LeetCode Challenge: Reverse Bits in Python

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

Day 24/100 – #100DaysOfCode 🚀 Solved LeetCode #747 – Largest Number At Least Twice of Others (Dominant Index) (Python). Today I practiced array traversal and comparison logic to find the dominant index in the array. Approach: 1) Find the largest element in the array and its index. 2) Traverse through the array. 3) For every other element, check if the largest is at least twice of it. 4) If any element violates this condition, return -1. 5) If all conditions are satisfied, return the index of the largest element. Time Complexity: O(n) Space Complexity: O(1) Learning how simple comparisons can solve array problems efficiently 💪 #LeetCode #Python #DSA #Arrays #ProblemSolving #100DaysOfCode

Day 75 Back to revisiting tree fundamentals. #Day75 🧩 100. Same Tree Solved it again today. What made it easier this time: • Convert the tree structure into string representation • Compare the left and right serialized strings • Python makes this approach very straightforward This problem is simple, but it reinforces an important idea: Sometimes a representation trick can simplify the comparison logic. Small revisions like this keep the patterns fresh. #LeetCode #DSA #Python #BinaryTree #Recursion #LearningInPublic #Consistency

Day 20/100 – #100DaysOfCode 🚀 Solved LeetCode #448 – Find All Numbers Disappeared in an Array (Python). Today I worked on an array problem to find all the numbers in the range [1, n] that are missing from the given array. Approach: 1) Convert the array into a set for quick lookup. 2) Traverse numbers from 1 to n. 3) Check if each number exists in the set. 4) If not present, add it to the result list. 5) Return the final list of missing numbers. Time Complexity: O(n) Space Complexity: O(n) Learning how sets help in fast lookup and simplify problems 💪 #LeetCode #Python #DSA #Arrays #HashSet #ProblemSolving #100DaysOfCode

🚀 Wrote a Hugging Face BPE Tokenizer visualizer library in python! Tokenization is the heart of modern LLMs, but looking at raw lists of integer IDs isn't exactly intuitive. That's why I created a lightweight Python library to help you visually inspect how Byte-Pair Encoding tokenizer break down your text. ✅ Easy to use ✅ Integrates seamlessly with Hugging Face tokenizers ✅ Perfect for debugging, model building, or just learning how BPE works under the hood! 📦 Try it out: pip install hf-tokenizer-visualizer 🔗 Github and Package links are in the comments #huggingface #tokenizer #deeplearning #python

Day 19 of my 20 Day Linkedin Challenge I recently learned about loops in Python. Here’s the simplest way to understand them: A loop tells the computer to repeat something. Instead of writing the same instruction multiple times you write it once and let the loop handle the repetition. For example: If you want to print numbers from 1 to 10, You don’t write 10 separate lines. You use a loop. This matters because computers are great at repetition. Loops allow you to: - save time - reduce errors - handle large tasks efficiently It’s one of those concepts that seems small but it’s actually very powerful. #AfricaAgility #ArtificialIntelligence #Python #MachineLearning #GIT20DayChallenge

Day 19/100 – #100DaysOfCode 🚀 Solved LeetCode #414 – Third Maximum Number (Python). Today I worked on an array problem to find the third distinct maximum number in the array. If it does not exist, return the maximum number. Approach: 1) Remove duplicates by converting the array into a set. 2) Convert it back to a list. 3) Sort the list in ascending order. 4) If the length is ≥ 3, return the third maximum element. 5) Otherwise, return the maximum element. Time Complexity: O(n log n) Space Complexity: O(n) Understanding how sets help remove duplicates efficiently 💪 #LeetCode #Python #DSA #Arrays #ProblemSolving #100DaysOfCode

🚀 Solved the First Bad Version problem using Binary Search in Python. Instead of checking every version sequentially, Binary Search helps reduce the number of checks from O(n) to O(log n) — making the solution highly efficient. Key idea: • If the middle version is bad → search the left side • If the middle version is good → search the right side This pattern is widely used in problems involving first occurrence, boundaries, and optimization. Always fascinating to see how a simple algorithm like Binary Search can solve real-world style problems so efficiently. #Python #BinarySearch #Algorithms #LeetCode #CodingJourney #SoftwareDevelopment

Poll Insight: Which data type does not allow duplicate values? The correct answer is Set ✅ A Set stores only unique elements, meaning duplicate values are automatically removed. That’s why sets are useful when you want to keep only distinct values in Python. 👉 Example use cases include removing duplicates from a list or storing unique items. #Python #LearnPython #CodingQuiz #ProgrammingBasics

LeetCode #235 – Lowest Common Ancestor of a Binary Search Tree | Python Implementation I implemented an iterative approach leveraging the BST property to find the LCA without recursion. Core Insight: BST ordering guarantees the LCA is where paths to p and q diverge. The iterative solution avoids recursion overhead by directly following the BST property until the split condition is met. Time: O(log h) where h = tree height | Space: O(1) #LeetCode #DataStructures #Python #BinarySearchTree #LCA #CodingInterview #SoftwareEngineering