Mehul Gupta’s Post

#DSAChallenge | #LeetCodeProgress 📌 Problem: 54.Spiral Matrix 📍 Platform: LeetCode Over the past few days, I’ve been exploring some classic yet challenging LeetCode problems — the kind that really test your understanding of logic, edge cases, and clean code design. 🧠 Problem Insight You're given a matrix, and the goal is to return all elements in spiral order — starting from the top-left corner and moving right → down → left → up, repeating this pattern until all elements are covered. ⚙️ Logic Breakdown (Step-by-Step): 1️⃣ Set boundaries: top, bottom, left, right 2️⃣ Traverse in order: → Move left to right (top row) → top++ → Move top to bottom (right column) → right-- → Move right to left (bottom row) → bottom-- → Move bottom to top (left column) → left++ 3️⃣ Continue while top <= bottom and left <= right 🧩 Key Learning Points: ✅ Strengthened my understanding of matrix boundaries and directions. ✅ Improved my ability to handle edge cases cleanly. ✅ Learned how to manage loop conditions efficiently. ⏱️ Time & Space Complexity Time Complexity: O(m × n) → Every element is visited once Space Complexity: O(1) → If output list is excluded 💡 Takeaway: In DSA, consistency beats intensity. Even small daily challenges — when done consistently — compound into stronger problem-solving intuition. 🔥 Next Up: Moving on to Rotate Image & Set Matrix Zeroes to deepen my matrix manipulation skills! #DSA #LeetCode #ProblemSolving #DataStructures #Algorithms #100DaysOfDSA #Matrix #SpiralMatrix #CodeNewbie #GrowthMindset #DailyLearning #NeverGiveUp #KeepCoding #LearningNeverStops

🚀 LeetCode Daily Challenge Complete! Just solved "Final Value of Variable After Performing Operations" - a simple yet elegant problem that teaches the value of finding shortcuts in pattern matching! 💡 Solution Approach: ✅ Iterate through all operation strings ✅ Check middle character (index 1) ✅ If '-': decrement, if '+': increment ✅ Return final value The key insight: Instead of comparing full strings ("--X", "X--", "++X", "X++"), we can observe that ALL operations have the operator in the middle position! By checking just op[1], we instantly know whether to increment or decrement. Pattern recognition: "--X" and "X--" both have '-' at index 1 → decrement "++X" and "X++" both have '+' at index 1 → increment This single-character check is more efficient than string comparisons! Time Complexity: O(n) | Space Complexity: O(1) #LeetCode #StringParsing #CPlusPlus #Algorithms #SoftwareEngineering #ProblemSolving #CleanCode #PatternRecognition

🚀 Just Completed a Deep Dive into Claude Code Here are my biggest takeaways: 💡 Smart Context Management Claude uses three-layer CLAUDE.md files (machine / project / local) to keep the right context at the right time — nothing more, nothing less. ⚙️ Plan Mode vs Thinking Mode Plan Mode → breadth: multi-step tasks across the codebase Thinking Mode → depth: complex logic and debugging Knowing when to use each is a real game-changer. 🔍 Hooks for Code Quality Automatic TypeScript or Python checks and duplicate code detection run after every edit — it’s like having a senior engineer reviewing changes in real time. 🧩 Extensibility MCP servers (like Playwright) and GitHub integration (e.g. @mentions in PRs) open up powerful automation workflows I’m still exploring. #AI #DeveloperTools #SoftwareEngineering #ClaudeCode #Productivity

LeetCode Daily Challenge 🌟 🔎 𝐏𝐫𝐨𝐛𝐥𝐞𝐦: 𝐂𝐡𝐞𝐜𝐤 𝐈𝐟 𝐃𝐢𝐠𝐢𝐭𝐬 𝐀𝐫𝐞 𝐄𝐪𝐮𝐚𝐥 𝐢𝐧 𝐒𝐭𝐫𝐢𝐧𝐠 𝐀𝐟𝐭𝐞𝐫 𝐎𝐩𝐞𝐫𝐚𝐭𝐢𝐨𝐧𝐬 𝐈 💡 Approach: I recently solved this LeetCode problem where we iteratively reduce the size of a string of digits by calculating the sum of each pair of consecutive digits modulo 10. We repeat this process until the string has exactly two digits and then compare those digits. Example: Input: s = "3902" Output: true Explanation: The digits reduce as follows: First iteration: [3, 9, 0, 2] → [2, 9] Second iteration: [2, 9] → [1] The final two digits are equal, so the result is true. Performance Considerations: • When handling larger inputs, such as those with constraints 10510^5105 (like a string of 100,000 digits), time complexity becomes crucial. In such cases: • Optimization: A direct approach where we repeatedly calculate and reduce the string would not be efficient for larger inputs. • Optimized Approach: We could use divide and conquer techniques or reduce the operations in a way that limits repeated calculations, such as reducing the string in batches or using modular arithmetic to streamline the process. 💬 For the LinkedIn Family: Let's challenge ourselves to think beyond brute force! When the constraints become larger (such as 10^5), optimization becomes key. How would you approach this problem with a more efficient solution? Share your thoughts in the comments! #LeetCode #CodingChallenge #ProblemSolving #Algorithm #DailyCoding #TechCommunity

🔥🔥 Day #51 of #100DaysOfDP When “#simple” problems teach you the most! Recently tackled a binary string problem on #Codeforces that looked deceptively straightforward: turn any binary string into its non-decreasing version with the least operations. Easy, right? 🚦 Here’s what actually happened… 🔸 Phase 1: The Indexing Mistake I dove in, started coding, and quickly realized my approach was totally off. I was using zero-based indexing, but the problem expected one-based! Small misunderstanding, huge confusion. 🔸 Phase 2: Brute Force Struggles Once I fixed indexing, I coded up the most straightforward brute-force recursive solution I could think of. It worked on sample tests… but promptly blew up with Time Limit Exceeded (TLE) on hidden cases! 🔸 Phase 3: Memoization (Hello, Memory Errors) Determined to make it work, I threw memoization at the recursion thinking, “This has to do it!” Instead, my program ate up all available memory almost instantly. Bye-bye, hope (and hello, MLE errors)! 🔸 Phase 4: The “Aha!” Moment After revisiting the problem, I realized the key wasn’t simulating every combination or caching every state. The real trick? Count only the segment transitions—specifically, the points where the string moves from 1 to 0, managing state with a simple flag. Elegant, efficient, and passed in a snap. ✨✨Key Takeaway: Sometimes the “optimal” approach means not simulating every possibility but understanding the underlying pattern and optimizing for it. Debugging and running into both TLE and MLE wasn’t a setback, it was part of the journey to deeper insight! #programming #learning #Codeforces #problemsolving #growthmindset

Title: 🔍 Day 4 of #LearningOfCode — Finding Duplicates in Code and in Mindset 💡 Body: Today’s challenge: LeetCode 287 – Find the Duplicate Number 🧠 At first, it looked like a simple array problem… but it taught me a deep lesson about optimization and focus. Here’s how I approached it 👇 ✅ Tried sorting — worked but costly (O(n log n)) ✅ Tried HashSet — faster, but used extra space (O(n)) 💬 What I learned: Finding “duplicates” in data is like finding distractions in life — once you identify them, the path becomes clearer and faster. Each optimized solution brings me closer to becoming a more efficient engineer — not just in code, but in thought. 💪 #LeetCode #ProblemSolving #DSA #LearnInPublic #SoftwareEngineering #CareerGrowth #100DaysOfCode #Motivation #CodingJourney

🚀 Day-77 of #100DaysOfCodeChallenge Today’s problem: Delete Nodes From Linked List Present in Array (LeetCode 3217) 💻 Working with linked lists always reminds me that sometimes in code — and in life — you have to remove what’s not needed to keep things moving efficiently ✂️ This challenge was about filtering out nodes whose values existed in a given array — a perfect mix of logic, pointer manipulation, and careful iteration 🔁 ✨ What I learned today: Efficiently handling deletions in a linked list without breaking the chain Using hash sets to make lookups faster on LeetCode. The importance of clean and minimal traversal logic Some problems aren’t just about syntax — they’re about clarity of thought 🧠 Another green check ✅ Another step closer to consistency 💪 #100DaysOfCode #LeetCode #ProblemSolving #CodingJourney #DataStructures #Algorithms #LinkedList #DeveloperLife #CodeEveryday #JavaDeveloper #TechJourney

#28 🚀 Problem Solved: Broken Calculator (LeetCode) 💻 Today I tackled the "Broken Calculator" problem — an interesting challenge that sharpens your greedy algorithm skills! 🧩 Problem Summary: Given two integers startValue and target, the goal is to determine the minimum number of operations needed to make startValue equal to target. The allowed operations are: 1️⃣ Multiply by 2 2️⃣ Subtract 1 🔍 Key Insight: Instead of moving from startValue → target, it’s more efficient to work backwards — from target → startValue. If target is even, divide it by 2; if odd, add 1. Continue until target <= startValue. Finally, add the difference to account for the remaining steps. ✅ Complexity: Time: O(log target) Space: O(1) Solving this reinforced how thinking in reverse can simplify what seems like a complex forward process — a useful mindset for both algorithms and life! 🌱 #LeetCode #Coding #ProblemSolving #Algorithms #C++ #GreedyAlgorithm #LearningEveryday

🚀 Just tackled an interesting string manipulation problem on LeetCode! 🚀 The challenge was to remove all digits from a string by repeatedly applying a specific operation: Find a digit. Delete that digit AND the closest non-digit character to its left. At first glance, it seems tricky to track which character to remove, but the solution becomes elegant with the right data structure. 💡 The Key Insight: This is a perfect use case for a stack! By processing the string from left to right, we can push non-digit characters onto the stack. Whenever we encounter a digit, we simply pop the top of the stack (the closest non-digit to the left). This efficiently handles the "deletion" in constant time. This problem is a great reminder of how choosing the right data structure can simplify complex-seeming tasks. The stack naturally models the "last-in, first-out" behavior we need for the deletion operation. #Coding #Algorithm #DataStructures #ProblemSolving #SoftwareEngineering #Tech #Programming #Developer

🔥 Day 152 of #1000DaysOfCode — Matrix Magic Continues! ⚡ 🔗 LeetCode Profile: leetcode.com/u/Rhythan 🔗 GitHub Day 152 Code:https://lnkd.in/gQEsVXQT 🧩 Problem: 867. Transpose Matrix Difficulty: Easy Category: Matrix | 2D Arrays 🧠 Concept Overview: The transpose of a matrix is one of the most foundational matrix operations in computer science and mathematics. It involves flipping the matrix over its main diagonal, which effectively means switching rows with columns. For example: Input: 1 2 3 4 5 6 7 8 9 Output: 1 4 7 2 5 8 3 6 9 ⚙️ Approach Summary: Create a new matrix where the number of rows becomes the number of columns, and vice versa. For each element (i, j) in the original matrix, place it in position (j, i) in the transposed one. The operation is simple yet powerful — a great refresher on index manipulation in 2D arrays. 🧮 Time Complexity: O(m × n) 💾 Space Complexity: O(m × n) 💡 Key Insights: Transposing is essential in matrix multiplication, image transformations, and graph adjacency matrices. Reinforces your understanding of row-column relationships. Also a great reminder that clarity in indexing is crucial when dealing with multidimensional structures. 🚀 Reflection: Day 152 showcases the elegance in simplicity — how a small transformation can entirely change the data perspective. Mastering such fundamentals builds strong intuition for complex data structures and algorithms. Each day adds another layer of depth to your problem-solving arsenal. 💪 Keep the streak alive — code, learn, repeat! 🔥 ✅ Progress: Day 152 of #1000DaysOfCode 💯 Mindset: Transform logic like matrices — systematically and beautifully. #1000DaysOfCode #Day152 #LeetCode #Matrix #Transpose #2DArrays #ProblemSolving #CodingChallenge #Java #Algorithms #RhythanCodes #CodeEveryday #LearnByDoing #ProgrammingJourney #DeveloperMindset #ConsistencyWins #CodingStreak #DailyPractice #DataStructures #MathematicsInCS #KeepBuilding #CodeToLearn #SoftwareEngineering #NeverStopCoding #Rhythan1000DaysChallenge