Happy Fourth of July! 🇺🇸 Just as the Founding Fathers wrote the code for a new nation, developers across the USA continue to write the code shaping our future🎇. Enjoy your holiday, from backyard barbecues to late-night debugging sessions. May your day be filled with fireworks—both in the sky and in your IDEs! 🎆💻
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B.Tech Computer Science and Engineering Student | Core Team Member at FOSS United Dehradun | Passionate Programmer & Software Developer | Technology Enthusiast
📸 “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. 🤝☕📚
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🎄🎅 Merry Christmas, Techies! 🎅🎄 Hey, fellow code wranglers, keyboard warriors, and debugging ninjas! 'Tis the season to be coding – fa la la la la, la la la ERROR! Wishing you all a Merry Christmas filled with joy, laughter, and bug-free code (fingers crossed!). May your holiday be as stress-free as deploying on a Friday afternoon. 🚀 Here's hoping Santa brings you a sleigh full of RAM, a stocking stuffed with solid-state drives, and a partridge in a pear-shaped algorithm that runs in O(log n) time. May your Christmas tree be fully optimized with lights and decorations, just like your favorite recursive function. And if you find yourself stuck under the mistletoe, don't worry – it's just a merge conflict waiting to be resolved. Remember, in the spirit of Christmas, forgiveness is divine, but version control is a close second. Let bygones be rollbacks! May your eggnog be creamy, your cookies be error-free, and your holiday be filled with joyous moments that compile into wonderful memories. So, raise your glasses (of code), toast to a year of successful deploys, and may your Christmas be merry, bright, and free from runtime errors. Merry Christmas, techies! May your holidays be full of joy, laughter, and plenty of semicolons! 🎉🎁
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I finally finished this CodeForces problem (https://lnkd.in/guhPPwvi). My implementation was quite long, but fortunately only O(n) time complexity, the same time complexity as the top solution.
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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
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Front-end Web developer || Proficient in (JAVA ◉ Python) || DSA || Learner and Mentor || Class Representative || #365daysofcode
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
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Your go-to Backend Engineer | Writer | Pythonista | Polyglot | Startups | OSS | I create comics sometimes
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. :)
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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
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Reliance Foundation Undergraduate Scholar || CP Enthusiast || Codechef 2 ⭐ Rating-1486 || Codeforces Newbie (980) ||
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.
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Day 17 of #100DaysOfCodeToday, I learned about function closures, arrow functions, and anonymous functions in Dart. These advanced function concepts help make code more concise and flexible. Check out my progress on GitHub: https://lnkd.in/d6G4ZfYA #Flutter #CodeNewbie #Closures #ArrowFunctions #AnonymousFunctions #100DaysOfCodeChallenge
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Low-code solution specialist at Neptune Software - Low-Code App Development that digitizes your enterprise
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
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