Remove Linked List Elements with Dummy Node Technique

🌟 Day 72 of #100DaysOfCode 🌟 🔗 Reverse Nodes in k-Groups in a Linked List – A Powerful Pattern in Linked List Manipulation! 🔹 What I Solved Today’s challenge was an advanced linked list problem where nodes must be reversed k at a time. This task strengthens understanding of pointer manipulation, group-based operations, and recursive thinking — all crucial for technical interviews. 🧩 Problem: Reverse Nodes in k-Groups 💡 Given: A linked list such as: 1 → 2 → 3 → 4 → 5 and a value k (e.g., k = 2). 💥 Goal: Reverse the nodes in chunks of size k and return the modified list. If the remaining nodes are fewer than k, keep them unchanged. 📌 Example 1 Input: head = [1,2,3,4,5], k = 2 Output: 2,1,4,3,5 📌 Example 2 Input: head = [1,2,3,4,5], k = 3 Output: 3,2,1,4,5 🧠 Concepts Used ➡️ Linked List Reversal Reversing pointers within a chunk of size k ➡️ Group Processing Reverse only when the group contains at least k nodes ➡️ Pointer Manipulation Careful handling of prev, curr, next, and group boundaries ➡️ Efficient Traversal Each node is visited once — O(n) time ➡️ Constant Extra Space Pointer-based solution — O(1) auxiliary space ⚙️ Approach 1️⃣ Check if the current segment has at least k nodes 2️⃣ If yes, reverse those k nodes 3️⃣ Recursively process the next segment 4️⃣ Connect the reversed segment with the remaining list 5️⃣ If remaining nodes < k, leave them unchanged 🚀 Learning This problem strengthened my understanding of: ✔️ Advanced linked list manipulation ✔️ Safe handling of multiple pointers ✔️ Breaking complex logic into smaller steps ✔️ Combining recursion with iteration ✔️ Writing efficient, interview-ready solutions #CodingLife #Programmer #LeetCode #GeeksForGeeks #LinkedList #ProblemSolver #CleanCoding

🌟 Day 71 of #100DaysOfCode 🌟 🔗 Count Occurrences of a Given Key in a Linked List – Simple but Powerful! 🔹 What I Solved Today’s challenge focused on traversing a singly linked list and counting how many times a specific value (key) appears. Even though it looks simple, this problem strengthens fundamentals like traversal, pointer handling, and space-optimized logic. 🧩 Problem: Count Occurrences of a Key in a Linked List 💡 Given: A singly linked list such as: 1 -> 2 -> 1 -> 2 -> 1 -> 3 -> 1 and a key, for example 1. 💥 Goal: Return the total number of times the key appears in the list. For the example above, the output is 4. 🧠 Concepts Used ➡️ Linked List Traversal Iterating through nodes one by one from head to end. ➡️ Constant Space Counting Only one integer counter is used — O(1) extra space. ➡️ Linear Time Complexity Each node is visited once — O(n). ➡️ Comparison Logic At each node, compare node.data with the key and increment the count. ⚙️ Approach 1️⃣ Start with the head of the list 2️⃣ Initialize a counter count = 0 3️⃣ Traverse each node using a loop 4️⃣ If node.data == key, increment count 5️⃣ Continue until the list reaches null 6️⃣ Return the counter 🚀 Learning This problem reinforced: ✔️ Clean linked list traversal ✔️ Handling edge cases (empty list, key not present) ✔️ Writing efficient and readable code ✔️ Understanding time & space optimization ✔️ Strengthening core DSA fundamentals #CodingLife #Programmer #LeetCode #GeeksForGeeks #JavaDeveloper #StudentCoder #LearnToCode #DeveloperCommunity #ProblemSolver #CleanCoding #LinkedList

Day 6 | Copy-Paste to Logic-Based 🚀 Today’s problem may look simple at first glance, but it strongly reinforces conditional thinking and clean control flow. Solved Fizz Buzz (LeetCode 412) — a classic problem that tests how well you understand conditions, order of checks, and edge cases, rather than just syntax. 🧠 Key Learnings: • The order of conditions matters more than the conditions themselves • Always check the most restrictive case first (divisible by both 3 & 5) • Writing readable conditional logic is as important as correctness • Simple problems help build strong foundations for complex logic later • Clean if-else chains improve both clarity and maintainability 🔍 Approach (Brief): • Iterate from 1 to n • Check divisibility by 3 and 5 in the correct order • Push the appropriate string into the result array ⏱️ Time & Space Complexity: • Time Complexity: O(n) • Space Complexity: O(n) FizzBuzz isn’t about difficulty — it’s about thinking clearly, structuring logic properly, and mastering fundamentals. Day 6 done. Consistency > Motivation. 💪📈 ⸻ #CopyPasteToLogicBased #Day6 #DSA #DataStructures #Algorithms #LeetCode #ProblemSolving #Coding #Programming #CPlusPlus #Cpp #SoftwareEngineering #ComputerScience #Developer #DevelopersOfLinkedIn #LearningInPublic #CodingJourney #DailyCoding #Consistency #Discipline #GrowthMindset #TechJourney #InterviewPreparation #PlacementPreparation #LogicBuilding #BeginnerToAdvanced #CodeNewbie #StudentDeveloper #FutureEngineer #TechCommunity

I spent 6 months writing "better prompts." Then I realized: I was solving the wrong problem. Prompts don't need to be better. They need to be DEBUGGED... 🐛 My prompts failed 60% of the time. Every failure, I'd start from scratch: "Let me try a completely different approach..." It was exhausting. Then I had a realization while debugging actual code: "Why am I not debugging prompts the same way I debug software?" I started treating prompt failures like bugs. The shift: ❌ Prompt fails → Rewrite everything → Hope it works ✅ Prompt fails → Isolate the bug → Fix ONE thing → Validate My success rate went from 40% to 94%. The breakthrough wasn't better prompts. It was systematic debugging. Here's the 5-step framework → #PromptEngineering #AItools #Debugging #SystematicThinking #ProductivityHacks #AIprompts #ProblemSolving #TechSkills #WorkSmarter #AIstrategy

Sometimes the most satisfying code isn’t the most complex , it’s the most intentional. Why this approach feels right: Elegant method chaining Pure function -> predictable & testable Self-documenting logic Edge cases handled without noise Reads like a data pipeline, not a puzzle Each step does one thing. No mutation. No magic. Just clarity. This is the kind of code you enjoy coming back to months later, and that's why I love functional programming... Readable code is a feature. #TypeScript #CleanCode #FunctionalProgramming #MethodChaining #SoftwareEngineering #AIEngineering #WebDevelopment #DeveloperLife

🌟 Day 73 of #100DaysOfCode 🌟 🔗 Move the Last Node to the Front – A Simple Yet Powerful Linked List Pattern! 🔹 What I Solved Today’s challenge focused on an elegant linked list manipulation where the last node is moved to the front. This problem may look simple, but it strengthens understanding of pointer traversal and last-node handling — techniques that are frequently tested in interviews. 🧩 Problem: Move Last Node to Front of Linked List 💡 Given: A linked list such as: 2 → 5 → 6 → 2 → 1 💥 Goal: Take the last node of the list and move it to the front, returning the modified list — all in O(n) time and O(1) space. 📌 Example 1 Input: 2 → 5 → 6 → 2 → 1 Output: 1 → 2 → 5 → 6 → 2 📌 Example 2 Input: 2 Output: 2 (Only one node → no change) 🧠 Concepts Used ➡️ Pointer Traversal Traverse the list to find the last and second-last nodes. ➡️ Re-linking Pointers Detach the last node and connect it before the head. ➡️ Edge Case Handling If the list has only one node, return it as-is. ➡️ Efficient Logic Single traversal → O(n) time No extra memory → O(1) space ⚙️ Approach 1️⃣ Traverse to the second-last node 2️⃣ Separate the last node 3️⃣ Point the last node to the current head 4️⃣ Update head to the last node 5️⃣ Return the modified list 🚀 Learning This problem helped improve my understanding of: ✔️ Boundary cases in linked lists ✔️ Precise pointer manipulation ✔️ Clean and optimal transformations ✔️ Writing space-efficient code ✔️ Thinking in terms of node connections #CodingLife #Programmer #LinkedList #GeeksForGeeks #JavaDeveloper #StudentCoder #ProblemSolving

LeetCode Problem of the Day — Back to Fundamentals Today’s POTD (LeetCode 3315: Construct the Minimum Bitwise Array II) serves as a valuable reminder that some of the most essential problem-solving skills in software engineering remain unchanged over the years. The task is straightforward: ans[i] | (ans[i] + 1) == nums[i] Minimize ans[i]. However, the real challenge lies in reasoning rather than coding. Here are a few insights that made a difference: - Even numbers are impossible: No matter what value you choose, x | (x + 1) is always odd. This observation resolves half the cases immediately. - Binary structure > brute force: Incrementing a number flips trailing bits. To achieve the minimum valid answer, you modify exactly one bit: the bit just before the first zero. - An O(1) solution emerges naturally: Once the pattern is clear, the solution simplifies to a single, clean bitwise operation. Why I appreciate problems like this is that they reward: - Understanding over memorization - Precision over trial-and-error - Fundamentals over frameworks These skills are crucial in real systems — performance tuning, debugging, and designing reliable software. Strong fundamentals compound over time. Problems like today’s POTD remind us to keep these skills sharp. On to the next one. Let's Connect : Harshit goel #LeetCode #ProblemOfTheDay #BitManipulation #DataStructures #Algorithms #ProblemSolving #SoftwareEngineering #CodingInterview #ComputerScience #LearningEveryDay

🚀 Consistency + Fundamentals = Results Just got an Accepted solution on LeetCode ✅ 💡 0 ms runtime 💯 Beats 100% submissions This problem looks simple on the surface, but it really tests how well you understand core data structures — especially stacks and matching logic. What this reinforced for me: Strong fundamentals matter more than fancy tricks Writing clean, readable logic often leads to optimal performance Small checks (like early exits) can make a big difference Every problem solved builds confidence and clarity for the next one. One step closer to becoming a better problem solver 🚀 If you’re grinding DSA right now — keep going. Progress compounds. 💪 #LeetCode #DSA #ProblemSolving #CPlusPlus #CodingJourney #Consistency #SoftwareEngineering #LearningEveryDay

🚀 Day 34/100 of My LeetCode Challenge – Solved "Construct the Minimum Bitwise Array I"! 🎯 Today’s problem presented an elegant bitwise puzzle that combined logical reasoning with systematic exploration. The task: Given an array of prime integers, construct another array where for each index i, the bitwise OR between ans[i] and ans[i] + 1 equals nums[i], while minimizing each ans[i]. If impossible, return -1. 🔍 Key Insight: The relationship x | (x + 1) == nums[i] has a special property: x | (x + 1) produces a binary number with consecutive ones from the least significant bit upward. For example: 1 | 2 = 3 (binary 01 | 10 = 11) 4 | 5 = 5 (binary 100 | 101 = 101) This means nums[i] must be of the form where all bits after the first zero from LSB are set to 1. ⚡ My Approach: I implemented a straightforward yet efficient solution: Iterate through each prime in the input array. For each, test all possible x from 1 to nums[i]-1. Check if x | (x + 1) equals the target. Track and return the smallest valid x (or -1 if none exists). ✅ Result: Runtime: 3 ms (faster than 57.02% of submissions) Memory: 46.79 MB (better than 49.12%) All 658 test cases passed! 💡 Takeaway: Sometimes the most direct approach—brute force within reasonable bounds—is both clean and effective, especially when constraints are small (nums[i] ≤ 1000). Understanding bitwise properties helps verify correctness and could inspire optimizations for larger scales. This daily practice continues to sharpen my problem-solving skills, reinforcing that even "simple" problems can teach valuable lessons about constraints, efficiency, and mathematical insight. #LeetCode #100DaysOfCode #BitwiseOperations #Algorithm #Java #CodingChallenge #ProblemSolving #TechJourney #SoftwareEngineering #LearnInPublic #CodingLife