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I Got Bored of LeetCode, so I Built a Coding RPG

https://dsa-life-simulator-frontend.vercel.app"I made a free tool to make DSA practice feel like an RPG — would like feedback from this community"Been grinding DSA for months and it never felt fun. So I built something. What it does: 🏟️ Real-time 1v1 Arena battles against other devs 🧪 Lab to create and publish your own challenges 🏘️ Community Hub to attempt others' challenges 📖 AI writes your weekly coding journey as a life story 🎮 XP, credits, levels, leaderboards Stack: React + Tailwind + Firebase + Node.js + Socket.IO + Groq AI Still early — would genuinely love feedback from people who've felt the pain of traditional DSA prep.

2026-06-11 原文 →
AI 资讯

Delete Node in a Linked List

Problem Link - https://leetcode.com/problems/delete-node-in-a-linked-list/ This is one of those interview questions that looks impossible at first. Normally, to delete a node from a Linked List, we need access to the previous node. But in this problem, we're only given the node that needs to be deleted. No head. No previous pointer. So how do we remove it? Let's understand the trick. Problem Statement Write a function to delete a node in a singly linked list. You are not given the head of the list. Instead, you are given only the node that needs to be deleted. Example Input: 4 -> 5 -> 1 -> 9 node = 5 Output: 4 -> 1 -> 9 Initial Thought Normally we delete a node like this: prev.next = node.next But here: We don't have prev We don't have head So the usual deletion approach is impossible. Key Observation Although we cannot delete the current node directly, we can make it look like it never existed. Consider: 4 -> 5 -> 1 -> 9 We need to delete: 5 Instead of removing node 5 , copy the value of the next node into it. 4 -> 1 -> 1 -> 9 Now remove the next node. 4 -> 1 -> 9 The original value 5 has disappeared. Mission accomplished. Intuition Copy the next node's value into the current node. Skip the next node. The current node now behaves as if it was deleted. Since the problem guarantees that the given node is not the tail node, a next node will always exist. Dry Run Input 4 -> 5 -> 1 -> 9 node = 5 Current node: 5 Next node: 1 Step 1 Copy next node value. node.val = node.next.val List becomes: 4 -> 1 -> 1 -> 9 Step 2 Skip next node. node.next = node.next.next List becomes: 4 -> 1 -> 9 Done. Optimal Java Solution class Solution { public void deleteNode ( ListNode node ) { ListNode cur = node . next ; node . val = cur . val ; node . next = cur . next ; } } Even Shorter Version class Solution { public void deleteNode ( ListNode node ) { node . val = node . next . val ; node . next = node . next . next ; } } Complexity Analysis Metric Complexity Time Complexity O(1) Space Comp

2026-06-10 原文 →
AI 资讯

A Record-Breaking Patch Tuesday for June 2026

Microsoft today released software updates to plug nearly 200 security holes across its Windows operating systems and supported software, a record number of fixes for the company's monthly Patch Tuesday cycle. Nearly three dozen of those bugs earned Microsoft's most dire "critical" rating, and exploit code for at least three of the weaknesses is now publicly available.

2026-06-10 原文 →
AI 资讯

Apple plays catch-up at WWDC

Apple spent much of its WWDC keynote highlighting fixes, performance improvements, and long-requested features before unveiling its upgraded AI-powered Siri, signaling that the company wants users to see AI as just one part of a broader effort to improve its software.

2026-06-09 原文 →