Python Control Flow and Decision Logic

𝗜 𝘀𝘁𝗼𝗽𝗽𝗲𝗱 𝘄𝗿𝗶𝘁𝗶𝗻𝗴 “𝗾𝘂𝗶𝗰𝗸 𝘀𝗰𝗿𝗶𝗽𝘁𝘀” 𝘁𝗵𝗲 𝗺𝗼𝗺𝗲𝗻𝘁 𝘁𝗵𝗲𝘆 𝘀𝘁𝗮𝗿𝘁𝗲𝗱 𝗹𝗶𝘃𝗶𝗻𝗴 𝗶𝗻 𝗽𝗿𝗼𝗱𝘂𝗰𝘁𝗶𝗼𝗻. 😅 If you’re doing Python automation, a few boring habits save you later: 𝗟𝗼𝗴 𝗹𝗶𝗸𝗲 𝘆𝗼𝘂’𝗹𝗹 𝗱𝗲𝗯𝘂𝗴 𝗶𝘁 𝗮𝘁 𝟮 𝗔𝗠: Timestamps, context, counts. 🧾 𝗔𝗱𝗱 𝗮 “𝗱𝗿𝘆-𝗿𝘂𝗻” 𝗺𝗼𝗱𝗲: Test before touching real data/files. 🧪 𝗠𝗮𝗸𝗲 𝗶𝘁 𝗶𝗱𝗲𝗺𝗽𝗼𝘁𝗲𝗻𝘁: Running twice shouldn’t double-create or double-send. 🔁 𝗩𝗮𝗹𝗶𝗱𝗮𝘁𝗲 𝗶𝗻𝗽𝘂𝘁𝘀 𝗲𝗮𝗿𝗹𝘆: Paths, env vars, credentials, expected formats. 🧱 𝗙𝗮𝗶𝗹 𝗹𝗼𝘂𝗱𝗹𝘆 + 𝗻𝗼𝘁𝗶𝗳𝘆: Don’t silently skip and “succeed”. 🧯 𝗔𝘂𝘁𝗼𝗺𝗮𝘁𝗶𝗼𝗻 𝗶𝘀𝗻’𝘁 𝗮𝗯𝗼𝘂𝘁 𝗰𝗹𝗲𝘃𝗲𝗿 𝗰𝗼𝗱𝗲. 𝗜𝘁’𝘀 𝗮𝗯𝗼𝘂𝘁 𝗽𝗿𝗲𝗱𝗶𝗰𝘁𝗮𝗯𝗹𝗲 𝗼𝘂𝘁𝗰𝗼𝗺𝗲𝘀. #python #automation #devtools #reliability #engineering

🚀 Day 3/100 — Enabling Repetition with Loops 🧠 “Efficiency emerges when logic learns to repeat itself.” Scalable systems rely on structured repetition to process continuous operations. Today, I explored how loops in Python enable efficient execution of repeated logic without redundancy. ⚙️ 🔧 Today’s focus areas: 🔁 for Loops — Iterating over structured sequences 🔄 while Loops — Executing logic based on conditions 🎯 Loop Control — Using break and continue 🧩 Iterative Thinking — Designing efficient execution flows 🎯 The objective was to build scalable execution patterns that reduce redundancy and improve efficiency. ✅ Day 3 complete: Iterative execution enabled. ▶️ Day 4: Structuring and organizing data using collections. Step by step. The system evolves. 🏗️ #100DaysOfCode #Python #DeveloperJourney #BackendDevelopment #SoftwareEngineering

Spent 30 minutes automating 30 minutes of monthly work. Bank statement PDF goes in, GPT categorizes every transaction, Excel workbook comes out with a sheet per category. ~100 lines of Python. $0.02 per run in API costs. Not everything needs to be a big project. #Python #Automation #PersonalFinance

Weekly Challenge 6: Two Pointers Reversing an array the "Senior" way (Two Pointers) Reversing a list in Python is easy. Just type `list.reverse()` or `list[::-1]` and you are done, right? But what if you are in a technical interview and you are asked to do it from scratch, using **zero extra memory**? For Week 6 of my coding challenge, I implemented the **Two Pointers Technique** to reverse an array *in-place*. **The Logic:** You place one pointer at the start (index 0) and one at the end. You swap their values, and then step both pointers toward the middle until they meet. **Why it matters:** By doing this *in-place*, our Space Complexity is $O(1)$ (Constant Space). We don't need to duplicate the array in our RAM. If you are processing massive datasets, this memory optimization is critical! Check out the console trace below to see how the pointers move step-by-step. Code on my GitHub: https://lnkd.in/eyZ4KqiN #Python #Algorithms #TwoPointers #CodingChallenge #Optimization #DataStructures

Small efficiency gains compound quickly in production. Inspired by #Blender3D Pie Menu, I built a custom radial tool in #Houdini using #Python. It saves seconds by: • Creating a referenced Null and Object Merge • Auto-organizing networks • Rendering a flipbook in one click The real value shows up in daily use. Used dozens of times daily, those seconds turn into hours and remove workflow friction. My technical approach: Identify repetition → remove steps → multiply efficiency. #TechnicalArt #GameDevelopment #Pipeline #Automation #ToolsDevelopment

🚀 Day 10/30 | LeetCode Problem: Reverse Linked List (206) Problem: Given the head of a singly linked list, reverse the list and return the new head. 💡 Approach (Iterative – Three Pointers) To reverse a linked list, we carefully change the direction of pointers. We use three variables: prev → previous node current → current node next_node → temporarily stores next node Steps: Store next node Reverse the current node’s pointer Move prev and current one step forward Repeat until the list ends Finally, prev becomes the new head. ⏱ Complexity Time Complexity: O(n) Space Complexity: O(1) 🧠 Python Code class Solution: def reverseList(self, head): prev = None current = head while current: nxt = current.next current.next = prev prev = current current = nxt return prev 📌 Example Input: [1,2,3,4,5] Output: [5,4,3,2,1] 🎯 Key Takeaway Linked List problems are all about pointer manipulation. Understanding pointer flow is more important than memorizing code. ✅ Accepted 🔖 Hashtags #LeetCode #30DaysOfLeetCode #Day10 #Python #LinkedList #DataStructures #Algorithms #ProblemSolving #CodingJourney #SoftwareEngineering

LeetCode 20 | Valid Parentheses 🔥 🔹 Data Structure: Stack 🔹 Concept: LIFO matching for bracket validation 🔹 Time Complexity: O(n) 🔹 Space Complexity: O(n) Push opening brackets, match closing brackets carefully 💡 Stack problems strengthen logical flow control and pattern recognition. #LeetCode #DSA #Stack #Python #CodingPractice #ProblemSolving

Just practiced Pandas and data cleaning hits different when you're working with real messy data. Covered data types, type conversion, handling missing values, replacing inconsistent entries, and using category dtype to save memory — FuelType column went from 11488 bytes to 1460 bytes just by changing the dtype. Notebook here 👉 https://lnkd.in/d3djYPvp #Python #Pandas #DataAnalysis #LearningInPublic

🚀 Day 44 of #100DaysOfCode 📌 Problem: 762 – Prime Number of Set Bits in Binary Representation Today I solved an interesting bit manipulation problem on LeetCode that combines binary representation with prime number logic. 🔎 Problem Summary: Given two integers left and right, count how many numbers in that range have a prime number of set bits (1s) in their binary form. 💡 Key Insight: The maximum number of set bits for numbers ≤ 10⁶ is 20. So we only need to check prime numbers up to 20: {2, 3, 5, 7, 11, 13, 17, 19} For each number, count the set bits using: Python Copy code num.bit_count() ✅ Python Solution: Python Copy code class Solution: def countPrimeSetBits(self, left: int, right: int) -> int: primes = {2, 3, 5, 7, 11, 13, 17, 19} count = 0 for num in range(left, right + 1): if num.bit_count() in primes: count += 1 return count ⏱ Time Complexity: O(n) 🧠 Concepts Used: Bit Manipulation | Prime Numbers | Set Every day I’m getting more comfortable with binary operations and optimization techniques. #LeetCode #Day42 #CodingJourney #Python #ProblemSolving #BitManipulation #100DaysOfCode

Today’s practice was focused on String and List Manipulation using Slicing. This session helped me understand how powerful slicing is in Python and how it can simplify many problems without using loops. Programs practiced today include: 🔹 Reverse a given string using slicing 🔹 Print every alternate character from a string 🔹 Extract the middle three characters from a string 🔹 Reverse a list of integers using slicing (without loops or built-in reverse functions) 🔹 Remove the first and last characters from a string using slicing Key learnings from today’s session: • Understanding slicing syntax: start : stop : step • Using negative indexing for reversing sequences • How [::-1] works internally • Extracting specific portions of strings efficiently • Writing clean and shorter code without unnecessary loops • Strengthening fundamentals in string and list operations Today’s practice showed me how Python provides elegant solutions for common problems. Instead of writing long logic with loops, slicing makes the code simpler and more readable. Step by step, improving both logic and code efficiency. Grateful for the continuous guidance and support from the #10kcoders team. #Day32 #Python #StringManipulation #ListOperations #Slicing #LogicBuilding #ProgrammingJourney #Consistency