Binary Search in Rotated Array with O(log n) Complexity

Another day another problem💪 In this problem, the array is sorted but rotated at an unknown pivot. Instead of using a linear search (O(n)), I applied a binary search strategy to maintain O(log n) time complexity. 💡 Key Learnings: Handling edge cases in rotated arrays Understanding how to calculate the real mid index after rotation Strengthening binary search fundamentals Consistent practice is helping me improve my problem-solving and DSA skills step by step. 💪 #LeetCode #DSA #BinarySearch #Java #Coding #ProblemSolving #SoftwareDeveloperJourney

Day 22 – Binary Search Implemented Binary Search on a sorted array to find the target efficiently. Key Learnings: • Using left, right, and mid pointers • Updating search boundaries correctly • Returning index or -1 based on condition #LeetCode #BinarySearch #Algorithms #Java #DSA

Day 2 #SDE Solved: • Search in Rotated Sorted Array II • Find Minimum in Rotated Sorted Array II Key Learning: Duplicates can break the usual binary search logic because it's harder to determine the sorted half. In such cases, shrinking the search space (low++ / high--) helps continue the search. #LeetCode #DSA #BinarySearch #SoftwareEngineering #Java

DSA daily streak |day-27 ✅Find Unique Binary String Today, I explored a problem where the goal is to generate a binary string that does not exist in a given list of binary strings. This problem introduces an interesting concept inspired by Cantor’s Diagonalization technique, which guarantees that the generated string will be unique. #DSA #Java #ProblemSolving #LeetCode #Algorithms #CodingJourney #LearningJourney

🚀 DSA Tip: Two Pointer Approach for Target Sum If your array is sorted, you don’t need nested loops to find a pair with a given target sum. 👉 Use the Two Pointer Technique — simple and O(n)! 💡 How it works: • Place one pointer at the start (left) • Place another at the end (right) • Compare their sum with the target ✅ If sum is small → move left++ ✅ If sum is large → move right-- ✅ If equal → 🎯 pair found ⏱️ Complexity: Time → O(n) Space → O(1) 🔥 When to use: ✔️ Target sum in sorted array ✔️ Pair problems ✔️ Remove duplicates ✔️ Reverse problems Mastering two pointers can save you from unnecessary O(n²) solutions! #DSA #CodingInterview #Java #TwoPointers #ProblemSolving #TechLearning

Day 26 – Find Peak Element Worked on finding a peak element using the “binary search on slope” technique. Key Learning: By comparing nums[mid] with nums[mid + 1], I determined the direction of the peak and narrowed the range in logarithmic time. #DSA #BinarySearch #LeetCode #Java #CodingPractice

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: 45/365 📌 LeetCode POTD: Special Binary String Hard Key takeaways/Learnings from this problem: 1. This one teaches that recognizing a balanced structure (like valid parentheses) inside a binary string is the real trick. 2. Big takeaway: recursion + sorting substrings smartly can turn a complex greedy-looking problem into something surprisingly elegant. #POTD #365DaysOfCode #DSA #Java #ProblemSolving #LearningInPublic #Consistency 🥷

🚀 Day 56 Out of #365DaysOfCode Github Link: https://lnkd.in/gGUy_MKZ Today I solved an interesting problem sort an integer array based on the number of 1’s in its binary representation. If two numbers have the same number of set bits, sort them in ascending numerical order. 🔎 Key Concept: Use Integer.bitCount() to count the number of 1’s. Apply a custom comparator with Arrays.sort(). Handle tie-breaking by normal ascending comparison. 💡 This problem is a great reminder of how powerful Java’s built-in methods and custom comparators can be for writing clean and efficient solutions. Time Complexity: O(n log n) It’s always exciting to explore bit manipulation problems—they strengthen logical thinking and deepen understanding of how numbers work internally. #Java #Coding #DataStructures #Algorithms #BitManipulation #ProblemSolving #Learning

Day 11/30 – 3Sum Closest 🚀 Problem: Find three numbers whose sum is closest to a given target. Approach: • Sort the array • Fix one element • Use Two Pointers • Track minimum difference using Math.abs() Key Idea: Instead of finding exact zero (like 3Sum), minimize the distance from target. Time Complexity: O(n²) Space Complexity: O(1) #30DaysOfCode #Java #DSA #TwoPointers #InterviewPrep