How does the container with most water problem use two pointers?

Place pointers at the leftmost and rightmost bars. Calculate the water area as the minimum of the two heights times the distance between them. Move the pointer at the shorter bar inward because moving the taller bar inward can only decrease or maintain the minimum height while also decreasing width. Moving the shorter bar has a chance of finding a taller bar that increases the constraining height enough to offset the lost width. Track the maximum area found. This runs in O of n time.