Why Java 26 is My Go-To Language for My Projects

Day 118 - LeetCode Journey 🚀 Solved LeetCode 98: Validate Binary Search Tree in Java ✅ Moving from searching a BST to validating one! This problem is a classic "Medium" for a reason—it teaches you that local properties (just checking a node against its children) aren't enough; you have to consider the global constraints of the entire tree. The Approach: 1) Beyond Local Checks: A common pitfall is only checking if root.left < root < root.right. However, every node in the left subtree must be smaller than the root, and every node in the right subtree must be larger. 2) Recursive Bounds (Min/Max): I implemented a helper function that passes down a range (minimum and maximum allowed values) for each node. ->When moving left, the max value is updated to the current node's value. ->When moving right, the min value is updated to the current node's value. 3) Integer Overflows: Instead of using Integer.MIN_VALUE or MAX_VALUE as initial bounds, I passed TreeNode objects (or used null checks) to handle edge cases where the node value itself is the integer limit. Key Takeaways: -> Global Constraints: Learned how to propagate constraints down the tree through recursion. -> 100% Runtime Efficiency: Achieved a 0 ms result by visiting each node exactly once (O(n) time complexity). -> Tree Traversal Mastery: Reinforced the concept of top-down validation where parent values dictate the boundaries for all descendants. Every validated node is another layer of logic mastered. Consistency is doing its work! 💪 #LeetCode #DSA #Java #BinaryTree #Algorithms #ProblemSolving #CodingPractice #InterviewPreparation #Consistency #DailyCoding

Today’s Java DSA challenge: Sort Colors. After conquering Two-Pointer problems like moving zeroes and reversing arrays, I tackled the classic Dutch National Flag algorithm on LeetCode to sort an array of 0s, 1s, and 2s. You could easily solve this with a counting sort approach (count the 0s, 1s, and 2s, then overwrite the array in a second pass). But we can optimize this to a single pass with O(1) space using three pointers! Here is how the magic works: • low pointer: Tracks where the next 0 should go. • mid pointer: The explorer scanning through the array. • high pointer: Tracks where the next 2 should go. The Logic: • If mid finds a 0: Swap it with low, then move both low and mid forward. • If mid finds a 1: It's already in the safe middle zone, so just move mid forward. • If mid finds a 2: Swap it with high, and move high backward. (The catch? Don't move mid yet, because you still need to check the newly swapped number!) Devs: The Dutch National Flag algorithm is an absolute masterpiece. What other algorithm completely blew your mind the first time? 👇 #DSA #Java #LeetCode #SortColors

#Post11 In the previous post(https://lnkd.in/dynAvNrN), we saw how to create threads in Java. Now let’s talk about a problem. If creating threads is so simple… why don’t we just create a new thread every time we need one? Let’s say we are building a backend system. For every incoming request/task, we create a new thread: new Thread(() -> { // process request }).start(); This looks simple. But this approach breaks very quickly in real systems because of below mentioned problems. Problem 1: Thread creation is expensive Creating a thread is not just creating an object. It involves: • Allocating memory (stack) • Registering with OS • Scheduling overhead Creating thousands of threads = performance degradation Problem 2: Too many threads → too much context switching We already saw this earlier(https://lnkd.in/dYG3v-vb). More threads does NOT mean more performance. Instead: • CPU spends more time switching • Less time doing actual work Problem 3: No control over thread lifecycle When you create threads manually: • No limit on number of threads • No reuse • Hard to manage failures This quickly becomes difficult to manage as the system grows. So what’s the solution? Instead of creating threads manually: we use something called the Executor Framework. In simple words consider the framework to be like: Earlier, we were manually hiring a worker (thread) for every task. With Executor, we have a team of workers (thread pool), and we just assign tasks to them. Key idea Instead of: Creating a new thread for every task We do: Submit tasks to a pool of reusable threads This is exactly what Java provides using: Executor Framework Key takeaway Manual thread creation works for learning, but does not scale in real-world systems. Thread pools help: • Control number of threads • Reduce overhead • Improve performance We no longer manage threads directly — we delegate that responsibility to the Executor Framework. In the next post, we’ll see how Executor Framework works and how to use it in Java. #Java #Multithreading #Concurrency #BackendDevelopment #SoftwareEngineering

Solved Find Peak Element -> LeetCode(162) Medium, Binary Search in Java. Till now I had solved around 10-11 binary search problems. But in all of them array was always sorted , even rotated ones had at least one sorted half. So I had one strong assumption , binary search only works on sorted arrays. This problem broke that assumption completely. No sorted half, no rotation logic working. I was stuck because none of my previous patterns were fitting here. Took help from Claude. It suggested slope based thinking , if right neighbour is greater, peak is on right side, go right. If left neighbour is greater, go left. If both neighbours are smaller, current element is peak. Then I questioned > what if slope breaks in between? Claude pointed me to re-read the problem. The key insight was that array has -∞ at both ends virtually, which guarantees at least one peak always exists. Applied my own logic from there and got it accepted Then Claude showed me a cleaner version , when low < high, at the point where low == high we already have our peak. No need for extra boundary checks. That gave me a second shorter solution. Also broke my second assumption > binary search doesn't always need while(low <= high). Sometimes while(low < high) is cleaner. Two wrong assumptions fixed in one problem 🙌 Took help but questioned it, understood it, then coded it myself. Github Repo link : https://lnkd.in/grF5ACw5 **Any other wrong assumption you had about binary search? Drop it below 👇** #DSA #Java #LeetCode #BinarySearch #LearningInPublic

Day 73 of #90DaysDSAChallenge Solved LeetCode 451: Sort Characters By Frequency Learned an important Java design concept today. Problem Overview: The task was to sort characters in a string based on descending frequency. What confused me initially: Why create a separate Freq class instead of just using HashMap and PriorityQueue directly? Key Learning: PriorityQueue stores one complete object at a time. For this problem, each item needs two pieces of data together: Character Frequency Example: Instead of storing: e and 2 separately We package them as: Freq('e', 2) That custom class acts like a container holding both values in one object, so PriorityQueue can compare and sort them correctly. Why this matters: This taught me that custom classes in Java are often not about complexity, they simply bundle related data into one manageable unit. Alternative approach: We can also use Map.Entry<Character, Integer> instead of creating a custom class, but building Freq makes the logic easier to understand while learning. Today’s takeaway: Not every class is for business logic — sometimes it exists just to package data cleanly. #Java #90DaysDSAChallenge #LeetCode #PriorityQueue #HashMap #CodingJourney #ProblemSolving

🔥 𝗗𝗮𝘆 𝟵𝟰/𝟭𝟬𝟬 — 𝗟𝗲𝗲𝘁𝗖𝗼𝗱𝗲 𝗖𝗵𝗮𝗹𝗹𝗲𝗻𝗴𝗲 𝟭𝟱𝟯𝟵. 𝗞𝘁𝗵 𝗠𝗶𝘀𝘀𝗶𝗻𝗴 𝗣𝗼𝘀𝗶𝘁𝗶𝘃𝗲 𝗡𝘂𝗺𝗯𝗲𝗿 | 🟢 Easy | Java Marked as Easy — but the optimal solution is pure binary search brilliance. 🎯 🔍 𝗧𝗵𝗲 𝗣𝗿𝗼𝗯𝗹𝗲𝗺 Given a sorted array, find the kth missing positive integer. Linear scan works — but can we do O(log n)? 💡 𝗧𝗵𝗲 𝗞𝗲𝘆 𝗜𝗻𝘀𝗶𝗴𝗵𝘁 At index i, the value arr[i] should be i+1 in a complete sequence. So missing numbers before arr[i] = arr[i] - 1 - i This lets us binary search on the count of missing numbers! ⚡ 𝗔𝗽𝗽𝗿𝗼𝗮𝗰𝗵 — 𝗕𝗶𝗻𝗮𝗿𝘆 𝗦𝗲𝗮𝗿𝗰𝗵 ✅ If arr[mid] - 1 - mid < k → not enough missing numbers yet, go right ✅ Else → too many missing, go left ✅ After the loop, left + k gives the exact answer 𝗪𝗵𝘆 𝗹𝗲𝗳𝘁 + 𝗸? After binary search, left is the index where the kth missing number falls beyond. left numbers exist in the array before that point, so the answer is left + k. No extra passes needed. ✨ 📊 𝗖𝗼𝗺𝗽𝗹𝗲𝘅𝗶𝘁𝘆 ⏱ Time: O(log n) — vs O(n) linear scan 📦 Space: O(1) This is a perfect example of binary searching on a derived condition, not just a value. A real upgrade from the naive approach. 🧠 📂 𝗙𝘂𝗹𝗹 𝘀𝗼𝗹𝘂𝘁𝗶𝗼𝗻 𝗼𝗻 𝗚𝗶𝘁𝗛𝘂𝗯: https://lnkd.in/gVYcjNS6 𝟲 𝗺𝗼𝗿𝗲 𝗱𝗮𝘆𝘀. 𝗧𝗵𝗲 𝗳𝗶𝗻𝗶𝘀𝗵 𝗹𝗶𝗻𝗲 𝗶𝘀 𝗿𝗶𝗴𝗵𝘁 𝘁𝗵𝗲𝗿𝗲! 🏁 #LeetCode #Day94of100 #100DaysOfCode #Java #DSA #BinarySearch #Arrays #CodingChallenge #Programming

📘 Day 25 – Unlocking the Magic of Java Casting Today I dove deep into non-primitive type casting in Java and had that haha moment! 💡 ✨ Upcasting – Treating a subclass object as a superclass reference. It makes my code cleaner, flexible, and ready for change. ⚡ Downcasting – Converting back safely to a subclass. Done wrong, it throws ClassCastException, but done right, it’s pure power. 🛡 instanceof operator – My safety net! It checks object type before casting, keeping runtime errors away. Seeing objects flow up and down the hierarchy revealed the true beauty of polymorphism, code that’s adaptable, maintainable, and future-proof. 💬 What really clicked: Java isn’t just about syntax; it’s about managing relationships between objects smartly. This makes every line of code safer, cleaner, and smarter. #Java #OOP #Polymorphism #Upcasting #Downcasting #ClassCastException #InstanceOf #DailyLearning #CodeBetter #ProgrammingJourney #DevLife

Day 74 of #100DaysOfLeetCode 💻✅ Solved #162. Find Peak Element problem in Java. Approach: • Used Binary Search technique to efficiently find the peak element • Set two pointers, left at start and right at end of the array • Calculated mid index using safe mid formula • Compared nums[mid] with nums[mid + 1] to determine direction • If mid element is smaller, moved search space to right half • Otherwise, moved search space to left half including mid • Continued until left and right pointers converged • Final position (left == right) represents the peak index Performance: ✓ Runtime: 0 ms (Beats 100.00% submissions) 🚀 ✓ Memory: 44.32 MB (Beats 25.49% submissions) Key Learning: ✓ Strengthened understanding of Binary Search on unsorted arrays ✓ Learned how to apply divide-and-conquer beyond traditional searching ✓ Improved intuition for peak finding using neighbor comparison ✓ Practiced optimizing search space instead of linear scanning Learning one problem every single day 🚀 #Java #LeetCode #DSA #BinarySearch #Arrays #ProblemSolving #CodingJourney #100DaysOfCode

🚀 Ever wondered what actually happens under the hood when you run a Java program? It’s not just magic; it’s the Java Virtual Machine (JVM) at work. Understanding JVM architecture is the first step toward moving from "writing code" to "optimizing performance." Here is a quick breakdown of the core components shown in the diagram: 1️⃣ Classloader System The entry point. It loads, links, and initializes the .class files. It ensures that all necessary dependencies are available before execution begins. 2️⃣ Runtime Data Areas (Memory Management) This is where the heavy lifting happens. The JVM divides memory into specific areas: Method/Class Area: Stores class-level data and static variables. Heap Area: The home for all objects. This is where Garbage Collection happens! Stack Area: Stores local variables and partial results for each thread. PC Registers: Keeps track of the address of the current instruction being executed. Native Method Stack: Handles instructions for native languages (like C/C++). 3️⃣ Execution Engine The brain of the operation. It reads the bytecode and executes it using: Interpreter: Reads bytecode line by line. JIT (Just-In-Time) Compiler: Compiles hot spots of code into native machine code for massive speed boosts. Garbage Collector (GC): Automatically manages memory by deleting unreferenced objects. 4️⃣ Native Interface & Libraries The bridge (JNI) that allows Java to interact with native OS libraries, making it incredibly versatile. 💡 Pro-Tip: If you are debugging OutOfMemoryError or StackOverflowError, knowing which memory area is failing is half the battle won. #Java #JVM #BackendDevelopment #SoftwareEngineering #ProgrammingTips #TechCommunity #JavaDeveloper #CodingLife