3Sum LeetCode Challenge: Sorting & Two-Pointer Approach

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

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 24/50 – LeetCode Solved the Reverse String problem using the two-pointer approach. The solution involves swapping characters from both ends of the array and moving inward until the pointers meet. This ensures an in-place reversal with O(1) extra space and O(n) time complexity. A straightforward problem that reinforces fundamental concepts. #LeetCode #50DaysOfCode #Java #DataStructures

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 9/100 – LeetCode Challenge Problem: 3Sum Today’s challenge was to find all unique triplets in an array whose sum equals 0. Approach: Sort the array to efficiently apply the two-pointer technique. Fix one element nums[i] and use two pointers (left and right) to find the remaining two numbers. If the sum equals 0, store the triplet. Skip duplicate elements to ensure only unique triplets are included. Key Idea: Sorting + Two Pointers helps reduce the brute-force O(n³) approach to O(n²). Concepts Practiced: Sorting Two-pointer technique Duplicate handling Array traversal optimization #100DaysOfCode #LeetCode #DSA #Java #TwoPointers #ProblemSolving #CodingJourney

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 50 – Decode String Worked on decoding encoded strings with patterns like k[encoded_string], including handling nested structures using stacks. Key Learnings: Learned how to use stacks to manage nested patterns effectively Understood how to process numbers, brackets, and characters step by step #DSA #Java #Stack #Strings #ProblemSolving #CodingPractice

🚀 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

🚀 Day 24 of #100DaysOfCode Solved 167. Two Sum II – Input Array Is Sorted on LeetCode 🔢👉👈 🧠 Key insight: Because the array is already sorted, we can avoid hash maps and use a two-pointer approach to find the target sum efficiently. ⚙️ Approach: 🔹Initialize two pointers at the start and end of the array 🔹Calculate the sum of values at both pointers 🔹If the sum is too large → move the right pointer left 🔹If the sum is too small → move the left pointer right 🔹Stop when the target is found ⏱️ Time Complexity: O(n) 📦 Space Complexity: O(1) #100DaysOfCode #LeetCode #DSA #TwoPointers #Java #ProblemSolving #LearningInPublic #CodingJourney