Reorder Linked List Challenge Solved in 50 Days of Code

🚀 Day 45/100 – LeetCode Challenge 🔗 Problem: Linked List Components (LeetCode 817) 💡 Difficulty: Medium Today’s problem was all about understanding connected components in a linked list using a smart approach. 🧠 Key Insight: Instead of checking every possible combination, I used a HashSet for quick lookups. A component is counted only when a sequence ends, i.e., when the current node is in nums but the next node is not. ⚡ Approach: Convert nums into a set for O(1) lookup Traverse the linked list Count only when a component ends 📈 Time Complexity: O(n) 📦 Space Complexity: O(n) ✨ What I learned: Sometimes, solving problems efficiently is about identifying where something ends rather than where it starts. Consistency is key 🔥 Halfway through the journey — staying committed! #Day45 #100DaysOfCode #LeetCode #DataStructures #LinkedList #CodingJourney #ProblemSolving #SoftwareEngineering

Day 4/100 of my LeetCode Journey Back on track after a short break, and today’s problem was “Kth Smallest Element in a Sorted Matrix.” At first, I didn’t approach it correctly. I was thinking in terms of simple indexing, but this problem required a different perspective. I learned how a min heap can be used to efficiently track the smallest elements across multiple rows. Instead of checking the entire matrix, the idea is to always pick the smallest available element and move forward from there. What I noticed today is that my logic is improving, but I still need to be careful with implementation. Small syntax and indexing mistakes can break the whole solution even when the approach is right. Takeaway: Understanding the approach is important, but writing clean and correct code is equally important. Making steady progress, one problem at a time. Question: Do you also find implementation harder than understanding the logic? #Day4 #LeetCode #DSA #LearningInPublic #Consistency

Day 28 of #100DaysOfCode 💻 Today’s focus: Subsets II (LeetCode 90) Building on yesterday’s Subset Sum (Problem 1), today I explored how the same recursion pattern works when duplicates are involved. 📌 What I focused on: At each step → either pick the element or skip it (same as Subset 1) Sorting the array to handle duplicates Skipping repeated elements at the same recursion level 💡 Realization: Subset II is not a completely new problem—it’s an extension of Subset 1 with an extra constraint. Understanding the base pattern made it much easier to adapt and solve this one. Felt good to see how concepts connect step by step. Still getting more comfortable with recursion and backtracking 🚀 #LeetCode #DSA #Recursion #Backtracking #CodingJourney

100 LeetCode problems solved ! and honestly, it's taught me more about thinking than coding. When I started, I used to jump straight to the solution. Now I've learned to slow down — understand the problem, think brute force first, then optimize. Patterns I've been focusing on: → Sliding Window → Two Pointers → Stack & Queue → Kadane's Algorithm → Prefix Sum → Merge Intervals → Linked List The biggest shift wasn't in the number of problems solved — it was learning to recognize WHY a particular approach works, not just HOW. Still a long way to go. But consistency is starting to show results. 100 done. More to come. 🎯 #DSA #LeetCode #CodingJourney #SoftwareEngineering #100Problems

🚀 Day 17/100 — LeetCode Challenge Solved Reverse Linked List At first, it feels tricky because we need to change the direction of links without losing the list. 💡 Approach (Iterative): 1) Use three pointers: prev, curr, next 2) Store the next node 3) Reverse the link (curr → prev) 4) Move all pointers forward 👉 Repeat until the list is fully reversed 💡 Key Insight: We’re not just moving through the list, we’re rebuilding it in reverse order. 🧠 Time Complexity: O(n) 💾 Space Complexity: O(1) 💡 What I learned: Linked List problems are less about logic, and more about careful pointer handling. One small mistake = broken list. #LeetCode #DSA #100DaysOfCode #Cpp #LinkedList #CodingJourney

🧠 Day 32/75: Solving for the "End" from the "Beginning" Linked Lists can be tricky because you can't access elements by index. So, how do you remove the Nth element from the back without knowing the total length? Today’s LeetCode session was a deep dive into Pointer Logic. By maintaining a specific distance between two pointers, I was able to identify and delete the target node in a single traversal. Current Progress: 📅 32 Days Consecutive 🎯 Focus: Advanced Linked List Patterns ⚡ Result: 100% Beats Runtime Another day, another problem solved. The second month is off to a strong start! 💻🔥 #Consistency #100DaysOfCode #DSA #CodingJourney #JavaDeveloper #TechGrowth #LeetCodeChallenge #LinkedInLearning

🚀 Day 80 of #100DaysOfCode Today, I solved LeetCode 61 – Rotate List, a problem focused on linked list manipulation and efficient pointer handling. 💡 Problem Overview: Given the head of a linked list, the task is to rotate the list to the right by k places. 🧠 Approach: ✔️ Calculated the length of the linked list ✔️ Connected the tail to the head to form a circular list ✔️ Found the new tail position using k % length ✔️ Broke the cycle to get the rotated list This approach avoids unnecessary rotations and ensures optimal performance. ⚡ Key Takeaways: Linked list problems require strong pointer manipulation Converting problems into circular structures can simplify logic Modulo operation helps handle large values of k efficiently 📊 Complexity Analysis: Time Complexity: O(n) Space Complexity: O(1) 🎯 Day 80 Insight: Consistency over time builds confidence in solving even complex problems efficiently. On to Day 100 🚀 #LeetCode #100DaysOfCode #DSA #LinkedList #ProblemSolving #CodingJourney #SoftwareDevelopment #Consistency #InterviewPrep

🚀 Day 15/100 — Coding Challenge Solved Search in Linked List The task is straightforward: 1) Traverse the linked list 2) Check each node’s value 3) Return true if found, else false 💡 Approach: 1) Start from head 2) Move node by node using next 3) Stop when element is found or list ends 🧠 Time Complexity: O(n) 💾 Space Complexity: O(1) 💡 What I realized: Unlike arrays, linked lists don’t allow direct access. You have to traverse sequentially, which makes understanding traversal very important. Focusing on strengthening fundamentals. #LeetCode #DSA #100DaysOfCode #Cpp #LinkedList #CodingJourney

🚀 Day 76 of #100DaysOfCode Today, I solved LeetCode 18 – 4Sum, a classic problem that extends the two-pointer technique to higher complexity. 💡 Problem Overview: Given an array, the task is to find all unique quadruplets that sum up to a target value. 🧠 Approach: ✔️ Sorted the array to simplify processing ✔️ Fixed two elements using nested loops ✔️ Applied the two-pointer technique for the remaining two elements ✔️ Carefully handled duplicates to ensure unique quadruplets This approach efficiently reduces unnecessary computations compared to brute force. ⚡ Key Takeaways: Sorting + Two Pointers is a powerful combination Avoiding duplicates is crucial in combination problems Breaking a complex problem into smaller parts simplifies logic 📊 Complexity Analysis: Time Complexity: O(n³) Space Complexity: O(1) (excluding output) Consistently pushing boundaries and solving more complex problems 🚀 #LeetCode #100DaysOfCode #DSA #TwoPointers #ProblemSolving #CodingJourney #SoftwareDevelopment #InterviewPrep