LeetCode 3713: Longest Balanced Substring in Java

🚀 Cracked LeetCode 189 – Rotate Array using an optimized in-place approach. Instead of brute force (O(n × k)), I implemented the reverse algorithm technique to achieve: ✅ Time Complexity: O(n) ✅ Space Complexity: O(1) ✅ 0 ms runtime (Beats 100%) Focused on improving optimization thinking and writing clean, efficient Java code. #LeetCode #Java #DataStructures #Algorithms #ProblemSolving #TechGrowth

Day 46: Binary Addition from scratch! 🔢 Problem 67: Add Binary Java's BigInteger is a thing, but manual bit manipulation is just more satisfying. My approach: 1. Padded both strings to the same length using character arrays. 2. Simulated schoolbook addition from right to left. 3. Used % 2 for the result bit and / 2 for the carry. It’s a clean O(N) solution that avoids overflow and keeps the logic transparent. No shortcuts, just pure logic. 🚀 #LeetCode #Java #Binary #Algorithms #CodingChallenge #ProblemSolving

🎯 Day 100 of #100DaysOfCode 🔥 What a way to wrap it up! Solved LeetCode #3666 – Minimum Operations to Equalize Binary String ✅ A problem that blends math, parity logic, and careful case analysis—not your usual binary flip question. Key takeaways: -> Count-based optimization over brute force -> Handling even/odd operation patterns smartly -> Early exits = cleaner & faster logic -> Thinking in terms of operations feasibility rather than simulation 🧠 Language: Java -> Runtime: 0 ms (Beats 83.87%) -> Memory: 47.90 MB 100 days. Countless problems. One habit built: consistency 💪 Onward to harder problems and deeper concepts 🚀 #LeetCode #Java #DSA #BinaryStrings #ProblemSolving #Consistency #100DaysChallenge

🚀 Day 98 of #100DaysOfCode Solved LeetCode #1356 – Sort Integers by The Number of 1 Bits ✅ A clean bit manipulation + sorting problem that rewards thinking beyond plain comparisons. Key Takeaways: -> Using Integer.bitCount() to count set bits efficiently -> Encoding sort priority directly into values for simplicity -> Leveraging built-in sorting for clean and fast solutions -> Small tricks can lead to elegant code ✨ Language: Java -> Runtime: 6 ms (Beats 93.70%) ⚡ -> Memory: 47.42 MB Almost there. Staying consistent till the end 💻🔥 #LeetCode #Java #BitManipulation #Sorting #ProblemSolving #DSA #100DaysOfCode

🚀 Day 90 of #100DaysOfCode | LeetCode Daily Solved LeetCode Problem #401 – Binary Watch ⌚💻✅ A fun problem that blends bit manipulation with simple iteration. The idea is to count set bits for hours and minutes and generate all valid time combinations. Key Takeaways: -> Smart use of Integer.bitCount() for counting LEDs -> Separating hours (0–11) and minutes (0–59) cleanly -> Formatting time output correctly (leading zeros matter!) -> Brute force done right with clear constraints Language: Java -> Runtime: 3 ms (Beats 91.19%) -> Memory: 43.45 MB (Beats 94.31%) Consistency builds confidence. 90 days in — still pushing forward. 🔥💻 #LeetCode #Java #BitManipulation #BinaryWatch #ProblemSolving #100DaysOfCode

Day 31/100 - LEETCODE Challenge ✅ Problem : Find First and Last Position of Element in Sorted Array Solved the Search for First and Last Position of Element in Sorted Array problem using an optimized Binary Search approach in Java. Instead of using a linear scan, I implemented two separate binary searches to efficiently find the first and last occurrence of the target element in O(log n) time complexity. This solution improves performance for large datasets and achieved 0 ms runtime (100% beats) on LeetCode. Problems like this help strengthen understanding of binary search variations and edge case handling in sorted arrays. #100DaysOfCode #java #Coding #SoftwareDeveloper

Day 65: The "One-Liner" Win 🎯 Problem 1784: Check if Binary String Has at Most One Segment of Ones Today was a lesson in simplifying logic. The challenge: check if a binary string contains more than one contiguous segment of '1's, given that the string starts with '1'. The Strategy: • Observation: If there are multiple segments of '1's, they must be separated by at least one '0'. • The Pattern: In a string starting with '1', any "new" segment of ones would look like "01" somewhere in the string. • The Execution: A simple !s.contains("01") handles the entire check. Sometimes we hunt for complex algorithms when a single string method is the ultimate counter. Clean, readable, and passed all test cases. 🚀 #LeetCode #Java #StringManipulation #Coding #Efficiency #DailyCode

Strings look simple until edge cases show up 👀 Day 8 / #100DaysOfCode 🚀 Solved: Valid Palindrome (LeetCode 125) 🔹 Approach: Used the two-pointer technique starting from both ends. Ignored non-alphanumeric characters and compared characters after converting the string to lowercase. Time Complexity: O(n) Space Complexity: O(1) ✔ Practiced string traversal with pointers ✔ Handled real-world edge cases (spaces, symbols, cases) ✔ Avoided extra space by working in-place 💡 Takeaway: Clean pointer logic often beats preprocessing with extra data structures. Building strong fundamentals in Java, one problem at a time. #LearnInPublic #100DaysOfCode #DSA #Java #Strings #ProblemSolving #SoftwareEngineer #CodingJourney #Consistency

🚀 Day 91 of #100DaysOfCode | LeetCode Daily Solved LeetCode Problem #693 – Binary Number with Alternating Bits 🔢💡✅ A clean and elegant bit manipulation problem. The trick is to XOR the number with its right shift and check if the result becomes all 1s — simple, fast, and effective. Key Takeaways: -> Using n ^ (n >> 1) to detect alternating patterns -> Leveraging highestOneBit() for boundary checks -> Writing concise, branch-free logic with bitwise ops -> Sometimes the smartest solutions are the shortest Language: Java -> Runtime: 0 ms (Beats 100%) -> Memory: 41.81 MB (Beats 97.08%) One more day, one more concept sharpened. 🔥 Onward to Day 92 💻🚀 #LeetCode #Java #BitManipulation #BinaryNumbers #ProblemSolving #100DaysOfCode

Day 2/100 – LeetCode Challenge 🚀 Problem: #48 Rotate Image   Difficulty: Medium   Language: Java   Approach: In-Place Rotation using Transpose + Row Reversal   Time Complexity: O(n²)   Space Complexity: O(1) 🔍 Key Insight: To rotate a matrix 90° clockwise without extra space: 1️⃣ First transpose the matrix (swap across diagonal). 2️⃣ Then reverse each row. This avoids creating a new matrix and satisfies the in-place constraint. 🧠 Solution Brief: Used nested loops to transpose the matrix by swapping arr[i][j] and arr[j][i]. Then reversed each row using two pointers (start and end). Combined both operations inside a rotate() method. Achieved full rotation with constant extra space. 📌 What I Learned: Matrix problems are more about pattern recognition than brute force. Understanding transformations (transpose + reverse) makes complex problems simple. Starting my 100 Days of LeetCode journey today 💪 Consistency > Motivation. #LeetCode #Day2 #100DaysOfCode #Java #DSA #Matrix #ProblemSolving #CodingJourney #MediumProblem