Markus Eisele’s Post

Optimizing Sorting Logic Using Bit Manipulation (Java) Today I solved a problem that required sorting integers based on the number of set bits in their binary representation. Instead of using a direct sorting approach, I implemented a custom comparator with a PriorityQueue and used Brian Kernighan’s algorithm to count set bits efficiently: n = n & (n - 1) This reduces time complexity for counting bits to O(number of set bits). 📊 Result: • 77/77 test cases passed • Runtime: 7ms • Beat 78% of submissions Key takeaway: Optimization is not just about solving — it’s about choosing the right approach and reducing unnecessary operations. Bit manipulation continues to be one of the most powerful tools in algorithm design. #DSA #Java #BitManipulation #LeetCode #ProblemSolving #SoftwareEngineering

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 🥷

Leetcode Problem | | Check If a String Contains All Binary Codes (1461) 🚀 Problem: Given a binary string s and integer k, check if all possible binary codes of length k exist in the string. Example: s = "00110", k = 2 Expected Output = true In substring and sliding window problems, boundary conditions matter more than logic sometimes. Consistent debugging > frustration. #DSA #Java #LeetCode #ProblemSolving #LearningInPublic

#day334 of #1001daysofcode problem statement (1009): Complement of Base 10 Integer 💡 Approach: To find the complement of a number, I converted it into binary and flipped every bit (0 → 1 and 1 → 0). After flipping the bits, the resulting binary string was converted back to decimal. Example: 5 → Binary: 101 Complement: 010 → 2 ⏱ Time Complexity: O(log n) 🧠 Space Complexity: O(log n) #1001DaysOfCode #DSA #Java #LeetCode #ProblemSolving Shivam Mahajan #leetcode

𝗝𝗮𝘃𝗮 𝟴 - 𝗖𝗼𝗹𝗹𝗲𝗰𝘁𝗼𝗿𝘀 I have covered Collectors, an important component of the Java 8 Stream API. Collectors are used to transform, group, partition, and summarize stream data into useful results. ✔ What is a Collector and how it works ✔ collect() method explained ✔ Common collectors like toList(), toSet(), toMap() ✔ groupingBy() and partitioningBy() ✔ mapping(), counting(), joining(), summarizing() ✔ Downstream collectors ✔ Custom collector basics ✔ Practical real-world examples ✔ Common mistakes and interview-focused scenarios Collectors make data processing more powerful, structured, and expressive in Java 8. #Java #Java8 #Collectors #StreamAPI #FunctionalProgramming #BackendDevelopment #SpringBoot #InterviewPreparation

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 🥷

#day328 of #1001daysofcode problem statement (1758): Minimum Changes To Make Alternating Binary String An alternating binary string can only follow two patterns: "010101..." or "101010...". I counted the number of changes required to convert the string into both patterns and returned the minimum of the two. ⏱Time Complexity: O(n) 🧠Space Complexity: O(1) Consistency check ✅ One LeetCode problem a day to sharpen problem-solving skills. #1001DaysOfCode #DSA #Java #LeetCode #ProblemSolving Shivam Mahajan #leetcode

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 🥷

#day314 of #1001daysofcode problem statement (0693): Binary Number with Alternating Bits Used XOR with right shift to transform alternating bits into a sequence of 1s, then applied the (x & (x + 1)) == 0 pattern(tocheck all bits are set bits or not). Sometimes bit tricks make problems beautifully simple. #1001DaysOfCode #DSA #Java #LeetCode #ProblemSolving Shivam Mahajan #leetcode

🚀 Day 73 / 100 Days of LeetCode Question: Check if Binary String Has at Most One Segment of Ones Given a binary string s, determine whether it contains at most one continuous segment of '1's. Solved in Java by checking whether the pattern "01" exists in the string. If "01" appears, it means a new segment of 1s starts after a 0, which violates the condition. This provides a clean and efficient solution. Consistency > perfection. Day 74 loading 🔥 #100DaysOfCode #LeetCode #Java #DSA #ProblemSolving