝐥𝐞 𝐡𝐚𝐛𝐢𝐭𝐬 𝐭𝐡𝐚𝐭 𝐠𝐨 𝐚 𝐥𝐨𝐧𝐠 𝐰𝐚𝐲: ▪️ Add meaningful log messages, not just “print(‘done’)” ▪️ Comment sections clearly so debugging is faster ▪️ Use structured logs for clarity when things go wrong ▪️ Treat your future self as another engineer who’ll be reading this code You don’t realise how powerful this is until you’re debugging something at 2 AM. #Python #DataEngineering #Debugging #DevOps #DataOps #Logging #SoftwareEngineering #CodeQuality #LearningInPublic

🌳 DSA Challenge – Day 100 🎉 Problem: Construct Binary Tree from Preorder and Inorder Traversal 🌲 We made it to Day 100 of the challenge! 💪🔥 Let’s end this journey with a classic Tree Construction problem — a perfect blend of recursion, hashing, and traversal logic. 🧠 Problem Summary: You’re given two integer arrays: preorder → preorder traversal of a binary tree inorder → inorder traversal of the same tree Your goal: Reconstruct and return the binary tree. ⚙️ My Approach: 1️⃣ The first element of preorder is always the root. 2️⃣ Use a hash map (index) to quickly find the root’s position in the inorder array. 3️⃣ Recursively build: The left subtree from elements before the root in inorder. The right subtree from elements after the root in inorder. 4️⃣ Reversing preorder allows efficient pop() operations from the end (O(1)). 💡 Why This Works: Preorder gives the root order, Inorder gives the structure (left–root–right). By combining both, we can rebuild the entire tree recursively in O(n) time. 📈 Complexity Analysis: Time Complexity: O(n) — Each node is processed once, and lookup is O(1) using a hash map. Space Complexity: O(n) — For recursion + hash map. ✨ Key Takeaway: Efficient recursion often comes from using traversal properties smartly — here, preorder identifies roots, while inorder defines subtree boundaries. 🌟 Milestone Moment: That’s Day 100 of DSA 🔥 From arrays to trees, recursion to heaps — what a journey! 🌱 Keep learning, keep building, and keep challenging yourself 💪 🔖 #Day100 #100DaysOfCode #DSA #BinaryTree #Preorder #Inorder #LeetCode #Recursion #Python #DataStructures #ProblemSolving #CodingChallenge #TechCommunity #Milestone