LeetCode 1009: Binary Complement of Integer

Output: [9, 1, 1, 25, 81] 🧠 filter() + map() chained together! Step 1 — filter() keeps only odd numbers: [3, 1, 1, 5, 9] Step 2 — map() squares each one: [9, 1, 1, 25, 81] ✅ A cleaner, Pythonic way using list comprehension: result = [x**2 for x in nums if x % 2 != 0] Same result, more readable! 🎉 #python #pythonprogramming #coding #programming

🚀 Day 66 / 200 – Consistency is the real power 💡 Today’s challenge: **Implementing Pow(x, n)** using an optimized approach ⚡ Instead of the brute-force method, I used **Binary Exponentiation (Fast Power)** to reduce the time complexity from O(n) to O(log n). This approach efficiently handles both positive and negative powers, making the solution scalable and interview-ready. 🔹 Key Learnings: * Breaking problems into smaller subproblems improves efficiency * Handling edge cases (like negative exponents) is crucial * Optimization matters just as much as correctness 📊 Result: ✔️ 307 / 307 test cases passed ⚡ Runtime: 1 ms 🧠 Memory Efficiency: Top 2% Small improvements every day lead to big results over time. Staying consistent and sharpening problem-solving skills one challenge at a time 💪 #Day66 #100DaysOfCode #200DaysChallenge #LeetCode #Python #DSA #CodingJourney #ProblemSolving #Consistency

🚀 #100DaysOfCode – Day 52 📌 Problem: Sort Colors (LeetCode 75) 💡 Problem Idea: You are given an array containing only 0s, 1s, and 2s, representing three colors. The task is to sort the array in-place so that all 0s come first, then 1s, and then 2s. ⚡ Key Insight: Instead of using a sorting algorithm, we can solve this in one pass using three pointers. We maintain three regions: Left pointer → position for next 0 Right pointer → position for next 2 Current pointer (i) → traverses the array 🎯 Approach (Dutch National Flag Algorithm): If element is 0 → swap with left, move both left and i If element is 1 → just move i If element is 2 → swap with right, decrease right 📈 Complexity: Time Complexity: O(n) Space Complexity: O(1) (in-place sorting) ✅ Efficient because the array is processed only once. 💭 Learning: This problem is a great example of how pointer techniques can optimize sorting problems without using built-in sort functions. #DSA #LeetCode #Python #CodingJourney #ProblemSolving #Algorithms #100DaysOfCode #LinkedInLearning

🚀 Day 11 of Consistency | #75DaysLeetCodeChallenge 🧠 Today’s Problem : Two Sum II – Input Array Is Sorted (#167) 💡 Key Learning: This problem highlights the power of the two-pointer technique on a sorted array, helping reduce time complexity efficiently compared to brute force. ⚡ Approach: Use two pointers → left (l) at start & right (r) at end → If sum == target → return indices If sum < target → move l++ If sum > target → move r-- 🧠 Why this works: Takes advantage of sorted array Reduces complexity → O(n) No extra space required → O(1) 🔥 Result : ✔️ Runtime: 0 ms (Beats 100%) 📈 Mastering patterns like two pointers is key to cracking medium & hard problems. Consistency is compounding. Keep going. 💪 #Day11 #LeetCode #DSA #CodingJourney #100DaysOfCode #Python #TwoPointers #Consistency

Solving Subarray Sums with Ease in Python with No Prior Experience Discover how to crack the Subarray Sum Equals Zero III challenge with ease. Understand the concept of prefix sums and its application in solving subarray sums. Read the full article 👉 https://lnkd.in/d3Mg-Sjy #PythonForFresher #DataProcessing #AlgorithmsAndProgramming #ITinterviewPreparation #TechLab Code. Learn. Build. — TechLab by Neeraj

Day 6/100 – #100DaysOfCode Solved the problem “Remove Element”, focusing on in-place array manipulation. Problem Summary: Remove all occurrences of a given value from an array without using extra space, and return the count of remaining elements. Approach: Used a two-pointer technique to efficiently overwrite unwanted elements while iterating through the array. Complexity: • Time: O(n) • Space: O(1) Key Insight: Efficient problem solving is not always about removing data — sometimes it's about rearranging it smartly in-place. Consistent progress > perfection. On to Day 7 #100DaysOfCode #Day6 #DSA #Python #LeetCode #ProblemSolving #CodingJourney

LeetCode Problem 3070: "Count submatrices with top left element and sum less than k": You are given a 0-indexed integer matrix grid and an integer k. Return the number of submatrices that contain the top-left element of the grid, and have a sum less than or equal to k. Approach: use Prefix sum to calculate the total value of submatrix including the top left element upto a given cell. This approach is optimal and straightforward. #Python #DSA #LeetCode #ProblemSolving #CompetitiveProgramming #DailyCoding

Day 67 of 365 Days of code the infamous brainrot number Qn 1) Reorder list You are given the head of a singly linked-list. The positions of a linked list of length = 7 for example, can intially be represented as: [0, 1, 2, 3, 4, 5, 6] Reorder the nodes of the linked list to be in the following order: [0, 6, 1, 5, 2, 4, 3] Notice that in the general case for a list of length = n the nodes are reordered to be in the following order: [0, n-1, 1, n-2, 2, n-3, ...] You may not modify the values in the list's nodes, but instead you must reorder the nodes themselves. approach: use 5 pointers(scary :")) #365daysOfCode #NeetCode  #leetcode  #DSA  #python #LeetCode #ProblemSolving  #Algorithms #365dayschallenge

Day 3 / 100 🚀 Solved “Reverse Integer” — a problem that looks simple but actually tests how carefully you handle edge cases. At first, reversing digits feels straightforward. But the real challenge is handling 32-bit overflow without using extra space. 💡 Key learning: Before updating the result, always check if multiplying by 10 will exceed the allowed range. Core idea: rev * 10 + digit must stay within [-2³¹, 2³¹ - 1] Highlights: • Time Complexity: O(log n) • Space Complexity: O(1) • Correctly handles negative numbers and overflow This problem reinforced a critical habit: Don’t just make the logic work — validate boundary conditions. #100DaysOfCode #LeetCode #DSA #Python #ProblemSolving #CodingInterview

Topic 10/100 🚀 🧠 Topic 10 — Partial Functions What if you could pre-fill some arguments of a function and reuse it later? 🤯 👉 What is it? Partial functions allow you to fix a few arguments of a function and generate a new function with fewer parameters. 👉 Use Case: Used in real-world applications for: Pre-configuring functions Simplifying repeated function calls Building reusable utilities 👉 Why it’s Helpful: Reduces repetition Makes code cleaner Improves readability 💻 Example: from functools import partial def multiply(x, y): return x * y double = partial(multiply, 2) print(double(5)) # Output: 10 🧠 What’s happening here? We fixed the value of x = 2, creating a new function (double) that only needs one argument. ⚡ Pro Tip: Use partial functions when you find yourself passing the same arguments repeatedly. 💬 Follow this series for more Topics #Python #BackendDevelopment #100TopicOfCode #SoftwareEngineering #LearnInPublic