LeetCode #98 – Validate Binary Search Tree | Python Implementation I implemented a recursive DFS approach with boundary tracking to validate BST properties. Core Insight: BST validation requires tracking allowable ranges, not just comparing immediate children. Each recursive call narrows the valid interval, ensuring global ordering throughout the entire subtree. Time: O(n) | Space: O(h) where h = tree height (recursion stack) #LeetCode #DataStructures #Python #BinarySearchTree #Recursion #DFS #CodingInterview #SoftwareEngineering
LeetCode 98: Validate Binary Search Tree with Python
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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
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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
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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
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When exploring a new dataset in Python, one simple command can save a lot of time: df.describe() It quickly shows key statistics for numerical columns — count, mean, standard deviation, min, max, and quartiles. Instead of manually checking distributions, this gives an instant snapshot of the data and often helps spot outliers or unusual values early in the analysis. Small habits like this make the data exploration phase much faster. #Python #DataAnalytics #MachineLearning #DataScience
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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
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A Python library that detects character encodings turned into the biggest open source fight of the month. The maintainer who kept it alive for twelve years used Claude to rewrite it and switched the license to MIT. The original author, gone since 2011, resurfaced to object. 244 comments. Most of them missed the point. I wrote about what actually went wrong — it's not what either side thinks.
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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
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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
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Day 12/100 – #100DaysOfCode 🚀 Solved LeetCode #217 – Contains Duplicate (Python). Today I worked on an array problem to determine whether any value appears at least twice in the array. Approach: 1) Sort the array first. 2) Traverse through the array from index 0 to n-2. 3) Compare each element with the next element. 4) If two adjacent elements are equal, a duplicate exists. 5) If no duplicates are found after traversal, return False. Time Complexity: O(n log n) due to sorting Space Complexity: O(1) Understanding how sorting can simplify duplicate detection in arrays 💪 #LeetCode #Python #DSA #Arrays #ProblemSolving #100DaysOfCode
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