LeetCode Challenge: Count Binary Substrings in O(n) Time

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 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 13/100 – LeetCode Challenge 🚀 Solved Count Binary Substrings using an optimized O(n) approach by grouping consecutive characters and comparing segment lengths. Key takeaway: Instead of generating all substrings, identify patterns and reduce the problem to counting adjacent groups. Consistency > Motivation 💪 #100DaysOfCode #DSA #Java #LeetCode

Day 34/100 – LeetCode Challenge 🚀 Problem: Container With Most Water Approach: Used two pointers (left & right) Calculated area using min(height[left], height[right]) × width Moved the pointer pointing to the smaller height Updated maximum area during traversal Time Complexity: O(n) Space Complexity: O(1) Key takeaway: When maximizing area between two boundaries, moving the smaller boundary is the only way to potentially increase the result. #LeetCode #100DaysOfCode #DSA #Java #TwoPointers #ProblemSolving

Solved "Container With Most Water" on LeetCode today using the Two-Pointer Technique. 🚀 The key insight is that the area formed by two lines depends on the shorter height and the distance between them. Starting with pointers at both ends of the array, we compute the area and move the pointer at the smaller height inward to potentially find a taller boundary and maximize the area. This approach efficiently reduces the problem from O(n²) brute force to O(n) time with O(1) space. Problems like this are a great reminder that the right observation can drastically optimize a solution. 💡 #LeetCode #DSA #TwoPointers #Java #ProblemSolving

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

🚀 Day 20 – #100DaysOfLeetCode Today’s problem: Product of Array Except Self (Medium) 🔹 Learned how to solve it in O(n) time 🔹 Without using division 🔹 Optimized space complexity The key idea was using prefix and suffix products instead of recalculating product every time. #Day20 #LeetCode #DSA #Java #CodingJourney

LeetCode Problem || Check if Binary String Has at Most One Segment of Ones(1784)🚀 we need to check: The string should have only one continuous block of '1's. After a '0' appears, '1' should never appear again. ✨ Insight: Instead of manually counting segments using loops, we can simply check if "01" exists in the string. 📌 Time Complexity: O(n) Practicing problems like these helps improve pattern recognition and problem-solving efficiency. #LeetCode #DSA #Java #CodingPractice #ProblemSolving

🚀 Day 42 of #100DaysOfCode 🌱 Topic: Linked List / Two Pointers ✅ Problem Solved: LeetCode 82 – Remove Duplicates from Sorted List II 🛠 Approach: Used a dummy node to simplify handling edge cases where the head might be removed. When two consecutive nodes had the same value, stored that value. Skipped all nodes with that duplicate value using a loop. Linked the previous node to the next distinct node. Continued traversal until reaching the end. This ensures that only unique elements remain in the sorted list. #100DaysOfCode #Day42 #DSA #LinkedList #TwoPointers #LeetCode #Java #ProblemSolving #CodingJourney #Consistency

Day: 71/365 📌 LeetCode POTD: Count Submatrices with Top-Left Element and Sum Less Than k Medium Key takeaways/Learnings from this problem: 1. Prefix sum is the hero here—once you precompute it, submatrix sum becomes instant lookup instead of recomputing every time. 2. Fixing the top-left corner simplifies things a lot, turning a 2D problem into something more manageable. 3. Instead of brute forcing all submatrices, you just expand and check smartly using precomputed sums. #POTD #365DaysOfCode #DSA #Java #ProblemSolving #LearningInPublic #Consistency 🥷