I don't think you have it. Say you have an input like this: [3,2,100,100,2,3] So you start on the outsides working in with your method, and you will stop at t

I don't think you have it. Say you have an input like this: [3,2,100,100,2,3] So you start on the outsides working in with your method, and you will stop at the two outermost 3s, because the 2 is less area, and never make it to the inner 100s. That's why the actual algorithm is to keep moving the pointers in until they meet, in a very specific way, keeping track of the maximum the whole time. It's not about actually finding the maximum and stopping early, it's about reducing the search space to O(n) from the naive O(n^2). EDIT: However! There is a condition where h[n] <= len(h), which actually makes your method work! But you never mentioned that, so I think you may have gotten it by accident. :D Like, with that condition, it becomes something like this: [1,1,6,6,1,1] and then stopping at the outer 1s is equivalent to making it in to the 6s. I guess my issue is that I didn't example the conditions well enough. There's was a huge clue in there!