Relive the #DevReach23 action by catching up with the awesome sessions and speakers we had the pleasure and honor of hosting this past October! That's right, all sessions from the conference are coming up for you to watch at your leisure! #dotnet, #javascript, #AI, #Design, #testing, #A11y and so much more! 👉 Grab the playlist from the explosive Day 1 and stay tuned for more videos from the dual tracks on Day 2 - which means more knowledge, more insights, and more great speakers: https://prgress.co/49l2DVb #softwaredevelopment
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We launched the new Syntax.fm website with TypeScript errors! And, hey... that's okay! Learn more in episode #693 Lessons Learned & Bugs Fixed from Launching Syntax.fm --> Scott and Wes discuss launching the new Syntax.fm site, including database timeouts from too much data, the importance of error monitoring, dark mode UI bugs, using AI for show notes, managing background jobs with serverless, launching with TypeScript errors, having fast local development, being mindful of payload sizes, taking advantage of new browser APIs, and how Wes helped improve the overall design. https://lnkd.in/g9dPUTHa #launching #database #typescript #errors #coding #webdev #programming #softwaredeveloper #webdeveloper #serverless #syntaxfm #website #nodeploysonfridays #bugs
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#LeetCodeMajorityElementChallenge Day 33 of LeetCode Challenge To solve this challenge, initialize "result" to "None" (It holds the majority element), and "freq" to 0 (It records how many times "result" has been spotted more than other elements). Iterate through the list and check if "freq" is equal to 0. If it is the case this implies, we have not found an element as the majority element up to now. So, we set the current "num" as the potential majority element and we start counting frequencies of the potential majority element at "freq = 1". Then check if the result is equal to the current "num" and increment "freq". If the "result" is not equal to the current "num" decrement "freq". The time complexity is O(n) and the space complexity is O(1). #LeetCodeChallenge #MajorityElement #DataStructures #Algorithms
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Data Enthusiast | Data Analyst | Data Science | ML/DL/AI | Analytics | Visualization | ETL | UI/UX | NFT | Power Apps | IT | Content Writer | Jobs/Recruitment | Quoran | Follow for more
🚀 Exciting News! Check out x-crawl - a versatile Node.js multifunctional crawler library that enables seamless, secure, and stable page, interface, and file crawling. With features like asynchronous/synchronous modes, dynamic page support, control over crawling operations, and more, it's a game-changer for obtaining network data using JavaScript/TypeScript. Don't forget to support x-crawl by giving its repository a star on GitHub! #NodeJS #CrawlerLibrary #DataCrawling #JavaScript #TypeScript #GitHub #xCrawl
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Hey LinkedIn! If you’re new to following me, I talk about #fullstackdeveloper, #fullstackjava, #fullstackwebdevelopment, #problemsolvingskills, and #algorithms. Let me know which topic interests you the most! ✒️#fullstackwebdevelopment
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Connecting with other developers in your area doesn't need to be that complicated. A quick dm as simple as, "hey wanna go grab a beer" can be all it takes to build your own local dev meetup. watch the full episode at https://lnkd.in/gFRmFy_g Yesterday's episode #796 Do We Need JS Frameworks × Are You Over-Engineering? × Webview vs Native >> In this potluck episode, Scott and Wes answer questions about over-engineering, generative AI, frameworks, meetups, and more. #overengineered #ai #framework #javascript #meetups #webdev #syntaxfm #devcommunity #meetup
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Front-end Web developer || Proficient in (JAVA ◉ Python) || DSA || Learner and Mentor || Class Representative || #365daysofcode
Day 78 of #365daysofcode Hello, Connections, Today I solved the LeetCode problem 41. First Missing Positive ✅ The Approach: The code uses Cycle Sort to rearrange positive integers within an array to their correct positions. It iterates through the array, placing each positive integer k at index k - 1. After rearrangement, it scans the array again to find the first element not in its correct position, which indicates the first missing positive integer. If all positive integers are in their correct positions, the code returns n + 1, where n is the array length. 🧠 The Algorithm: > Initialize n as the length of the input array nums. > Rearrange positive integers within the array using Cycle Sort: Iterate through the array: For each positive integer k at position I: While k is within the valid range (greater than 0 and less than or equal to n) and k is not in its correct position: Swap the element at position i with the element at the position k - 1. > Find the first missing positive integer: Iterate through the modified array: Check if each element is in its correct position (element k at index k - 1). > If an element is not in its correct position, return k, which is the first missing positive integer. > If all positive integers are in their correct positions, return n + 1 as the first missing positive integer. > The algorithm rearranges positive integers to their correct positions using > Cycle Sort and then finds the first missing positive by checking for the first element not in its correct position. > If all positive integers are correctly positioned, the code returns n + 1. 🔍 The time and Space complexity of this solution is:- Time Complexity: O(n). Space Complexity: O(1). Please do let me know if you know a much better and optimized code for this problem 👩💻. #coders #engineers #hustlers #leetcode #365daysofcode #nevergiveup #consistency
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Check out this article on CRDTs - it's a fascinating read and lets you even try out CRDTs live in the browser! After going through it, I'm eager to try it out in one of my projects. Personally, I'm more interested in exploring operation-based CRDTs. If you're curious about CRDTs too (or don’t know what this acronym stands for), give this article a read! #CRDTs #distributedsystems #coding
An Interactive Intro to CRDTs | jakelazaroff.com
jakelazaroff.com
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🚀 #Day3 Dived deep into the art of rendering data to templates, merging them seamlessly, and extending their functionality. 💻Passing data, played around with template variables and tags, and mastered the craft of passing IDs into URL parameters. ✨linked dynamic context to my URLs, creating a more personalized user experience. One of the highlights was unraveling the mystery of URL parameters and seamlessly linking dynamic context to my URLs. #30daychallenge #LearningJourney🚀 #codechallenge #coding
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🚀 New Project Alert! I’ve developed a React application that allows users to visualize various sorting algorithms, and I'm excited to share it with you all! 🎉 Here are some features of the application: - Real-time visualizations of sorting processes - Comparison of different sorting algorithms - Dark Theme Code: https://lnkd.in/geQPDY4n #ReactJS #SortingAlgorithms #TechInnovation #Coding #DataVisualization #WebDevelopment #LearningTools #Material-UI
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Front-end Web developer || Proficient in (JAVA ◉ Python) || DSA || Learner and Mentor || Class Representative || #365daysofcode
Day 75 of #365daysofcode Hello, Connections, Today I solved the LeetCode problem 33. Search in Rotated Sorted Array ✅ The Approach: This code implements a modified binary search algorithm to find the index of a target element in a rotated sorted array. It maintains low and high pointers, and as it iterates, it compares the middle element with the target. Based on the rotation and the comparison, it adjusts the pointers to efficiently search for the target element. This approach allows for finding the target with a time complexity of O(log n) in the rotated array. 🧠 The Algorithm: > Initialize low and high pointers to cover the entire array. > Enter a while loop while low is less than or equal to high. > Calculate the middle index (mid) within the current range. > If nums[mid] equals the target, return mid as the result. > Determine whether the left half or right half of the current range is sorted by comparing nums[low] and nums[mid. > If the left half is sorted and the target falls within it, adjust high to mid - 1; otherwise, adjust low to mid + 1 to focus the search on the right half. > If the right half is sorted and the target falls within it, adjust low to mid + 1; otherwise, adjust high to mid - 1 to focus the search on the left half. > Continue this process until low becomes greater than high, indicating that the target element is not in the array. In this case, return -1. 🔍 The time and Space complexity of this solution is:- Time Complexity: O(logn). Space Complexity: O(1). Please do let me know if you know a much better and optimized code for this problem 👩💻. #coders #engineers #hustlers #leetcode #365daysofcode #nevergiveup #consistency
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