Bubble Sort Efficiency Analysis

Solved Search in a Rotated Sorted Array — where the array isn’t fully sorted… but still searchable in O(log n). Key insight? At every step, one half is always sorted. Identify it → check target range → eliminate half. 🎯 This problem was a reminder that: Binary Search isn’t just a template — it’s about thinking in ranges. #DSA #BinarySearch #Java #Coding #ProblemSolving #LearningDaily

Recently solved "Search a 2D Matrix", and it was a nice reminder that sometimes a problem that looks 2D is actually 1D in disguise. Since each row is sorted and the first element of every row is greater than the last element of the previous row, the matrix can be treated like a single sorted array — making Binary Search the most efficient approach with O(log m×n) time complexity. Small insight, but a good lesson in carefully reading constraints before jumping into coding. 🚀 Problems like this remind me that understanding constraints is more important than memorizing patterns. #LeetCode #DSA #BinarySearch #CodingInterview #ProblemSolving #Java

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 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

𝗬𝗼𝘂 𝗵𝗮𝘃𝗲 𝗯𝗲𝗲𝗻 𝘂𝘀𝗶𝗻𝗴 𝗕𝗶𝗻𝗮𝗿𝘆 𝗦𝗲𝗮𝗿𝗰𝗵 𝘄𝗿𝗼𝗻𝗴. It is not just for sorted arrays. The real definition — "Eliminate HALF the search space with each decision." That one shift in thinking unlocks 20+ LeetCode problems. Swipe to see the template, live code, and 7 problems you can now solve with one pattern. 👇 Save this. 🔖 #DSA #BinarySearch #Java #LeetCode #CodingInterview #DebugWithPurpose

Day 35/100 – LeetCode Challenge 🚀 Problem: 3Sum Approach: Sorted the array Fixed one element and used two pointers for the remaining two Skipped duplicates to ensure unique triplets Used early stopping when the current number became positive Time Complexity: O(n²) Space Complexity: O(1) Key takeaway: Many 3-element sum problems reduce to sorting + two-pointer pattern. #LeetCode #100DaysOfCode #DSA #Java #TwoPointers #ProblemSolving #InterviewPrep

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

Day 36/100 – LeetCode Challenge 🚀 Problem: 3Sum Closest Approach: Sorted the array Fixed one element and applied the two-pointer technique Tracked the closest sum by comparing absolute differences Returned immediately if an exact match was found Time Complexity: O(n²) Space Complexity: O(1) Key takeaway: Many optimization problems are variations of classic patterns. Understanding 3Sum deeply makes solving 3Sum Closest straightforward. #LeetCode #100DaysOfCode #DSA #Java #TwoPointers #ProblemSolving #InterviewPrep

Day 25 – Find Minimum in Rotated Sorted Array Solved this rotated array problem using a modified Binary Search approach in O(log n). Key Learning: By comparing mid with right, I identified which half contains the minimum and reduced the search space efficiently. #DSA #BinarySearch #RotatedArray #Java #InterviewPrep

Stop writing code that “just works.” Day 45 / 100 Days of DSA Problem: Find a pair from two arrays whose sum is closest to a target. First approach: Nested loops. Better approach: Two pointers (since arrays are sorted). Move intelligently based on the sum. Time Complexity: O(n + m). Same answer. Smarter thinking. Day 45 lesson: Don’t stop at “it works.” Push for optimal. #Day45 #100DaysOfCode #DSA #Java #ProblemSolving