🔥 LeetCode 15: 3Sum Explained! | Two-Pointer Approach | Java Solution with Step-by-Step Breakdown 🔥 скачать в хорошем качестве

🔥 LeetCode 15: 3Sum Explained! | Two-Pointer Approach | Java Solution with Step-by-Step Breakdown 🔥

🚀 3Sum is a fundamental LeetCode Medium problem that challenges your ability to find unique triplets in an array that sum to zero. This is a common question in FAANG coding interviews, testing both brute-force thinking and optimization skills. In this video, I will walk you through the complete approach, including: ✅ Understanding the Problem Statement (Why this problem is tricky) ✅ Brute Force vs. Optimized Approach (O(N³) vs. O(N²)) ✅ Sorting & Two-Pointer Strategy (Why sorting matters!) ✅ Skipping Duplicates (How to ensure unique triplets) ✅ Step-by-Step Implementation in Java ✅ Time & Space Complexity Analysis 💻 Full Code Implementation on GitHub: https://github.com/jsherbert95/LeetCo... 🔍 Problem Statement Recap: Given an integer array nums, return all unique triplets [nums[i], nums[j], nums[k]] where: nums[i] + nums[j] + nums[k] == 0 No duplicate triplets allowed Solution should be optimized beyond brute-force O(N³) 💡 Approach: Sorting + Two-Pointer Technique The brute-force approach involves checking all triplets (O(N³)), which is too slow for large inputs. Instead, we use: 1️⃣ Sorting the Array (O(N log N)) This helps with duplicate detection and enables two-pointer traversal. 2️⃣ Fixing One Element (nums[i]) and Finding Two Others (nums[left], nums[right]) Iterate i through nums.length - 2 Define left = i + 1 and right = nums.length - 1 Use a two-pointer strategy: If nums[i] + nums[left] + nums[right] == 0, we found a triplet If the sum is too small, move left++ If the sum is too large, move right-- 3️⃣ Skipping Duplicates Efficiently If nums[i] == nums[i - 1], skip to avoid duplicate triplets If nums[left] == nums[left - 1], skip to prevent repeating triplets If nums[right] == nums[right + 1], skip to remove duplicate results This results in an O(N²) solution, significantly faster than brute force. 📊 Time & Space Complexity Analysis ✅ Sorting the array: O(N log N) ✅ Iterating through the array (i): O(N) ✅ Two-pointer search (left and right): O(N) 🔹 Overall Complexity: O(N²) (Much faster than brute force O(N³)) 📌 Space Complexity: O(1), since we modify nums in-place O(N) for storing the results 🔗 Links & Additional Resources 📌 LeetCode Problem Link: 🔗 https://leetcode.com/problems/3sum/de... 📌 More Java Solutions & Tutorials: 🔗 GitHub Repo: https://github.com/jsherbert95/LeetCo... LeetCode Profile: https://leetcode.com/u/jsherbert95/