Congrats, Team India Women! 🏆💙 World Champions — you’ve brought the joy and pride home! Every inning, every victory, every moment made the nation proud🇮🇳 #WorldChampions #indwvssaw

Writing perfect TypeScript types for 3 hours... then fixing one error with as any 💀

Why are Indian https://t.co/Yi7uVTqzvp colleges obsessed with Turbo C? You open VS Code and suddenly you're the villain of the syllabus 😭 Let me code in peace, not in 1995. Is your college like this, too, or is it just mine?

Node.js built the web. 🌐 Bun is rebuilding it — faster. ⚡ Node = stable, proven Bun = fast, all-in-one (runtime + bundler + test runner) Same goal. New speed. Which one are you using? 👇 #JavaScript #NodeJS #Bun

I am stuck in this loop. How should I break it?

IDE auto-complete is killing real learning. You think you “know” code until the interview asks you to write syntax from scratch 💀

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Just installed Cursor free version 😎 Any tips on using it efficiently, or should I go premium? Let me know your thoughts 👇

Which is the best source of learning Machine learning online? Please give some #suggestions

This keeps counts aligned with the original array, even though merge sort keeps rearranging elements. Final complexity → O(n log n), space O(n). Clean, reliable way to solve the problem using merge sort in Java.

To keep results in the correct order, maintain an indexes[] array that stores each element’s original position. When you update counts, use that original index: counts.set(indexes[left], counts.get(indexes[left]) + rightTaken);

During the merge: If rightVal < leftVal, move the right element and increase a counter (rightTaken). When placing a left element, add rightTaken to its result because those many smaller elements came from the right side.

For LeetCode 315 – Count of Smaller Numbers After Self, the merge-sort approach counts smaller elements as it merges. Here’s how it works in Java 👇

This keeps counts aligned with the original array, even though merge sort keeps rearranging elements. Final complexity → O(n log n), space O(n). Clean, reliable way to solve the problem using merge sort in Java.

To keep results in the correct order, maintain an indexes[] array that stores each element’s original position. When you update counts, use that original index: counts.set(indexes[left], counts.get(indexes[left]) + rightTaken);

During the merge: If rightVal < leftVal, move the right element and increase a counter (rightTaken). When placing a left element, add rightTaken to its result because those many smaller elements came from the right side.

Does anyone else absolutely struggle to solve LeetCode 315 'Count of Smaller Numbers After Self' with merge sort? 😵‍💫 If merge sort made you want to pull your hair out, how did you finally crack it? Share your tips

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Delete LinkedIn. Delete Insta. Make friends on X. Start coding 👨‍💻 Am I right???

My code right now: Literally collapsing Me: adds try-catch block Everything’s fine 😭