Reverse Integer LeetCode Solution and Lessons Learned

#Day92 Of Problem Solving Today’s LeetCode problem turned out to be simpler than I expected — but only after looking at it from the right angle. At first, I was thinking of solving it using loops or recursion. But then I realized there’s a much cleaner way using a small mathematical trick to check if a number is a power of three. That shift in thinking made the solution not just easier, but also more efficient. Result: Runtime: 7 ms (beats 100%) Memory: 46 MB Moments like this remind me that problem solving isn’t always about writing more code, it’s about finding better approaches.Trying to stay consistent and improve a little every day. #LeetCode #Java #DSA #ProblemSolving #CodingJourney #LinkedIn

There was a time when even starting a problem felt confusing. Reading the question multiple times, not knowing where to begin, and getting stuck often. With consistent practice, things started to change — patterns became familiar, approaches became clearer, and solving started to feel more structured. That consistency has now led to solving 150 problems on LeetCode. Focused on building a strong foundation through arrays, strings, hashing, sliding window, and binary search — spending more time understanding the logic rather than just aiming for accepted submissions. This journey improved: * Problem breakdown and thinking approach * Pattern recognition across questions * Code clarity and structure Currently working on improving speed and taking on more challenging problems. 🔗 LeetCode Profile: https://lnkd.in/gkf7m6wp #LeetCode #DSA #Java #ProblemSolving #Coding #SoftwareEngineering

“What I learned after solving 300+ LeetCode problems 👇” At the beginning, I made a mistake. I kept solving random problems every day thinking: “More problems = more skill” But I was wrong. I realized that solving problems randomly is mostly a waste of time if you don’t understand the pattern behind them. Everything changed when I started focusing on patterns instead of problems. Here are some important patterns I learned: • Sliding Window • Two Pointers • Prefix Sum • Binary Search • Recursion & Backtracking • Linked List Patterns • Stack & Monotonic Stack • Hashing / Frequency Count Once you understand these patterns: You don’t need to solve 1000 problems. You start recognizing solutions instantly. Now when I see a problem, I don’t panic. I ask: 👉 “Which pattern does this belong to?” That one question changed everything for me. Still learning, but now with direction 🚀 #DSA #LeetCode #Java #BackendDevelopment #CodingJourney

🚀 Solved LeetCode Problem #58 – Length of Last Word Today I worked on a simple yet insightful string manipulation problem that emphasizes attention to edge cases. 🔍 Problem Insight: Given a string containing words and spaces, the goal is to find the length of the last word, ignoring any trailing spaces. 💡 Approach Used: Instead of using built-in methods like split(), I implemented an optimized approach by: Traversing the string from the end Skipping trailing spaces Counting characters until the next space is encountered This approach avoids extra space usage and improves efficiency. 🧠 Key Learning: Importance of handling edge cases like trailing spaces How reverse traversal can simplify string problems Writing memory-efficient solutions 📈 Complexity: Time: O(n) Space: O(1) ✨ Problems like this help strengthen: String manipulation skills Logical thinking Writing clean and optimized code Consistency is key—one step closer to mastering DSA! 💪 #LeetCode #DSA #StringManipulation #Coding #ProblemSolving #Java #TechJourney

🚀 Solved “Search Insert Position” using Binary Search on LeetCode! Today I worked on an interesting problem where the goal is to find the index of a target element in a sorted array. If the target is not present, we return the index where it should be inserted to maintain sorted order. 🔍 Approach: Instead of using a linear search (O(n)), I implemented an efficient Binary Search (O(log n)) approach. 💡 Key Learning: One important detail is calculating the middle index safely: mid = low + (high - low) / 2 This avoids potential integer overflow compared to (low + high) / 2 and ensures correct behavior even for large inputs. ⚙️ Logic: If target == arr[mid] → return mid If target > arr[mid] → search right half Else → search left half If not found → return ‘low’ as the correct insert position 📈 Result: Achieved 100% performance on LeetCode 🚀 This problem reinforced my understanding of: ✔ Binary Search fundamentals ✔ Edge case handling ✔ Writing optimized and safe code Looking forward to solving more problems and improving problem-solving skills! #LeetCode #BinarySearch #Java #DataStructures #Coding #ProblemSolving

Most developers are not slowed down by bad code. They are slowed down by bad thinking. Not syntax. Not framework choice. Not whether the project uses Java, Go, or Python. The real damage usually comes earlier: - no clear boundaries - too many dependencies in one request path - retries added without thinking - APIs designed around convenience instead of failure - teams optimizing for feature speed over system clarity That’s why some codebases feel heavy even before they get big. The problem is not always technical debt. Sometimes it’s decision debt. And that is much harder to fix. Debate: What does more long-term damage to software teams? A) bad code B) bad architecture C) bad product decisions D) bad debugging habits My vote: B first, C second. What’s yours? #Java #SoftwareArchitecture #Microservices #DistributedSystems #BackendEngineering

🚀 Solved LeetCode Problem #22 – Generate Parentheses Today I worked on an interesting recursion + backtracking problem that really strengthens understanding of constraint-based generation. 🔍 Problem Insight: Given n pairs of parentheses, generate all combinations of well-formed parentheses. 💡 Key Idea: Instead of generating all combinations and filtering invalid ones, we build only valid strings by: Adding '(' only when we still have some left Adding ')' only when it won’t break validity (i.e., open > close) 🧠 This ensures we never create invalid sequences, making the solution efficient. ⚙️ Approach Used: Backtracking (DFS) with two counters: open → number of '(' used close → number of ')' used 📈 Complexity: Time complexity follows Catalan Numbers → grows approximately as O(4^n / √n) ✨ Key Learning: This problem highlights how smart constraints in recursion can avoid unnecessary computation and lead to optimal solutions. 📌 Problems like this are great for mastering: Recursion Backtracking Combinatorial generation #LeetCode #Coding #DSA #Recursion #Backtracking #Java #ProblemSolving #TechJourney

🔥 Day 3 of my 50 Days Wild Coding Kickoff! 🔥 💡 Problem 3: Valid Parentheses (Easy) Given a string containing just '(', ')', '{', '}', '[' and ']', determine if the input string is valid. A string is valid if: ✔ Open brackets are closed by the same type ✔ Open brackets are closed in the correct order Example 1: Input: s = "[]" Output: true Example 3: Input: s = "[(])" Output: false 🚀 Approach (Optimized without Stack class): Instead of using Java’s built-in Stack, I used a char array to simulate a stack for better performance. Created a char[] as stack Used top pointer to track elements Push opening brackets On closing bracket → pop and compare If mismatch or stack empty → invalid #100DaysOfCode #50DaysOfCode #CodingChallenge #JavaDeveloper #DataStructures #Stack #Algorithms #DSA #CodingJourney #InterviewPrep #LeetCode #ProblemSolving #DeveloperLife #CodingDaily #CodePractice

🚀 Day 88 – DSA Journey | Binary Tree Postorder Traversal Continuing my daily DSA practice, today I tackled one of the trickier tree traversals using an iterative approach. 📌 Problem Practiced: Binary Tree Postorder Traversal (LeetCode 145) 🔍 Problem Idea: Traverse a binary tree in postorder — Left → Right → Root — and return the node values. 💡 Key Insight: Postorder traversal is not straightforward iteratively. A clever trick is to reverse a modified preorder traversal (Root → Right → Left) to achieve the correct order. 📌 Approach Used: • Use a stack to process nodes • Traverse in Root → Right → Left order • Insert elements at the beginning of the result list • This effectively reverses the order to get Left → Right → Root 📌 Concepts Strengthened: • Tree traversal (Postorder) • Stack usage • Iterative traversal tricks • Reversing traversal logic ⏱️ Time Complexity: O(n) 📦 Space Complexity: O(n) 🔥 Today’s takeaway: Sometimes solving a problem becomes easier when you reverse the perspective — small tricks can simplify complex logic. On to Day 89! 🚀 #Day88 #DSAJourney #LeetCode #BinaryTree #Stack #Java #ProblemSolving #Coding #LearningInPublic #Consistency

🚀 LeetCode Challenge 7/50 🔍 Problem: Valid Anagram Today’s problem was about checking whether two strings are anagrams of each other efficiently. 💡 Approach: I used the Character Frequency Count method: First checked if both strings have equal length Used an array of size 26 to track character frequencies Incremented counts for one string and decremented for the other If all values are zero, both strings are anagrams ⚡ This approach avoids sorting and ensures optimal performance. 📊 Complexity Analysis: Time Complexity: O(n) Space Complexity: O(1) 📚 Key Learning: Using frequency arrays is a powerful technique for string problems. It helps achieve constant space complexity and improves efficiency compared to sorting-based approaches. Step by step, getting better at problem solving 💪 #LeetCode #Algorithms #DataStructures #ProblemSolving #CodingJourney #Java #100DaysOfCode #StudentDeveloper #Learning