Binary Alternating Bits Solution with XOR and Bit Tricks

#day316 & #day317 of #1001daysofcode problem statement (0762): Prime Number of Set Bits in Binary Representation problem statement (0868): Binary Gap Tracked positions of consecutive 1s using bit manipulation and computed the maximum distance between them. #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 🥷

🚀 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

🚀 Day 24/180 | #180DaysOfCode 📍 LeetCode | 💻 Java Solved: 2859. Sum of Values at Indices With K Set Bits Used bit manipulation to count the number of set bits in each index and summed the values whose index contains exactly k set bits. ⏱️ Time Complexity: O(n log n) 📦 Space Complexity: O(1) Strengthening understanding of bitwise operations and binary representation through consistent practice. 💪 Consistency continues 🚀 #DSA #LeetCode #Java #CodingJourney #Consistency

#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

#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

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

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 12/100 – LeetCode Challenge Problem: Middle of the Linked List Today’s problem focused on finding the middle node of a singly linked list. Approach: Used the Two-Pointer Technique (Slow & Fast pointers). slow moves one step at a time fast moves two steps at a time When fast reaches the end of the list, slow will be at the middle node Complexity: Time: O(n) Space: O(1) Concepts Practiced: Linked List traversal Two-pointer technique Efficient single-pass solution #100DaysOfCode #LeetCode #DSA #Java #LinkedList #ProblemSolving #CodingJourney

#day329 of #1001daysofcode problem statement (1784): Check if Binary String Has at Most One Segment of Ones Approach 1: If the pattern "01" appears in the string, it means the sequence of '1's was broken Approach 2: Traverse the string and count segments of consecutive '1's. Whenever a '1' is found, skip the entire sequence of continuous '1's and increase the segment count. If the number of segments of '1's is exactly one, the condition is satisfied. --both approaches have same time and space complexity. ⏱ Time Complexity: O(n) 🧠 Space Complexity: O(1) #1001DaysOfCode #DSA #Java #LeetCode #ProblemSolving Shivam Mahajan #leetcode