Rishu Kumar’s Post

Day 10 of the Geek Streak – 30 Days POTD Challenge is done! 🚀 Today's problem was a medium-level challenge involving a 2D matrix and palindromes. It was an interesting problem that required a good understanding of both matrix manipulation and palindrome logic. 💻 Every problem is a new opportunity to learn and grow, and I'm enjoying every step of this journey. Excited to keep pushing forward and see what the next challenge brings! 🌟 #geekstreak2024 #Day10 #CodingChallenge #LearningJourney #ProblemSolving

New Unity tutorial! Persistent Spatial Anchors with the PICO SDK. Persist your well planned out Spatial anchors into the local disk so you can come back for endless fun! Jokes aside, check out the video to learn more. https://lnkd.in/gsdTx_wa

✅ Day #6 of the Geek Streak – 30 Days POTD Challenge! Problem Of The Day - Topic: Rat in maze - Difficulty Level: Medium GeeksforGeeks #geekstreak2024 #codingchallenge #dynamicprogramming #recursion

💪Day 49-50/ #leetcodechallenge 🔄This method simulates Conway's Game of Life on a given 2D board. 🔄 It iterates through each cell, counting live neighbors and updating cell states accordingly. 🔢 Directions array defines neighboring cell offsets. 🔍 Counts live neighbors for each cell. 🔄 Applies game rules to update cell states. 🔄 Encodes next states with special values. 🔄 Updates board based on encoded states. 🔁 The process iterates until the desired number of generations. 🕒 The time complexity of the algorithm is O(n), where n is the total number of cells on the game board. ✅ Solved #dailycoding #challengeaccepted

🚀 Day 58: #gfg160 #geekstreak2024 #CodingChallenge ✨ Solved :-[Longest substring with distinct characters] . Approach:-Use a sliding window with two pointers (start and end). Track the last seen position of each character in a map. If a character repeats, move start to exclude the previous occurrence. Update the maximum length as end - start + 1. Process each character once for efficiency. 💻 Let’s keep the streak alive! #CodingJourney

💥 **Day #32/100 of the DrGViswanathan Challenge ✔️**. The Nim Game is a classic problem in combinatorial game theory. It asks whether the first player can win, given n piles of stones where players take turns removing any number of stones from one pile. The player who removes the last stone wins. Problem Understanding: You are given an integer n, representing the number of stones in a pile. The goal is to determine if the first player can win the game assuming both players play optimally. Simple Approach (Without XOR): Instead of using XOR (as is done in the optimal game theory solution), let's look at the game mechanics more simply. In the normal approach, there’s a straightforward pattern based on turns: If n % 4 == 0, the first player will lose, assuming both players play optimally. If n % 4 != 0, the first player can force a win. Why does this work? The game's outcome is directly related to the number of stones and the number of piles. The reason this works is because of the way players take turns and the number of stones they can remove. Here's the intuition: If n (the total number of stones) is divisible by 4, no matter what the first player does, the second player can always play in a way that the number of stones left after their turn is still divisible by 4. This forces the first player into a losing position. If n % 4 != 0, the first player can always reduce the number of stones in such a way that the second player faces a pile where the number of stones is a multiple of 4, ultimately leading to a win for the first player. Explanation: If n % 4 == 0: The second player wins with optimal moves. If n % 4 != 0: The first player can force a win. Time Complexity: O(1): This solution is constant time since it only requires a simple modulus operation. #Day32 #100DaysOfCode #DrGViswanathan #CodingChallenge #CodingJourney #TechLearning #NeverSettle #KeepGrowing #SoftwareEngineering #GrowthMindset #ProblemSolving

Day 13 of #100DaysOfLeetCode 🎮 Problem 3232: Find if Digit Game Can Be Won Today’s challenge was an exciting game theory problem between Alice and Bob. The goal was to find out if Alice could win by choosing either all single-digit numbers or all double-digit numbers from the array. Alice wins if the sum of her numbers is strictly greater than Bob's. 🧠 Solution Breakdown 1️⃣ Two Sums: We maintain two sums—one for single-digit numbers and one for double-digit numbers. 2️⃣ Quotient Check: For each number in the array, divide it by 10 to check if it’s a single or double-digit number. 3️⃣ Sum the Values: Based on the quotient, add the number to either the single or double-digit sum. 4️⃣ Comparison: At the end, compare the sums. If they’re equal, Alice cannot win. Otherwise, Alice wins. Alice’s game was all about finding the right balance 🎲💥. This was a rewarding challenge and a fun exercise. 🚀 Can't wait for tomorrow's problem! #GameTheory #ProblemSolving #CodingFun #HappyCoding

Cracked the elimination game! This code finds the winner in a circular game where players are skipped k-1 times before elimination. Used a queue to simulate the process and track the remaining player. Solution(GitHub) : https://lnkd.in/deXcRfJi #algorithms #problemsolving

Sizzle reel - 50% A.I. - 90 mins from concept to deployment. (Also, it was a lot of fun to make...) Tools: GPTo1, ElevenLabs, RunwayML Gen3 Alpha, Artlist, Descript

C++, no game engine, raylib. The math on this one was a bit tricky, after alot of collision detection attempts I eventually had to be satisfied with using radial distances and plane intersects for approximating the collisions and subsequent wave functions. Its slightly inaccurate and wont detect collision if you hit just the tip of one of the vertices of each square, but im satisfied with the current results.