To solve this problem efficiently, you should try to do a single pass through the array. The idea is to use 2 pointers which represent the walls of a bucket. The pointers start on the left and right sides of the array.

The trick is to realize that you will never get a larger area of water by moving the taller pointer inwards, like this:

This means you should only move the shortest of the 2 pointers inwards - this is the only way to get a larger area. You can solve the problem by repeatedly moving the shortest pointer inwards, until the pointers cross.

Here's an example of moving shorter pointer inwards:

To code this up, you have to start a pointer on the left and right sides of the array. You should keep moving the shortest pointer inwards, until the 2 pointers cross. Each time you move a pointer, you have to update the max area you've seen so far maxArea. See the spoiler for details on how to compute the amount of water in a bucket.

Open Spoiler

The water in the bucket is always in the shape of a rectangle. This means we can get the area of water in the bucket by multiplying its height and width area = height*width.

The water's height will go up to the shortest of the walls of the bucket min(arr[i], arr[j]). The width will be equal to j - i.

Here's the code: