Yes @seeker001 . The logic you are using is perfect. You can also use **Debrief** section to verify. Inorder to get the max area, either we have to increase the width or we have to increase the height. We already start with maximum width so we can’t increase width anymore. But inorder to increase the area we keep on changing the lower side (either of start or end) till both crosses each other.

Now, let’s understand why this will always work. Assume Height at index 0 is 5 and index 7 is 15. Right now, how much water we can hold by considering just this 2 heights? We can hold `(end-start)*min(height[start], height[end]) = (7-0) * min(5,15) = 70`

. Now think will it help to get any other heights even more than 15 by keeping the same height of another line that is 5 here? The answer is No it won’t help as it does not matter as we can only save up to `min(height[start], height[end])`

height and extra water will fall away. Now think will it help to find any other line with height more than 5 and keep 15’s line the same? The answer is it might be helpful if its height is more than 5 then we may save more water as `min(height[start], height[end])`

will be more now.

I hope the explanation is clear now. If not try it with some examples on the paper to understand better. Feel free to post again if it is still unclear or needs more explanation.