Adding Two Numbers in Reverse Linked List

#Day28 of #365DaysOfCode Today’s LeetCode practice: 🔹 Word Search (LeetCode 79) Solved a backtracking problem where I had to check whether a given word exists in a 2D board by moving horizontally or vertically through adjacent cells. 💡 Key Insight: Start exploring from every cell. If the character matches the current letter of the word, move in all 4 possible directions. Mark the cell as visited and revert the change if the path doesn’t lead to a solution. On to Day 29 🔥 #LeetCode #Java #DSA #ProblemSolving #CodingJourney #Backtracking #Recursion

LeetCode Grinding 🚀 Today was focused on strengthening Two Pointers mastery and improving problem-solving speed through practice. 🔹 DSA & LeetCode Practice Solved multiple problems by applying the Two Pointers pattern: LeetCode 11 — Container With Most Water Learned how pointer movement helps maximize area efficiently LeetCode 125 — Valid Palindrome Applied two pointers for string validation after preprocessing LeetCode 2824 — Count Pairs Whose Sum is Less than Target Understood how sorted arrays + two pointers help count pairs efficiently LeetCode 26 — Remove Duplicates from Sorted Array Strengthened in-place array manipulation using pointer technique 📌 Key takeaway: The Two Pointers pattern is powerful for: Sorted arrays Pair problems String validation Space-optimized solutions The more I practice patterns, the faster I recognize where to apply them. #DailyGrind #DSA #LeetCode #TwoPointers #ProblemSolving #Java #Consistency #LearningInPublic #InterviewPrep

Day 16 | LeetCode Daily Solved LeetCode Problem #173 – Binary Search Tree Iterator Concept: • Inorder traversal using stack • Iterator design over tree structures Key Learnings: • BST inorder traversal naturally gives sorted order • Using a stack allows lazy traversal with O(1) average time per operation Keeping the daily streak alive and strengthening tree traversal patterns. Submission link: https://lnkd.in/gMPEhX_m #LeetCode #DSA #BinarySearchTree #Trees #Java #ProblemSolving

Day #42/100 Days of Code🔥 Solved #LeetCode 61: 𝐑𝐨𝐭𝐚𝐭𝐞 𝐋𝐢𝐬𝐭 𝐭𝐨𝐝𝐚𝐲  Focused on understanding linked list manipulation, edge cases, and optimizing the approach to 𝐎(𝐧) 𝐭𝐢𝐦𝐞 & 𝐎(1) 𝐬𝐩𝐚𝐜𝐞. Key takeaways: Always handle 𝐤 % 𝐥𝐞𝐧𝐠𝐭𝐡 in rotation problems Turning the list into a circular linked list simplifies logic #100DaysOfCode #LeetCode #DSA #LinkedList #Java #ProblemSolving #Consistency #LearningJourney

Day 8/100 – LeetCode Challenge 🚀 Problem: Implement strStr() Approach: Checked every possible starting index in haystack Matched characters one by one with needle Time Complexity: O(n × m) Space Complexity: O(1) Key takeaway: Brute-force solutions are fine when constraints allow them. Understanding the basic approach comes before optimization. #LeetCode #100DaysOfCode #DSA #Java #ProblemSolving #100DaysOfLeetCode

Day 20/100 – LeetCode Challenge 🚀 Problem: Subsets (Power Set) Approach: Started with an empty subset For each number in the array, created new subsets by adding the number to all existing subsets Appended the newly formed subsets to the result list Time Complexity: O(n × 2ⁿ) Space Complexity: O(n × 2ⁿ) (to store all subsets) Key takeaway: Many combinatorial problems can be solved by building results incrementally, expanding previously generated subsets. #LeetCode #100DaysOfCode #DSA #Java #ProblemSolving #Backtracking #100DaysOfLeetCode

Day 12/100 – LeetCode Challenge 🚀 Problem: Guess Number Higher or Lower Approach: Applied Binary Search Narrowed the search space based on feedback from the API Time Complexity: O(log n) Space Complexity: O(1) Key takeaway: Whenever the problem gives “higher or lower” feedback, binary search should be your first instinct. #LeetCode #100DaysOfCode #DSA #Java #ProblemSolving #InterviewPrep #100DaysOfLeetCode

🚀 Day 19 of #100DaysOfCode Solved 2. Add Two Numbers on LeetCode ➕🔗 (in-place linked list approach) 🧠 Key insight: Instead of creating a new list, we can reuse the longer linked list and perform addition in place, handling carry carefully across nodes. ⚙️ Approach: 🔹Calculate the size of both linked lists 🔹Use the longer list as the result list 🔹Traverse both lists, adding values with carry 🔹Continue processing remaining nodes of the longer list 🔹Append a new node if a carry remains at the end ⏱️ Time Complexity: O(max(n, m)) 📦 Space Complexity: O(1) (no extra list created) #100DaysOfCode #LeetCode #DSA #LinkedList #Java #ProblemSolving #LearningInPublic #CodingJourney

🚀 Day 88 of #100DaysOfCode | LeetCode Daily Solved LeetCode Problem #67 – Add Binary ➕💻✅ A clean fundamentals problem that reinforces how low-level operations really work under the hood. Simulating binary addition digit by digit with carry is simple, efficient, and elegant. Key Takeaways: -> Binary addition using carry logic -> Traversing strings from right to left -> Using StringBuilder for efficient string construction -> Handling edge cases when carry remains at the end Language: Java -> Runtime: 1 ms (Beats 99.65%) -> Memory: 43.90 MB Strong basics make complex problems easier. One day, one win. 🔥💻 #LeetCode #Java #Binary #Strings #ProblemSolving #100DaysOfCode

🚀 Day 77 of #100DaysOfCode Today’s challenge was a classic two-pointer optimization problem — LeetCode: Container With Most Water 💧 📌 Problem Summary You’re given an array where each element represents the height of a vertical line. The task is to find two lines that together with the x-axis form a container that holds the maximum amount of water. 🧠 Approach Used: Two Pointers Instead of checking all possible pairs (which would be inefficient), I used a two-pointer strategy: Start with pointers at both ends of the array Calculate the area using: area = min(height[left], height[right]) × (right − left) Move the pointer pointing to the smaller height, because that’s the only way to possibly increase the area This greedy logic drastically reduces unnecessary comparisons. ⚙️ Complexity ⏱ Time: O(n) 💾 Space: O(1) 🔥 Key Learnings Brute force isn’t always the answer — patterns like two pointers can cut complexity instantly Greedy decisions work well when backed by solid reasoning Pointer movement logic is just as important as the formula itself Another strong problem solved 💪 On to Day 78 🚀 #100DaysOfCode #LeetCode #TwoPointers #Java #DSA #ProblemSolving #CodingJourney