Longest Common Prefix Challenge on LeetCode

problem #15 🙂 , 3 sum problem from #leetcode, unlike 2 sum problem, here we have to use 3 pointers to reach the target, this was little bit complicated than the 2 sum problem, of course we must avoid the duplicate numbers. TIME COMPLEXITY - o(n^2) #problemsolving #datastructuesandalgorithms #arrays #hardproblem #computerscience #3sumproblem #java #geeksforgeeks #leetcode #softwaredevelopment

Solved the Valid Parentheses problem using a stack approach. Instead of storing opening brackets, the solution pushes expected closing brackets and matches them directly while traversing the string. This simplifies comparison and avoids extra checks. Time Complexity: O(n) Space Complexity: O(n) #Java #DSA #Stack #LeetCode #Coding

🚀 Day 2/30 — LeetCode Challenge Solved the Reverse Integer problem on using Java. At first glance, reversing digits looks straightforward. But the real challenge is handling integer overflow within 32-bit limits — that’s where most implementations fail. ✅ Key takeaway: Correctness isn’t just about logic — it’s about respecting constraints and edge cases. Continuing to focus on writing solutions that are not just working, but reliable. #LeetCode #Java #ProblemSolving #DataStructures #Algorithms #Consistency

🚀100 Days of Code Day-26 LeetCode Practice – Remove Duplicates from Sorted Array Solved a classic problem using the Two Pointer Technique 💡 📌 Problem: Given a sorted array, remove duplicates in-place and return the number of unique elements. 🔍 Key Idea: Since the array is sorted, duplicates are adjacent. Using two pointers helps efficiently overwrite duplicates without extra space. ⚡ Complexity: Time → O(n) Space → O(1) 💻 Clean and optimized approach makes this problem a great example of in-place array manipulation! #LeetCode #Java #DataStructures #CodingPractice #ProblemSolving #100DaysOfCode

#Day72 Of Problem Solving Solved today’s LeetCode Daily Challenge ✅. Problem: Check if Binary String Has at Most One Segment of Ones Sometimes, the simplest logic gives the best results. Instead of overthinking, I went with a clean approach—just checking if "01" exists in the string. If it does, that means more than one segment of 1’s… and that’s it! ✔️ All test cases passed ⚡ Runtime: 0 ms 📊 Beat 100% in performance This problem was a good reminder that not every solution needs complexity—clarity matters more. Consistency > Complexity 🚀 On to the next challenge! #LeetCode #DailyChallenge #Java #ProblemSolving #CodingJourney #100DaysOfCode #LinkedIn

🚀 Day 6/30 — LeetCode Challenge Solved "Divide Two Integers" on LeetCode using Java. This problem restricts the use of division, multiplication, and modulus — forcing an alternative approach using "bit manipulation and subtraction logic". Handled tricky edge cases like overflow (Integer.MIN_VALUE / -1) to ensure correctness. ✅ Key takeaway: Constraints push you to think beyond standard solutions and understand how operations work at a deeper level. #LeetCode #Java #Algorithms #ProblemSolving #Consistency

🚀 Day 4/30 — LeetCode Challenge Solved the "Palindrome Number" problem on LeetCode using Java. The logic is simple, but the key is choosing the right approach — reversing only half of the number instead of the entire value to avoid unnecessary operations. ✅ Key takeaway: Even simple problems have smarter solutions if you focus on efficiency. Continuing to build consistency and improve problem-solving skills. #LeetCode #Java #Algorithms #ProblemSolving #Consistency

Solved the Remove Duplicates from Sorted Linked List problem using iterative traversal. The idea is to traverse the list and compare current node with next node. If values are same, skip the duplicate node by adjusting pointers; otherwise move forward. Since the list is sorted, duplicates are always adjacent, making it efficient. Time Complexity: O(n) Space Complexity: O(1) #Java #DSA #LinkedList #LeetCode #Coding

Day 92/100 Completed ✅ 🚀 Solved LeetCode: Median of Two Sorted Arrays (Java) ⚡ Used a two-pointer approach to efficiently find the median without fully merging the arrays. Focused only on tracking the required middle elements, reducing unnecessary space usage. 🧠 Key Learnings: • Applying two-pointer technique on sorted arrays • Handling edge cases when one array is exhausted • Understanding median logic for even vs odd total length • Optimizing space by avoiding full array merge 💡 This problem highlighted how tracking only what’s necessary can lead to better performance and cleaner solutions. Profile: https://lnkd.in/gaJmKdrA #leetcode #datastructures #algorithms #java #twopointers #array #coding #problemSolving

🚀 Day 35 of my #100DaysOfCode Journey Today, I solved the LeetCode problem: Valid Anagram Problem Insight: Given two strings, check if one is an anagram of the other. Approach: • First, check if the strings have the same length; if not, return false • Convert both strings to character arrays • Sort both arrays • Compare the sorted arrays — if equal, the strings are anagrams Time Complexity: • O(n log n) — due to sorting the arrays Space Complexity: • O(n) — for the character arrays Key Learnings: • Sorting is a simple and effective way to compare character compositions • Edge cases like different lengths should be handled first • Breaking the problem into small steps makes it easy to reason about Takeaway: Sometimes, sorting can reduce a seemingly complex problem into a simple comparison. #DSA #Java #LeetCode #100DaysOfCode #CodingJourney #ProblemSolving #Strings