HEMAMALINI M’s Post

💻 Day 52 of #100DaysOfCodeChallenge Today’s problem: LeetCode 203 – Remove Linked List Elements 🧩 I worked on a Linked List manipulation problem where the goal is to remove all nodes that match a given value. This problem helped strengthen my understanding of pointer handling and dummy node usage in linked lists. 🧠 What I Learned: Handling edge cases where the head itself might need to be removed. Using a dummy node before the head to simplify pointer operations. How current and next pointers work together to modify a linked list safely. 🧩 Java Solution: class Solution {   public ListNode removeElements(ListNode head, int val) {     ListNode dummy = new ListNode(0);     dummy.next = head;     ListNode current = dummy;     while (current.next != null) {       if (current.next.val == val)         current.next = current.next.next;       else         current = current.next;     }     return dummy.next;   } } ⏱ Complexity: Time: O(n) Space: O(1) Every day, I’m getting more confident with data structures and linked list traversal logic.  On to the next challenge 🚀 #Day52 #LeetCode #Java #100DaysOfCode #CodingChallenge #LinkedList #ProblemSolving

💻 Day 52 of #100DaysOfCodeChallenge Today’s problem: LeetCode 203 – Remove Linked List Elements 🧩 I worked on a Linked List manipulation problem where the goal is to remove all nodes that match a given value. This problem helped strengthen my understanding of pointer handling and dummy node usage in linked lists. 🧠 What I Learned: Handling edge cases where the head itself might need to be removed. Using a dummy node before the head to simplify pointer operations. How current and next pointers work together to modify a linked list safely. 🧩 Java Solution: class Solution {   public ListNode removeElements(ListNode head, int val) {     ListNode dummy = new ListNode(0);     dummy.next = head;     ListNode current = dummy;     while (current.next != null) {       if (current.next.val == val)         current.next = current.next.next;       else         current = current.next;     }     return dummy.next;   } } ⏱ Complexity: Time: O(n) Space: O(1) Every day, I’m getting more confident with data structures and linked list traversal logic.  On to the next challenge 🚀 #Day52 #LeetCode #Java #100DaysOfCode #CodingChallenge #LinkedList #ProblemSolving

📌 Day 27/100 - Generate Parentheses (LeetCode 22) 🔹 Problem: Given n pairs of parentheses, generate all combinations of well-formed (balanced) parentheses. 🔹 Approach: Use backtracking to build strings character-by-character. Keep two counters: open = number of '(' used, close = number of ')' used. You may add '(' while open < n. You may add ')' while close < open (to maintain balance). When the current string length reaches 2*n, add it to the result list. 🔹 Complexity: Time: proportional to the number of valid combinations (Catalan number) — roughly O(4ⁿ / n^(3/2)). Space: output-sensitive — O(n * Cn) for storing results, and O(n) extra for recursion depth. 🔹 Key Learning: Backtracking is ideal when you need to enumerate valid combinations subject to constraints. Always enforce constraints early (here: close < open) to prune invalid branches. Think in terms of state (open/close counts) rather than raw string manipulation. #100DaysOfCode #LeetCode #Java #ProblemSolving #DSA

🚀 Day 116 of 120 – #DSAwithJava Challenge Hey LinkedIn fam! 👋 Today’s problem focused on Linked List manipulation — a classic and essential concept for every developer to master! ⚡ ✅ Problem : Remove Nth Node From End of List (LeetCode #19 – Medium) 🎯 Objective: Given the head of a linked list, remove the nth node from the end and return the updated list. 🧠 Key Insight: By using the two-pointer (fast & slow) technique, we can remove the Nth node in a single pass. Move fast pointer n steps ahead. Then move both pointers together until fast reaches the end. The slow pointer will now be just before the node to delete. This elegant trick eliminates the need to calculate the list’s length first! 💡 Example: Input: [1,2,3,4,5], n = 2 Output: [1,2,3,5] 🕒 Time Complexity: O(L) 📦 Space Complexity: O(1) 🔥 Takeaway: Sometimes, optimizing your approach isn’t about new data structures—it’s about smart pointer movement. The two-pointer method shines again in making linked list problems clean and efficient! ✨ #120DaysOfCode #DSAwithJava #LeetCode #CodingChallenge #Java #LinkedList #ProblemSolving #TechJourney

🚀 Just Solved LeetCode #1367 — Linked List in Binary Tree 📘 Problem: Given the head of a linked list and the root of a binary tree, determine whether all the elements of the linked list appear as a downward path in the binary tree. Downward means starting from any node and moving only to child nodes. Example: Input → head = [4,2,8], root = [1,4,4,null,2,2,null,1,null,6,8,null,null,null,null,1,3] Output → true Explanation → Nodes in blue form a subpath in the binary tree. 🧠 My Approach: I used recursion to check whether the linked list sequence can be found starting from any node in the binary tree. 1️⃣ For each node in the tree, check if the list starting from `head` matches the downward path from that node. 2️⃣ If not, recursively check the left and right subtrees. 3️⃣ Used a helper function `checkPath()` to verify if a valid path continues as the recursion goes deeper. 💡 What I Learned: ✅ How to combine linked list and binary tree traversal logic ✅ How recursion can efficiently check multiple starting points ✅ Improved my understanding of tree path traversal and backtracking logic #LeetCode #Java #DSA #BinaryTree #LinkedList #CodingUpdate #LearningByDoing

LeetCode Question #199 — Binary Tree Right Side View Thrilled to share another Accepted Solution (100% Runtime 🚀) on LeetCode! This problem focuses on Binary Trees — specifically, how to capture the view of a tree from its right side 🌳➡️. I implemented a Modified Pre-Order Traversal (Root → Right → Left) in Java, ensuring that the first node at each level (from the right) gets recorded. 💡 Key Idea: Use recursion with a level tracker — when the current level equals the list size, it means we’re seeing the rightmost node for the first time. Here’s the performance snapshot: ⚙️ Runtime: 0 ms (Beats 100% of Java submissions) 💾 Memory: 42.4 MB Every problem like this sharpens my understanding of tree traversal patterns and depth-first search optimization. #LeetCode #Java #DataStructures #BinaryTree #ProblemSolving #CodingJourney #DSA #100PercentRuntime

23/30 days✅ Solved LeetCode Problem : #70 – Climbing Stairs The challenge was to determine how many distinct ways one can reach the top of a staircase with n steps, where at each step you can either climb 1 or 2 steps. The logic behind the problem is similar to the Fibonacci sequence, where each state depends on the sum of the previous two states. I implemented the solution in Java, using an iterative approach with three variables to optimize space complexity. Instead of using recursion or an array, the approach updates values in constant space (O(1)) while maintaining linear time complexity (O(n)). Here’s the logic in brief: Initialize a = 0, b = 1, and c = 1. For each step, compute c = a + b, then shift values forward (a = b, b = c). Return c as the total number of distinct ways to reach the top. #LeetCode #Java #DynamicProgramming #ProblemSolving #CodingJourney

🚀 Day 25 of 100 Days of LeetCode 📘 Problem: Remove Nth Node from End of List 💻 Language: Java ✅ Status: Accepted — Runtime: 0 ms ⚡ (Beats 100%) Today’s challenge was about working with Linked Lists, one of the most fundamental yet tricky data structures in DSA. The problem required removing the Nth node from the end — which tested how well I could use two-pointer techniques efficiently. ✨ Key Learnings: The two-pointer approach simplifies traversal logic beautifully 🎯 Dummy nodes are game-changers for clean code and edge case handling Linked Lists demand visualization — drawing it out helps a lot! Every accepted solution sharpens my ability to reason through edge cases and optimize thought flow 🔄 #Day25 #100DaysOfCode #LeetCode #Java #LinkedList #ProblemSolving #CodingJourney #DSA #SoftwareDevelopment #CodeEveryday #KeepLearning