🚀 Day 16/30 – List Comprehension & Generators Today I learned how to write cleaner and more efficient Python code. 📌 List Comprehension A shorter way to create lists using a single line. Python Copy code squares = [x*x for x in range(1,6)] print(squares) 📌 Generators Use yield to generate values one by one instead of storing everything in memory. 💡 Key Takeaway: List comprehension improves code readability, while generators improve memory efficiency. Day 16 complete ✅ #Python #30DaysChallenge #LearningInPublic #ProgrammingJourney #Consistency Aditya Chaturvedi JECRC University

Day 1 was about probability foundations: sample space, events, union, intersection, complement, and the basic axioms. Main lesson: if the sample space is not defined clearly, the rest of the solution cannot be trusted. Also, an outcome is not the same as an event. That distinction matters. Wrote small Python scripts as well. Good reminder that coding exposes weak understanding quickly. The goal is not to move fast. The goal is to build correctly. #Probability #Python #Statistics

🐍 Day 22 of My 30-Day Python Learning Challenge Today I improved my Log File Analyzer by allowing user input (file name) instead of hardcoding. 📌 Code: filename = input("Enter file name: ") with open(filename, "r") as file: content = file.read().lower() print(content[:100]) # preview first 100 characters 📌 Why this matters? • Makes the program flexible • Works with any file • Closer to real-world usage 📊 Quick Question What happens if the user enters a wrong file name? A) Program crashes B) Empty output C) None D) Skips execution Answer tomorrow 👇 #Python #MiniProject #UserInput #LearningInPublic #SoftwareDeveloper

Learn in Public — Day 14 Today I solved the Subset Array Problem using three different approaches in Python. Problem: Check whether array b is a subset of array a. Approaches I implemented: 1️⃣ Brute Force Check each element of b in a Time Complexity: O(m × n) 2️⃣ Sorting + Two Pointers Sort both arrays and compare Time Complexity: O(m log m + n log n) 3️⃣ Hash Set (Optimal) Convert array a into a set Check membership in O(1) Time Complexity: O(m + n) Key Learning: Whenever fast lookup is needed, hashing is often the best approach. #LearnInPublic #Python #DSA #CodingJourney

Excited to Share My New Python Library: 𝗽𝘆𝗶𝗺𝗴𝟮𝗮𝘀𝗰𝗶𝗶 I just released 𝘃𝟬.𝟱 of pyimg2ascii, a Python library I built to convert images into ASCII art. I created it to experiment with image processing and make coding a bit more fun and creative.  1. 𝗽𝗶𝗽 𝗶𝗻𝘀𝘁𝗮𝗹𝗹 𝗽𝘆𝗶𝗺𝗴𝟮𝗮𝘀𝗰𝗶𝗶 You can try it yourself and see your images come alive as ASCII art! I’d love to hear your feedback and see what you create. Check it out here: https://lnkd.in/g86iPSU4 #Python

🎮 Rock Paper Scissors – Python Another step forward in my Python learning journey. This time I built a Rock Paper Scissors game in Python where the user plays against the computer in the terminal. The game keeps track of the score, validates user input, and allows the player to continue playing multiple rounds. I also focused on making the interaction smoother by guiding the user whenever an invalid input is entered. 🧠 Concepts practiced in this project: • Python lists • Conditional logic • While loops • Random module • Input validation and user interaction 🎮 Try the game: 🔗 Live Demo (Replit): https://lnkd.in/gwJ6C9tP 💻 GitHub Repository: https://lnkd.in/gjkJwvEX Always learning, one small program at a time. 💻 #Python #CodingJourney #LearningToCode #BeginnerProgrammer #100DaysOfCode

💫A simple but powerful Python logic: Palindrome Check🌟 💥A string is a palindrome if it reads the same forward and backward. Examples: radar level madam Python makes this extremely simple: word = "radar" if word == word[::-1]: print("Palindrome") else: print("Not a Palindrome") The trick here is: [::-1] reverses the string. So we simply compare: original string == reversed string This concept is commonly used in: Coding interviews String manipulation problems Algorithm practice Sometimes the smartest solution is also the simplest. What was the first string problem you solved in Python? #Python #Programming #Coding #Learn ToCode #PythonLearning #Developer #Software Engineering

Learn in Public — Day 11 Today I explored multiple ways to compute the Greatest Common Divisor (GCD) in Python. Implemented several approaches: • Brute force approach using iteration • Recursive subtraction-based GCD • Optimized recursive version • Euclidean algorithm using modulo • Python's built-in math.gcd() function Key takeaway: The Euclidean Algorithm is significantly more efficient than the naive approach because it reduces the problem size quickly using modulo operations. This exercise helped me understand how the same problem can be solved with different algorithmic strategies — each with different time complexities and performance trade-offs. Consistently learning and improving every day. #LearnInPublic #Python #Algorithms #DataStructures #CodingJourney #SoftwareEngineering #ProblemSolving

My team and I recently built a Maze Solver GUI application using Python. The program lets users generate random mazes and solve them using both Breadth-First Search (BFS) and Depth-First Search (DFS), with a real-time visualization of how each algorithm explores the maze. It also tracks useful performance metrics such as path length, number of visited nodes, and runtime, making it easy to compare how BFS and DFS behave. In our implementation, BFS ensures the shortest path to the goal, while DFS is used to explore different possible routes through the maze. I collaborated on this project with Saif Al Sabashneh and Hai Sieu Cao, and it was a great opportunity to apply algorithm concepts while building an interactive tool. #CSUF #ComputerScience #Python #Algorithms #SoftwareEngineering

Tried solving a sliding window problem today in Python Found the maximum average subarray of size k without recalculating everything again and again — just shifting the window and updating the sum. Simple logic, but really powerful for optimization! nums = [1,12,-5,-6,50,3] n = len(nums) k = 4 max_avrg = float("-inf") window = sum(nums[:k]) max_avrg = window/k for j in range (k,n): window = window - nums[j-k] + nums[j] if max_avrg < window/k: max_avrg = window/k print(max_avrg) Learning DSA step by step #Python #DSA #SlidingWindow