Find Missing Number in Java 8 Array

Leetcode Problem || Complement of Base 10 Integer(1009) 🚀 The Bitwise Complement problem using Java. 🔹 Idea: The complement of a number means flipping all bits in its binary representation (0 → 1 and 1 → 0). Instead of manually converting the number to binary, I used a bit manipulation trick: 1️⃣ Find the highest set bit using Integer.highestOneBit(n) 2️⃣ Create a mask with all bits set to 1 up to that position 3️⃣ Use XOR (^) with the mask to flip the bits #LeetCode #Java #BitManipulation #CodingPractice #ProblemSolving #SoftwareDevelopment

📢 Stop writing messy Lambdas ❗ When I first started with Java 8, I thought Lambda expressions were the ultimate "clean code" hack. Then I discovered Method References, and it changed the game. If your code looks like this: list.forEach(s -> System.out.println(s)); It could look like this: list.forEach(System.out::println); It’s cleaner, more readable, and honestly? It just looks more professional. I’ve put together a quick 7-slide guide to the 4 types of Method References you need to know for your next MNC interview (and for your daily PRs!): 1️⃣ Static Methods (Integer::parseInt) 2️⃣ Instance Methods of a Particular Object (System.out::println) 3️⃣ Instance Methods of an Arbitrary Object (String::toLowerCase) 4️⃣ Constructor References (ArrayList::new) Mastering these is a small change that makes a huge difference in how your "Java Full-Stack" skills are perceived. Which one do you find yourself using the most? Let’s discuss in the comments! #Java8 #CleanCode #BackendDeveloper #FullStack #ProgrammingTips #JavaDeveloper #CodingLife #SoftwareEngineering

Before Java 8, we spent a lot of time writing boilerplate loops just to filter a list. With Streams and Lambdas, Java shifted toward declarative programming, making our code more readable, maintainable, and expressive. This quick reference breaks down the essential flow: Lambdas & Method References: Clean shorthand to keep your logic concise. The Pipeline: Understanding the difference between Intermediate (lazy) and Terminal (eager) operations is key to avoiding "ghost" code that never executes. Short-Circuiting: Tools like findFirst() or limit() are performance lifesavers when dealing with large datasets. #Java #CleanCode #FunctionalProgramming #SoftwareEngineering #CodingTips

Just published a clean and efficient walkthrough of LeetCode 169 – Majority Element using the Boyer–Moore Voting Algorithm in Java. In this breakdown, I explain: 🔹 Why the algorithm works in O(n) time & O(1) space 🔹 How to identify the majority element intuitively 🔹 A clear Java implementation with step‑by‑step reasoning If you're preparing for coding interviews or sharpening your DSA fundamentals, this one’s a must-watch! Watch here: https://lnkd.in/gyB-6DAx Let’s grow together — one problem at a time. 💙💻 #LeetCode #Java #DSA #CodingInterview #ProblemSolving #TechCareer #Algorithms #LearningEveryday

🚀 Day 46/180 | #180DaysOfCode 📍 LeetCode | 💻 Java Solved: 2315. Count Asterisks Used a boolean flag approach to track whether the current position is inside a pair of '|' and count only the valid '*' outside those sections. ⏱️ Time Complexity: O(n) 📦 Space Complexity: O(1) Strengthening understanding of state-based string traversal and conditional counting. 💪 Consistency continues 🚀 #DSA #LeetCode #Java #CodingJourney #Consistency

Day 94/100: #LeetCodeChallenge – Grid Partitioning in Java 🧩💻 Another day, another algorithmic deep dive! Today’s problem was about determining whether a grid can be partitioned in a specific way. While the problem may seem straightforward at first, the real challenge lies in handling edge cases, optimizing for efficiency, and ensuring clean, maintainable code. 🔍 Key takeaways from today’s solution: Understanding 2D array traversal and prefix sums Handling edge cases like 1x1 grids early Writing readable code that can scale with larger test cases Even though the sample output shows "You must run your code first," the process of thinking through the logic, testing edge cases, and refining the approach is where the real growth happens. Every problem adds another tool to the problem-solving toolkit. 🚀 Consistency > Intensity. Day 94 is in the books. On to the final sprint! #100DaysOfCode #LeetCode #Java #CodingChallenge #ProblemSolving#GridPartitioning #DataStructuresAndAlgorithms #TechJourney#SoftwareEngineering #CodeNewbie #DeveloperLife #AlgorithmDesign#ConsistencyIsKey

Day 62 of #100DaysOfLeetCode 💻✅ Solved #219. Contains Duplicate II problem in Java. Approach: • Used two nested loops to compare elements within range k • For each element, checked next k elements • If any duplicate is found within distance k, returned true • If no such pair exists, returned false Key Learning: ✓ Practiced checking duplicates within a given range ✓ Strengthened understanding of array traversal with conditions ✓ Learned the importance of optimizing nested loop solutions Learning one problem every single day 🚀 #Java #LeetCode #DSA #Arrays #ProblemSolving #CodingJourney #100DaysOfCode

LeetCode Problem || Find Unique Binary String (1980)🚀 Today I solved the problem "Find Unique Binary String" using Java. 🔹 Problem: We are given an array of n binary strings, each of length n. The goal is to return a binary string of length n that does not exist in the array. 🔹 Approach (Diagonal Flip Technique): The idea is simple but powerful: Traverse the array using index i. Look at the i-th character of the i-th string (nums[i][i]). Flip the bit (0 → 1, 1 → 0). Append it to a result string. 💡 Time Complexity: O(n) Practicing problems like this strengthens logical thinking and problem-solving skills. #LeetCode #Java #CodingPractice #ProblemSolving #DSA

Just dropped a new video on Comparator Interface in Java — covering custom sorting logic using multiple approaches (class, anonymous class and lambda). A clean, visual explanation to make sorting truly click. #Java #Comparator #Coding #LearningEveryday Watch here: https://lnkd.in/geW_UKh9

🚀 Day 73 / 100 Days of LeetCode Question: Check if Binary String Has at Most One Segment of Ones Given a binary string s, determine whether it contains at most one continuous segment of '1's. Solved in Java by checking whether the pattern "01" exists in the string. If "01" appears, it means a new segment of 1s starts after a 0, which violates the condition. This provides a clean and efficient solution. Consistency > perfection. Day 74 loading 🔥 #100DaysOfCode #LeetCode #Java #DSA #ProblemSolving