Mastering Linked Lists on LeetCode

🚀 Day 11 of My LeetCode Journey — Deep Dive into Linked Lists Today was all about mastering patterns in Linked Lists: 🔹 Linked List Cycle (LeetCode 141) 🔹 Palindrome Linked List (LeetCode 234) 💡 Problem 1: Linked List Cycle Tried two approaches: ✅ Using Set Store visited nodes If node already exists → cycle detected ⏱️ Time: O(n) | 📦 Space: O(n) ✅ Fast & Slow Pointer (Floyd’s Algorithm) 🔥 Slow → 1 step Fast → 2 steps If they meet → cycle exists ⏱️ Time: O(n) | 📦 Space: O(1) 👉 This approach is elegant and optimal! 💡 Problem 2: Palindrome Linked List Again explored two approaches: ✅ Convert to Array Store values in array Compare from both ends ⏱️ Time: O(n) | 📦 Space: O(n) ✅ Optimal Approach (In-place) 🔥 Find middle node Reverse second half of linked list Compare both halves ⏱️ Time: O(n) | 📦 Space: O(1) 👉 This one really tested pointer manipulation skills! 🧠 What I Learned: One problem can have multiple valid approaches Optimal solutions often reduce space complexity Linked Lists = pointer mastery + careful thinking 🔥 Key Takeaways: Always try brute force first → then optimize Fast & slow pointer is a game-changing technique In-place solutions are highly valued in interviews Big thanks to Namaste DSA and Akshay Saini 🚀 for the guidance Day 12 loading… 💪 #LeetCode #DataStructures #Algorithms #CodingJourney #100DaysOfCode #SoftwareEngineering #Programming #InterviewPrep #JavaScript #CodingLife #TechGrowth #ProblemSolving #Developers #LearnToCode #LinkedList #TwoPointers #DSA #NamasteDSA

🚀 Day 3 of My LeetCode Journey — Consistency > Motivation Today’s problem: Merge Sorted Array (LeetCode 88) At first glance, it looks simple — just merge two arrays, right? But the real challenge is doing it in-place without extra space 🤯 💡 Key Learning: Instead of merging from the front (which causes overwriting), the optimal approach is to: 👉 Use two pointers from the end 👉 Fill the array backwards This small shift in thinking makes a huge difference: ⏱️ Time Complexity: O(m + n) 📦 Space Complexity: O(1) 🔥 What I’m realizing: It’s not about solving problems — it’s about learning how to think differently Every problem is teaching me: How to optimize How to avoid brute force How to think like an interviewer On to Day 4 💪 #LeetCode #DataStructures #Algorithms #CodingJourney #100DaysOfCode #SoftwareEngineering #Programming #InterviewPrep #JavaScript #CodingLife #TechGrowth #ProblemSolving #Developers #LearnToCode #Consistency #CodeDaily

🚀 Day 4 of my Coding Challenge Today's problem was simple but a great reminder of string fundamentals in JavaScript. Problem Statement Given a string s, return the last character of the string. Example Input: "learncoding" Output: "g" Any language Approach Use string indexing with length - 1 to access the last character. function getLastCharacter(s) { return s[s.length - 1]; } console.log(getLastCharacter("learncoding")); // g Key Learning:- Strings are index-based length - 1 always points to the last character Small problems help strengthen core programming logic Consistency matters more than complexity. One problem every day 📚 #CodingChallenge #Day4CodingChallenge #JavaScriptDeveloper #DSA #DataStructures #Algorithms #ProblemSolving #CodingJourney #CodeDaily #LearnToCode #DevelopersLife #SoftwareDeveloper #FrontendDeveloper #FullStackDeveloper #TechCommunity #ProgrammingLife #CodeNewbie #CodingPractice #DeveloperJourney #CodingSkills #TechLearning #JavaScriptCoding #WebDeveloper #ProgrammingChallenge #DailyCoding #CodeMotivation #CareerInTech #DeveloperMindset #CodingGrind #TechGrowth

🚀 Day 29/128 – LeetCode Challenge Today’s problem: Combination Sum (Medium) This problem was a great exercise in understanding backtracking and recursion. The goal was to find all unique combinations of numbers that add up to a target, with the twist that each number can be used multiple times. 🔍 Key Learnings: Backtracking is all about exploring possibilities and undoing choices Using a start index helps avoid duplicate combinations Recursive thinking becomes much easier with practice Important to handle base cases properly (target reached / exceeded) 💡 What clicked today: Instead of thinking “which combination is correct?”, I focused on trying every possible path and letting recursion filter valid ones. 🧠 Pattern recognized: This is a classic DFS + Backtracking problem — a pattern that shows up frequently in coding interviews. 📈 Progress update: Consistency is starting to pay off. Problems that once felt complex are becoming more intuitive. On to Day 30 🔥 #LeetCode #128DaysOfCode #DSA #CodingJourney #Backtracking #JavaScript #ProblemSolving

🚀 Day 12 of My LeetCode Journey — Linked List Patterns Getting Stronger Today’s problems: 🔹 Intersection of Two Linked Lists (LeetCode 160) 🔹 Remove Linked List Elements (LeetCode 203) 💡 Problem 1: Intersection of Two Linked Lists Explored two approaches: ✅ Brute Force Traverse headA and check every node in headB ⏱️ Time: O(m × n) ✅ Using Set Store all nodes of headB in a Set Traverse headA and check for intersection ⏱️ Time: O(m + n) | 📦 Space: O(n) 👉 Helped me understand how hashing can optimize comparisons. 💡 Problem 2: Remove Linked List Elements Solved using a Sentinel (Dummy) Node 🔥 👉 Create a dummy node before head 👉 Use prev pointer to skip unwanted nodes: prev.next = prev.next.next This approach simplifies edge cases like: Removing head node Multiple consecutive deletions 🧠 What I Learned: Sentinel nodes make linked list problems cleaner Hashing can reduce time complexity significantly Thinking in terms of nodes (not values) is key 🔥 Key Takeaways: Always look for ways to reduce nested loops Dummy nodes = lifesaver in linked list problems Practice is making pointer manipulation more intuitive Grateful for the guidance from Namaste DSA and Akshay Saini 🚀 Day 13 loading… 💪 #LeetCode #DataStructures #Algorithms #CodingJourney #100DaysOfCode #SoftwareEngineering #Programming #InterviewPrep #JavaScript #CodingLife #TechGrowth #ProblemSolving #Developers #LearnToCode #LinkedList #DSA #NamasteDSA #AkshaySaini

🚀 Day 24 of My Coding Journey — Power of Two Today’s problem was “Power of Two” — a simple-looking question that really highlights the beauty of bit manipulation. 🔍 What I learned: Instead of using loops or recursion, I explored how binary representation works. A power of two always has only one set bit (1) in its binary form — and that insight leads to a super efficient solution. 💡 Key trick: n & (n - 1) === 0 This removes the lowest set bit, and if the result is zero, the number is a power of two. ⚡ Takeaway: Sometimes the most optimized solutions come from understanding how data is represented internally, not just from writing more code. 📈 Progress: Day by day, I'm getting more comfortable with problem-solving patterns and thinking beyond brute force approaches. #128DaysOfCode #LeetCode #JavaScript #CodingJourney #ProblemSolving #BitManipulation

𝐃𝐚𝐲 𝟏𝟐/𝟏𝟓 𝐨𝐟 𝐦𝐲 𝐉𝐚𝐯𝐚𝐒𝐜𝐫𝐢𝐩𝐭 𝐉𝐨𝐮𝐫𝐧𝐞𝐲 🚀 Not gonna lie… Today’s topic felt a bit confusing. But I didn’t skip. 💡 What I learned: Promises in JavaScript .then() and .catch() Handling asynchronous code 🧠 What I understood: JavaScript doesn’t always run things step by step. Some tasks take time… Like fetching data from an API. Promises help handle those situations. At first, I didn’t understand why we even need promises… But after learning about async tasks, it started making sense. 📌 My biggest takeaway: Not everything happens instantly in coding… And that’s okay. Learning how to handle delays is part of becoming a developer. Still learning… Still pushing through confusion… See you on Day 13 🚀 #JavaScript #CodingJourney #LearningInPublic #Day12 #Promises #WebDevelopment #Consistency #Programming

🚀 Day 14 of My LeetCode Journey — Linked List Confidence Growing Today’s problems: 🔹 Add Two Numbers (LeetCode 2) 🔹 Odd Even Linked List (LeetCode 328) 💡 Problem 1: Add Two Numbers This problem simulates addition like we do manually: 👉 Traverse both linked lists 👉 Add corresponding digits + carry 👉 Create new nodes for the result 👉 Handle remaining carry at the end 🔥 A great mix of linked list traversal + math logic 💡 Problem 2: Odd Even Linked List This one is all about rearranging nodes: 👉 Separate nodes into odd index and even index 👉 Maintain two pointers (odd & even) 👉 Finally connect odd list with even list ⚡ No extra space needed — done in-place! 🧠 What I Learned: - Handling carry properly is crucial in problems like addition - Rearranging pointers without losing references is key - Linked List problems are getting more intuitive with practice 🔥 Key Takeaways: - Break problems into smaller steps (traverse, compute, connect) - Always track pointers carefully to avoid losing nodes - Practice is making complex problems feel simpler Grateful for the learning journey with Namaste DSA and Akshay Saini 🚀 Day 15 loading… 💪 --- #LeetCode #DataStructures #Algorithms #CodingJourney #100DaysOfCode #SoftwareEngineering #Programming #InterviewPrep #JavaScript #CodingLife #TechGrowth #ProblemSolving #Developers #LearnToCode #LinkedList #DSA #NamasteDSA #AkshaySaini

Leetcode Day 1 : Problem 1 (Merge Sorted Array) Just solved my first LeetCode problem. It was the classic "Merge Sorted Array" sounds simple, right? But here's what I actually learned: JavaScript's .sort() doesn't sort numbers by default, it sorts them as strings. [1, 10, 2] becomes [1, 10, 2] not [1, 2, 10]. A one-line bug that would fail silently. Reassigning an array variable (arr = []) doesn't modify the original. LeetCode wants in-place changes. That took me a moment. I was filtering out zeros thinking they were empty slots, but 0 is a perfectly valid element. Hidden test cases would've caught me. Three bugs. One "easy" problem. Tons of learning. The real lesson? Passing the visible test cases doesn't mean your code is correct. Always think about edge cases. If you've been putting off starting DSA Interview questions like I was just open problem 1 and start. The first step is the hardest. #DSA #LeetCode #JavaScript #CodingJourney #Programming

🚀 Day 13 of My LeetCode Journey — Refining Linked List Skills Today’s problems: 🔹 Remove Duplicates from Sorted List (LeetCode 83) 🔹 Remove Nth Node From End of List (LeetCode 19) 💡 Problem 1: Remove Duplicates from Sorted List Since the list is already sorted: 👉 Just compare current node with next node 👉 If curr.val === curr.next.val → skip the duplicate 👉 Else → move forward Simple logic, but very effective due to the sorted property! 💡 Problem 2: Remove Nth Node From End of List Solved using two approaches: ✅ Two-Pass Approach First pass → calculate length Second pass → remove (length - n) node ⏱️ Time: O(n) ✅ One-Pass Approach (Optimized) 🔥 Use two pointers (fast & slow) Move fast ahead by n steps Move both together → when fast reaches end, slow is at target 👉 This approach is cleaner and more efficient! 🧠 What I Learned: Sorted data can simplify problems significantly Two-pointer technique continues to be super useful There’s always a way to reduce passes in linked list problems 🔥 Key Takeaways: Look for patterns (sorted, reversed, etc.) Optimize from two-pass → one-pass when possible Linked Lists are all about pointer precision Big thanks to Namaste DSA and Akshay Saini 🚀 for the guidance Day 14 loading… 💪 #LeetCode #DataStructures #Algorithms #CodingJourney #100DaysOfCode #SoftwareEngineering #Programming #InterviewPrep #JavaScript #CodingLife #TechGrowth #ProblemSolving #Developers #LearnToCode #LinkedList #TwoPointers #DSA #NamasteDSA #AkshaySaini