LeetCode POTD: Longest Balanced Substring

Day: 43/365 📌 LeetCode POTD: Binary Number with Alternating Bits Easy Key takeaways/Learnings from this problem: 1. This one shows how clean bit tricks can replace messy string checks when working with binary patterns. 2. Big takeaway: sometimes a small observation (like using XOR or shifting) turns a pattern problem into a super neat one-liner. #POTD #365DaysOfCode #DSA #Java #ProblemSolving #LearningInPublic #Consistency 🥷

Day: 40/365 📌 LeetCode POTD: Add Binary Easy Key takeaways/Learnings from this problem: 1. This one’s a reminder that binary addition is just normal addition with a carry you manage manually, nothing fancy. 2. Big takeaway: when strings represent numbers, think digit-by-digit from the end — clean logic > converting to integers. #POTD #365DaysOfCode #DSA #Java #ProblemSolving #LearningInPublic #Consistency 🥷

Day: 41/365 📌 LeetCode POTD: Reverse Bits Easy Key takeaways/Learnings from this problem: 1. This problem really strengthens your bit manipulation basics—shifting and masking is all you need. 2. Big takeaway: when dealing with fixed 32 bits, just loop consistently and build the answer bit by bit without overthinking it. #POTD #365DaysOfCode #DSA #Java #ProblemSolving #LearningInPublic #Consistency 🥷

Day: 44/365 📌 LeetCode POTD: Count Binary Substrings Easy Key takeaways/Learnings from this problem: 1. The real trick is noticing you don’t need to check every substring — just track lengths of consecutive 0s and 1s. 2. Big takeaway: many string problems become simple once you focus on grouping patterns instead of brute force counting. #POTD #365DaysOfCode #DSA #Java #ProblemSolving #LearningInPublic #Consistency 🥷

Day: 52/365 📌 LeetCode POTD: Minimum Operations to Equalize Binary String Level: Hard Key takeaways/Learnings from this problem: 1. This one really makes you think in terms of mismatched positions instead of raw characters. 2. Greedy feels tempting, but you need to carefully compare operation costs before deciding what to fix first. 3. It’s a good reminder that binary string problems often reduce to counting patterns smartly. 4. Biggest takeaway: always try to simplify the string into a difference representation — the solution usually becomes much clearer. #POTD #365DaysOfCode #DSA #Java #ProblemSolving #LearningInPublic #Consistency 🥷

#Day39 of #365DaysOfCode Solved Set Matrix Zeroes on LeetCode. Implemented an in-place matrix approach using the first row and column as markers to track zero positions. ✨ Lesson: Smart space optimization and careful traversal help solve matrix problems efficiently. #CodingJourney #DSA #Java #LeetCode #Consistency

Day 17 – Daily DSA Practice Solved LeetCode 401: Binary Watch today. A simple yet insightful problem focused on bit manipulation and careful formatting. Iterated through all valid hour and minute combinations, used Integer.bitCount() to filter cases, and ensured correct time representation. A good reminder that when constraints are small, a clean brute-force approach is often the most practical solution. #LeetCode #DSA #Java #ProblemSolving #Consistency

🚀 Day 20/180 | #180DaysOfCode 📍 LeetCode | 💻 Java Solved: 350. Intersection of Two Arrays II Used sorting + two-pointer technique to efficiently find common elements appearing in both arrays while maintaining correct frequency. ⏱️ Time Complexity: O(n log n + m log m) 📦 Space Complexity: O(min(n, m)) (for storing the intersection) Strengthening understanding of array traversal and two-pointer pattern through consistent practice. 💪 Consistency keeps the progress moving 🚀 #DSA #LeetCode #Java #CodingJourney #Consistency

🚀Day 11 of #Leetcode75 LeetCode #392 – Is Subsequence | Two Pointer Approach Today I solved “Is Subsequence” (Easy) and revised an important concept — the Two Pointer Technique. 🔎 Problem Summary: Given two strings s and t, determine whether s is a subsequence of t. A subsequence maintains the relative order of characters but doesn’t need to be continuous. 💡 Key Insight: Instead of using extra space or complex logic, we can solve this efficiently using two pointers: ✔ One pointer for s ✔ One pointer for t ✔ Move forward and match characters in order ⏱ Time Complexity: O(n) 📦 Space Complexity: O(1) This problem reinforces how powerful simple logic can be when applied correctly. Consistency > Complexity 💪 Excited to keep improving step by step! #LeetCode #DSA #Java #CodingPractice #ProblemSolving #100DaysOfCode #TechJourney

LeetCode Problem || Check if Binary String Has at Most One Segment of Ones(1784)🚀 we need to check: The string should have only one continuous block of '1's. After a '0' appears, '1' should never appear again. ✨ Insight: Instead of manually counting segments using loops, we can simply check if "01" exists in the string. 📌 Time Complexity: O(n) Practicing problems like these helps improve pattern recognition and problem-solving efficiency. #LeetCode #DSA #Java #CodingPractice #ProblemSolving