"Day 99: Sum of Two Integers in Java using Bit Manipulation"

📌 Day 20/100 – Add Binary (LeetCode 67) Today’s challenge was about adding two binary strings and returning the result in binary format — just like simulating manual binary addition with carry logic. 🔹 Problem: Given two binary strings a and b, return their sum as a binary string without converting them into decimal. 🔹 Approach: Start from the end of both strings (like column-wise addition). Use a carry variable to handle overflow (0 or 1). Keep appending the result (sum % 2) to a StringBuilder. Reverse the string at the end since we append from LSB to MSB. 🔹 Key Learnings: StringBuilder is more efficient than string concatenation in Java. Handling indices carefully avoids edge-case bugs. Binary addition logic is similar to decimal addition — just base changes, logic stays. #100DaysOfCode #Day20 #LeetCode #Java #BinaryMath #DSA #ProblemSolving #CodingJourney #KeepLearning

#Day-69) LeetCode 2536: Increment Submatrices by One using a 2D difference array technique in Java — a powerful approach for handling multiple range updates efficiently. 🔧 Instead of brute-force iteration over each query, I used a prefix sum strategy to apply all updates in constant time per query, followed by a cumulative pass to build the final matrix. 🧠 Highlights: Efficient submatrix updates using difference matrix Prefix sum accumulation for final values Clean, scalable Java implementation 📌 This kind of problem is a great reminder: smart preprocessing beats brute force. Let’s connect if you’ve explored similar matrix tricks or want to brainstorm more Java optimizations! #Java #LeetCode #DSA #MatrixOptimization #CodingInPublic #ProblemSolving #TechJourney #LinkedInTech

🚀Day 96/100 #100DaysOfLeetCode 🧩Problem: Reverse Linked List II✅ 💻Language: Java 💡 Approach: 1️⃣ First, use a dummy node to handle edge cases where reversal starts at the head. 2️⃣ Traverse to the node just before the left position — call it prev. 3️⃣ Reverse the sublist between left and right using standard pointer manipulation. 4️⃣ Reconnect the reversed portion back into the original list. 🔑Key Takeaways: 🔹Dummy nodes simplify linked list edge cases. 🔹In-place reversal reduces memory overhead. 🔹Careful pointer tracking ensures list integrity. ⚙️Performance: ⏱️Runtime: 0 ms(beats 100.00%) 💾Memory: 41.29 MB(beats 70.37%) #100DaysOfLeetCode #Java #LinkedList #CodingJourney #ProblemSolving #DSA #LeetCode #CodingChallenge

✅ Day 57/100 — LeetCode Challenge Problem: Transpose Matrix Language: Java Today I worked on a classic matrix operation — transposing a matrix. The goal is to convert rows into columns and vice-versa. This problem reinforced my understanding of 2D arrays and index manipulation in Java. 💡 Key Idea: If matrix has dimensions m x n, the transpose will have dimensions n x m, and each element at (i, j) becomes (j, i). 📌 Approach: Calculate row & column length Create result matrix with swapped dimensions Iterate & swap positions accordingly #100DaysOfCode #LeetCode #Java #DSA #Matrix #TechJourney #CodingChallenge #SoftwareEngineering #LearningEveryday

Java Bit Manipulation Practice – Finding the Rightmost Set Bit Today, I wrote a simple Java program to find the position of the rightmost set bit in a number’s binary representation. 🔹 First, the number is converted into binary using a while loop. 🔹 Then, I scan the binary string from right to left to locate the first '1'. 🔹 Finally, the program prints its position (counting starts from 1). Input: 21 Binary: 10101 Output: 1 (rightmost set bit position) It’s a great exercise for understanding bitwise logic and binary representation in Java! 💻🔥 #Java #Coding #BitManipulation #DSA #LearningJourney

🚀Day 88 – DSA in Java Solved “Hand of Straights” (LeetCode – Medium) Approach: ➡️ Sorted the array ➡️ Used HashMap to track frequencies ➡️ Formed consecutive groups of size k ➡️ Decremented counts as groups formed ⚡ Runtime: 24 ms (Beats 89.95%) 💾 Memory: 45.2 MB (Beats 89.4%) Explored how HashMap + sorting can simplify grouping problems and also learned about TreeMap for maintaining sorted order dynamically. #Day88 #LeetCode #Java #DSA #ProblemSolving #LearningEveryday

🚀 Day 137/160 – Maximize Partitions in a String Learned how to split a string into the maximum number of substrings such that no two parts share a common character. ✨ Key Concepts: Greedy Algorithm, Character Tracking, String Partitioning 💻 Language: Java #GeeksforGeeks #DSA #Java #ProblemSolving #CodingChallenge #KeepLearning #160DaysOfDSA

#Day-68) LeetCode #3228 – Maximum Number of Operations to Move Ones to the End (Java Edition) Tackled this neat string problem using a greedy approach in Java. The challenge? Move '1's to the end of the string under specific movement rules, maximizing the number of valid operations. 🧠 Core Idea: Track how many '0's we've seen and how many '1's we've already moved. Only move a '1' if there's enough '0's to justify it and the next character is also '1'. 💻 Java Strategy: Loop through the string Use counters to manage '0's and '1's Let me know how you'd tweak this or if you see a more optimal path! #Java #LeetCode #GreedyAlgorithm #StringProblems #DSA #CodingChallenge #LinkedInTech #ProblemSolving

Day 50 of #75DaysDSAChallenge Problem: 7. Reverse Integer Difficulty: 🟠 Medium Platform: LeetCode 🧩 Problem Statement Given a signed 32-bit integer x, return x with its digits reversed. If reversing x causes it to go outside the signed 32-bit integer range [-2³¹, 2³¹ - 1], return 0. Example: Input: x = 123 Output: 321 💡 Approach 1️⃣ Extract the last digit using x % 10. 2️⃣ Remove the last digit using x / 10. 3️⃣ Add the digit to the reversed number. 4️⃣ Before updating, check for overflow conditions. 5️⃣ Continue until all digits are processed. #LeetCode #Java #DSA #CodingChallenge #75DaysDSAChallenge #ProblemSolving #TechLearning #CodingJourney