Devpost’s Post

📸 “Code, Chai, and Cheers”: Our after-class hangouts were a mix of lines of code and cups of chai. From debugging sessions to celebrating small wins, we wrote our own algorithms for friendship. 🤝☕📚

Well, the battleship game is almost finished. It sets all the ships, and the user can click to see if in the clicked title there was a ship or water. Now comes the difficult part: make 2 boards, one for a human the other for the machine, and be able to put ships into the human's board. All the code in my gitHub: https://lnkd.in/dXZTMVEv

Day 53 of #365daysofcode Hello, Connections, Today I solved the LeetCode problem 52. N-Queens II ✅ The Approach: The provided code solves the N-Queens problem, aiming to count the total number of valid solutions for placing N queens on an NxN chessboard without attacking each other. The approach involves a recursive backtracking algorithm. It iterates through each row, attempting to place a queen in each column while ensuring it does not conflict with previously placed queens. If a valid solution is found (when the row index equals the board's size N), it increments a count variable sol. The isSafe function checks for conflicts in rows, columns, and diagonals. The algorithm explores all possible queen placements and accumulates the count of valid solutions, returning it as the final result. 🧠 The Algorithm: > Initialize an empty chessboard (2D array) of size NxN to represent the placement of queens. > Define a recursive function nQueen that takes two parameters: the current row and the chessboard. > Inside the nQueen function: > Check if the current row equals N (the size of the chessboard). > If true, return 1 (indicating a valid solution). > Iterate through each column in the current row: > Check if it's safe to place a queen in the current cell using the isSafe function. > If safe, mark the cell with a queen ('Q') and recursively call nQueen for the next row. > After the recursive call, remove the queen from the cell ('.') to backtrack and explore other possibilities. > Define the isSafe function to check for queen conflicts in rows, columns, and diagonals. > Initialize a variable sol to 0 to keep track of the total number of valid solutions. > In the totalNQueens function: > Initialize the chessboard. > Call the nQueen function starting from the first row. > Return the value returned by nQueen, which represents the total number of valid solutions. 🔍The time and Space complexity of this solution is:- Time Complexity: O(n^n). Space Complexity: O(n^2). Please do let me know if you know a much better and optimized code for this problem 👩💻. #coders #engineers #hustlers #leetcode #365daysofcode #nevergiveup #consistency

We’re back with the second session of our 2 credit extracurricular course! Join us on 1st May as we take a deep dive into the basics of Go, an introduction to backend development - mux, handlers, Helloweb, Interfaces and more. Learn how to implement your backend with Go and finally learn about Go - Concurrency and Goroutines. You do not want to miss this! 🔗: https://dscv.it/2cc-go-api

🚨 Another new video is now available! I've recorded a video for my blog post on exploring new codebases. In this video, I provide five tips that help you get up and running in a new codebase quickly while having fun and getting to know your teammates https://lnkd.in/eer6ajA5

Sharing some funny codes from top devs. 1. Software comes from heaven when you have good hardware. – Ken Olsen 2. There is always one more bug to fix. – Ellen Ullman 3. If debugging is the process of removing bugs, then programming must be the process of putting them in. – Sam Redwine 4, Sometimes it pays to stay in bed on Monday, rather than spending the rest of the week debugging Monday’s code. – Dan Salomon 5. Computers are fast; developers keep them slow. – Anonymous 6. Talk is cheap. Show me the code. ― Linus Torvalds 7. If, at first, you do not succeed, call it version 1.0. ― Khayri R.R. Woulfe

India significantly shines on the global coding landscape, as evidenced by the GitHub heat map showcasing a substantial density of skilled developers. 🌐 This noteworthy presence highlights the nation's commitment to excellence in software development, contributing significantly to technological advancements on a global scale. 🚀🇮🇳 #Devnation #IndiaTechExcellence #GlobalCodingHub #DeveloperCommunity #InnovateFromIndia #GitHubLeadership #TechAdvancements #CodingNation 🌐💻🇮🇳

My go-to to build any API for my side projects has been Flask, it is easy to setup, easy to use, and most importantly it is lightweight. You can also use gin (Golang), it is also lightweight. PoCs should always be built on Flask. :)

Yesterday, I participated in Codeforces Round 961 (Div. 2). In this contest, I solved only one question. I had an idea for the second question but couldn't implement it. My rank was 11844, and my rating increased from 811 to 825.  Here are the approaches: A. Diagonals: In this problem, we have an n*n chessboard and k chips to place on it. Cells with coordinates (i, j) that have the same sum (i+j) lie on the same diagonal. We need to find the minimum number of diagonals that have at least one chip. The approach is simple: 1. By observation, first, we place n chips on the biggest diagonal of the board. 2. Then, we place the remaining chips symmetrically on the left and right sides of the biggest diagonal. 3. The size of these diagonals goes from n to 1, as n is the size of the biggest diagonal. By following this method, we ensure that we use the minimum number of diagonals to accommodate all k chips.

Q1 is almost closed and Easter is behind the door! So just wanted to wish a really Happy Easter, Buona Pasqua, Glad Påsk to everyone out there! Have a Great Easter weekend, enjoy the time off and get some quality time with family and friends to load the batteries. And let's not forget to eat a lot of Easter eggs.. Here are some of mine.. See u in Q2 to talk more about #lowcode #nocode #neptunesoftware