Reorder List: Java Solution with Slow & Fast Pointers

🚀 Day 16 of #100DaysOfCode Today’s problem: Minimum Absolute Difference 📊 🔍 Problem Understanding: Given an array of distinct integers, find all pairs with the minimum absolute difference. 🧠 Approach: First, I sorted the array Then compared adjacent elements (since minimum difference will always be between neighbors in sorted order) Tracked the minimum difference and stored all valid pairs 👉 Key Insight: Sorting reduces unnecessary comparisons and simplifies the problem ⚙️ Concepts Used: Sorting (Arrays.sort) Greedy observation (check only neighbors) List handling for storing pairs 📊 Complexity: ⏱️ Time Complexity: O(N log N) (due to sorting) 💾 Space Complexity: O(N) (for storing result pairs) 📚 Key Learnings: Learned how sorting simplifies problems Understood how to reduce complexity from brute force O(N²) → O(N log N) Improved thinking in terms of patterns and optimization 💯 Result: ✔️ Accepted (All test cases passed) ✔️ Runtime: 1 ms (100%) 🔥 ✔️ Clean and efficient solution Sometimes the best optimization is just sorting + observation 💡 Let’s keep building consistency 🚀 #Day16 #100DaysOfCode #Java #DSA #LeetCode #ProblemSolving #Sorting

Day 46 – Length of Last Word Worked on a problem to find the length of the last word in a string while handling extra spaces efficiently. Key Learnings: Traversing a string from right to left to avoid unnecessary operations Skipping trailing spaces before processing the actual word Using a single pass approach for optimal performance #DSA #Java #Strings #ProblemSolving #CodingPractice

🚀 Day 58 of #100DaysOfCode Solved 165. Compare Version Numbers on LeetCode 🔗 🧠 Key Insight: Version strings are split by "." into multiple revisions. We compare each revision numerically (not lexicographically). Example: "1.01" = "1.1" (leading zeros don’t matter) ⚙️ Approach (Split + Compare): 1️⃣ Split both versions using "." 🔹 version1 → s1[] 🔹 version2 → s2[] 2️⃣ Traverse till max length of both arrays 3️⃣ For each index i: 🔹 num1 = i < s1.length ? parseInt(s1[i]) : 0 🔹 num2 = i < s2.length ? parseInt(s2[i]) : 0 4️⃣ Compare: 🔹 if num1 < num2 → return -1 🔹 if num1 > num2 → return 1 5️⃣ If all equal → return 0 ⏱️ Time Complexity: O(n + m) 📦 Space Complexity: O(n + m) (for split arrays) #100DaysOfCode #LeetCode #DSA #Strings #TwoPointers #Java #InterviewPrep #CodingJourney

🚀 Day 6 of #100DaysOfCode | Pattern: Two Pointers Today I tackled a classic Linked List problem: 👉 Remove Nth Node From End of List At first, I thought of a brute force approach: Traverse the list to find length Then remove (length - n) node ❌ Works, but requires 2 passes 💡 Then I learned the optimal approach (Two Pointers): Use fast and slow pointers Move fast n steps ahead Move both together until fast reaches end Delete the node using slow ✅ Only 1 pass ✅ Cleaner and efficient ⚠️ Mistakes I made (important learning): Forgot to use a dummy node → failed for head deletion Got confused between moving n vs n+1 steps Faced null pointer issues in edge cases 🔥 Key Takeaway: Two pointer technique is not just for arrays, it’s super powerful in Linked Lists too 📈 Progress Update: Consistency building day by day Next week: Starting a new pattern Along with revision of Two Pointers #100DaysOfCode #DSA #Java #LinkedList #CodingJourney #ProblemSolving #LearnInPublic #PatternWise

Day 31/50 🚀 — Reverse Vowels of a String (LeetCode 345) Today’s problem was a great reminder that sometimes the simplest approaches are the most efficient. 🔹 Used the two-pointer technique 🔹 Focused on in-place swapping 🔹 Optimized for both time (O(n)) and space (O(1)) Key takeaway: Instead of overthinking, break the problem into smaller checks—identify vowels, move pointers smartly, and swap only when needed. Clean and efficient 💡 Happy to see this solution performing well: ⚡ Runtime: 2 ms (faster than 99%+) 📦 Space: Decent optimization #Day31 #LeetCode #DSA #Java #CodingJourney #50DaysOfCode #ProblemSolving #SoftwareEngineering

🚀 Day 29 of #75daysofLeetCode 2095 – Delete the Middle Node of a Linked List Just solved an interesting linked list problem that perfectly demonstrates the power of the two-pointer technique (slow & fast pointers). 🔍 Problem Insight: Given a linked list, delete its middle node where the middle is defined as ⌊n/2⌋ (0-based indexing). 💡 Key Idea: Instead of calculating the length, we can efficiently find the middle using: 🐢 Slow pointer (1 step) ⚡ Fast pointer (2 steps) When the fast pointer reaches the end, the slow pointer will be at the middle node! 🛠 Approach: ✔ Handle edge case (single node → return null) ✔ Traverse using slow & fast pointers ✔ Keep track of previous node ✔ Remove the middle node in one pass ⏱ Time Complexity: O(n) 💾 Space Complexity: O(1) 🔥 Why this matters? This pattern is widely used in: Finding middle of linked list Detecting cycles Splitting lists Mastering this unlocks many problems! #LeetCode #DSA #LinkedList #Java #CodingInterview #ProblemSolving #TechLearning

🚀 Day 55 of #100DaysOfCode Solved 147. Insertion Sort List on LeetCode 🔗 🧠 Key Insight: We apply the classic Insertion Sort, but on a linked list instead of an array. The challenge is handling pointer manipulation efficiently. ⚙️ Approach: 1️⃣ Create a dummy node to act as the start of the sorted list 2️⃣ Traverse the original list node by node 3️⃣ For each node: Find its correct position in the sorted part Insert it there by updating pointers 🔁 This builds a sorted list incrementally ⏱️ Time Complexity: O(n²) 📦 Space Complexity: O(1) #100DaysOfCode #LeetCode #DSA #LinkedList #Sorting #Java #InterviewPrep #CodingJourney

⏰ Temporary Field: When instance variables only show up for certain operations, cluttering your class interface with null-initialized fields. Here's how Extract Class refactoring transforms your code: ✅ Move temporary fields into dedicated classes ✅ Eliminate null-initialized field clutter ✅ Create clearer class interfaces ✅ Make object state predictable at every stage 🎯 Key takeaway: Extract Class refactoring organizes temporary state into focused objects that only exist when they're needed. What's your experience with refactoring Temporary Fields? Share your favorite techniques in the comments. #CleanCode #Refactoring #Java #TemporaryField #DeveloperTips https://lnkd.in/gJQeWPJ9

🚀 Day 7 – Exception Handling: More Than Just try-catch Today I focused on how exception handling should be used in real applications—not just syntax. try { int result = 10 / 0; } catch (Exception e) { System.out.println("Error occurred"); } This works… but is it the right approach? 🤔 👉 Catching generic "Exception" is usually a bad practice 💡 Better approach: ✔ Catch specific exceptions (like "ArithmeticException") ✔ Helps in debugging and handling issues more precisely ⚠️ Another insight: Avoid using exceptions for normal flow control Example: if (value != null) { value.process(); } 👉 is better than relying on exceptions 💡 Key takeaway: - Exceptions are for unexpected scenarios, not regular logic - Proper handling improves readability, debugging, and reliability Small changes here can make a big difference in production code. #Java #BackendDevelopment #ExceptionHandling #CleanCode #LearningInPublic

🚀 𝗗𝗮𝘆 - 51/60 Problem: URLify a Given String 🔍 Learned: Replaced all spaces in the string with "%20" to make it URL-friendly. This can be done using string traversal or built-in methods. 😅 Struggles: Initially overthought the problem expecting complex logic, but realized it’s just about handling characters efficiently. 🧠 Key Learning: Simple problems test clarity, not complexity. Can use replace() for clean code, or manually build using StringBuilder for better control. Important to maintain O(n) time complexity. 📦 Concepts Used: #Strings #StringBuilder #BasicLogic #TimeComplexity #DSA Not every problem needs complexity—sometimes simplicity is the real skill. 🚀 #Java #CodingJourney #ProblemSolving #DSA