LeetCode 11 - Container With Most Water Explained in C++ | Two Pointer Approach

In this video, we solve LeetCode Problem 11: Container With Most Water using C++. We cover: Two Pointer Technique Greedy Thinking Area Calculation Logic Step-by-step Dry Run Optimized O(n) Solution Time and Space Complexity Analysis Example Input: height = [1,8,6,2,5,4,8,3,7] Output: 49 Because: The lines with heights 8 and 7 form the container that holds the maximum amount of water. Key Idea The area is calculated as: Area = min(height[left], height[right]) × width Instead of checking every pair, we use two pointers: One at the beginning One at the end At each step: Calculate the current area Move the pointer with the smaller height This greedy observation reduces the solution from O(n²) to O(n). Problem LeetCode 11 - Container With Most Water Language Used C++ What You'll Learn ✅ Two Pointer Pattern ✅ Greedy Optimization ✅ Area Calculation Logic ✅ Interview Problem-Solving Techniques If you enjoyed the video, make sure to: 👍 Like 💬 Comment 🚀 Subscribe More LeetCode videos coming daily! #leetcode #cpp #dsa #twopointers #greedy #arrays #programming #codinginterview #leetcode11 #containerwithmostwater #cplusplus #algorithms #interviewprep