"Majority Element solved with sorting in 7 ms"

🎯 Day 81 of #100DaysOfCode 🎯 🔹 Problem: Reverse Only Letters – LeetCode ✨ Approach: Used a two-pointer strategy to reverse only the alphabetic characters while keeping all non-letter characters in their original positions. By moving pointers inward and swapping only when both characters are letters, the solution stays efficient and elegant! 🔄✨ 📊 Complexity Analysis: ⏱ Time Complexity: O(n) — each character visited at most once 💾 Space Complexity: O(n) — due to character array construction ✅ Runtime: 0 ms (Beats 100% 🚀) ✅ Memory: 42.96 MB 🔑 Key Insight: Sometimes, all you need is smart pointer movement — not everything needs to be reversed, only the right pieces. #LeetCode #100DaysOfCode #DSA #TwoPointers #StringManipulation #ProblemSolving #CleanCode #AlgorithmDesign #CodingChallenge

🕵️♀️ Day 67 of #100DaysOfCode 🕵️♀️ 🔹 Problem: The Two Sneaky Numbers of Digitville – LeetCode ✨ Approach: Designed a simple nested loop comparison to uncover the two “sneaky” duplicate numbers hiding in the array. By checking every pair and exiting early once both are found, the solution ensures clarity and efficiency with minimal extra logic! ⚡ 📊 Complexity Analysis: Time Complexity: O(n²) — double traversal for pair comparison Space Complexity: O(1) — constant space for result storage ✅ Runtime: 1 ms (Beats 100.00%) ✅ Memory: 45.58 MB 🔑 Key Insight: Sometimes, the best detective work in code isn’t about fancy tricks — it’s about precision, early exits, and keeping things simple yet sharp. 🕶️ #LeetCode #100DaysOfCode #ProblemSolving #DSA #AlgorithmDesign #Array #LogicBuilding #CodingDaily #CodeJourney #Efficiency #ProgrammingChallenge

Cracking the "Container With Most Water" Problem in O(n)! The "Container With Most Water" (LeetCode 11) is a fantastic problem that demonstrates the power and elegance of the Two-Pointer technique. While a brute-force approach checks every pair of lines in O(n^2), the optimal solution is a O(n) algorithm. The Two-Pointer Strategy: The area of the container is defined by: Area = min(height_L, height_R) x (index_R - index_L) Maximize Width First: Start the left pointer (L) at the beginning and the right pointer (R) at the end. This guarantees the maximum possible width. Iterative Optimization: We move the pointers inward, always aiming to potentially find a taller constraining line. The Key Insight: To find a larger area, we must increase the minimum height, because the width is guaranteed to shrink. Therefore, we always move the pointer associated with the shorter line. This simple rule ensures we only check meaningful configurations, leading to a linear time complexity. #Algorithms #DataStructures #CodingInterview #TwoPointers #LinearTime #SoftwareDevelopment #100daysofcode #leetcode

🔥 Day 69 — “Strings, Structure & Strategy.” 🧩💡 Theme: Strings × Dynamic Programming Today’s challenges tested precision in pairing logic with layers — where every substring, sequence, and selection demanded careful calculation. Dynamic Programming once again became the art of 'balancing time, value, and possibility.' 💻 What I Solved: 🔹 Count Repetitions — (Hard) ✅ 150/150 — finding hidden rhythm within repeated patterns. 🔹 Optimized Washing — (Hard) ✅ 150/150 — time and efficiency woven through clean DP logic. 🔹 Removal of String — (Hard) ⏳ 38/150 — untangling overlaps one substring at a time. 🔹 Maximum Number of Events That Can Be Attended II — (Hard) ⏳ 113/150 — strategizing choices with weighted intervals. 🎯 Key Learnings: * Every DP decision carries a cost — 'choose wisely, gain maximally.' * In Strings, removing one character can change the entire state. * Optimization isn’t speed alone — it’s clarity in constraint. 📊 Leaderboard Update: 🏆 Points: 31,183+ 🏅 Rank: 122 💬 From the coder’s desk: > “Where strings connect and choices collide, > logic finds its rhythm — and patterns come alive.” 🎶💻 Join me on Unstop 👉 https://lnkd.in/gRFmWUyA #100DaysOfCode #Day69 #CodingJourney #Unstop

⚡ Day 9 of #100 Days Of DSA Challenge Today’s focus: Remove Outermost Parentheses | Reverse Polish Notation 🔹 LeetCode 1021 – Remove Outermost Parentheses Keep a counter for open and close parentheses. Add characters to the result only when depth > 0. Avoid using a stack by tracking balance of brackets directly. Time: O(n) | Space: O(1) 🔹 LeetCode 150 – Evaluate Reverse Polish Notation (Postfix Expression) Use a stack to evaluate expressions. For each token: If number → push to stack If operator → pop two numbers, perform operation, push result back Final stack top gives the answer. Time: O(n) | Space: O(n) 💡 Key takeaway: Stack logic can often be optimized or even replaced with simple counters and pointer techniques when operations follow predictable patterns. #DSA #Stack #Queue #LeetCode #CodingChallenge #100DaysOfCode #100DaysOfDSAChallenge

Day 52 of #100DaysOfCode 🧩 Problem: 3321. Find X-Sum of All K-Long Subarrays II Difficulty: Hard This version builds on the easier “Find X-Sum” problem, but the constraints are massive — meaning brute force dies instantly. The goal is to maintain the sum of top x most frequent elements for every subarray of size k, with efficient updates as the window slides. 💡 What I learned: Balancing efficiency and correctness in sliding window problems is tricky. Frequency tracking with ordered structures (like multiset or heap) keeps updates O(log n). Rebalancing the “top x” elements correctly after each update is crucial to avoid stale counts. 🧠 Key takeaway: Optimization isn’t about shaving milliseconds; it’s about understanding how to reuse work instead of starting over every time. Every “Hard” problem eventually boils down to patience, debugging, and a little bit of emotional damage. #LeetCode #C++ #DSA #ProblemSolving #100DaysOfCode #CodingJourney

🔢 Day 69 of #100DaysOfCode 🔢 🔹 Problem: Maximum Count of Positive and Negative Integers – LeetCode ✨ Approach: A simple yet powerful one-pass solution! Iterated through the array, counting positives and negatives separately — and returned whichever count was greater. Clean, direct, and efficient. ⚡ 📊 Complexity Analysis: Time Complexity: O(n) — single traversal of the array Space Complexity: O(1) — no extra data structures used ✅ Runtime: 1 ms (Beats 12.27%) ✅ Memory: 46.89 MB 🔑 Key Insight: Sometimes, simplicity wins — clarity in logic often outperforms complex tricks. Counting right leads to coding right! 💡 #LeetCode #100DaysOfCode #ProblemSolving #DSA #Array #AlgorithmDesign #LogicBuilding #CodeJourney #ProgrammingChallenge #CodingDaily #Efficiency

🧠 Day 52 of #100DaysOfCode Today I built a Magic Square Generator in C! ✨ Given an odd number n\geq 3, the program constructs an n\times n grid where the sum of every row, column, and diagonal equals the magic constant. Used the Siamese method to place numbers smartly—moving up and right, wrapping around, and stepping down when blocked. Loved the clean logic and the satisfying symmetry in the final output. Perfect for showcasing matrix manipulation and modular arithmetic! 🔢 Magic Constant: n(n^2+1)/2 📐 Output is neatly formatted for visual clarity. #CProgramming #CodeSymmetry #100DaysOfCode #Day52 #CodingChallenge #LearnByBuilding

Day 52 of the #90DaysWithDSA challenge is complete! Today continued our tree traversal journey with a problem that builds directly on yesterday's height calculation concepts. Today's Problem: Balanced Binary Tree (LeetCode 110 - Easy) The challenge: Determine if a binary tree is height-balanced. A height-balanced binary tree is defined as one where the left and right subtrees of every node differ in height by no more than 1. This problem is another perfect application of post-order traversal with early termination: The Approach: Use DFS to calculate heights of left and right subtrees At each node, check if the height difference between left and right subtrees is ≤ 1 If any node violates this condition, propagate a failure signal upward Return the height of the current node for parent calculations, or a special value (-1) to indicate imbalance This optimized approach runs in O(n) time and avoids redundant calculations by checking balance during height computation. Key Takeaway: Many tree validation problems can be solved efficiently by combining the computation (like height) with the validation check during traversal. The early termination optimization is crucial for efficiency - once we detect imbalance, we can stop further computation. 52 days of consistent coding. The recursive patterns for tree problems are becoming second nature! 🚀 Let's keep balancing our tree knowledge together! 👉 Want to master tree validation and optimization techniques? JOIN THE QUEST! Comment "Balancing with you!" below and share what tree problem you're solving today. 👉 Repost this ♻️ to help other developers discover this challenge and improve their recursive thinking skills. What's your favorite tree validation problem? #Day52 #BinaryTree #BalancedTree #DFS #Recursion #CodingInterview #Programming #SoftwareEngineering #Tech #LearnInPublic #Developer #LeetCode #ProblemSolving

🌟 Day 71 of #100DaysOfCode 🌟 💡 Problem: Matrix Diagonal Sum – LeetCode ✨ Approach: Traversed both primary and secondary diagonals in a single loop — adding elements smartly while avoiding double-counting the center in odd-sized matrices. Simple logic, elegant optimization! ⚡ 📊 Complexity Analysis: Time Complexity: O(n) — single traversal through the matrix rows Space Complexity: O(1) — constant auxiliary space ✅ Runtime: 0 ms (Beats 100%🔥) ✅ Memory: 46.26 MB 🔑 Key Insight: In coding, clarity wins. The shortest paths often come from the clearest logic. 🎯 #LeetCode #100DaysOfCode #ProblemSolving #DSA #AlgorithmDesign #ProgrammingChallenge #CodeJourney #LogicBuilding #Efficiency #DeveloperGrowth #CodingDaily