[igor47]

you could solve this with a simple state machine; there's no reason to resort to parlour tricks...

https://gist.github.com/igor47/7228586

[acchow]

Your solution doesn't work when there is more than one "puddle":

https://gist.github.com/alexchow/7228746

Try the case: ([2,5,1,2,3,4,7,7,6,3,5], 12)

[igor47]

it's not "more than one puddle"; it's that i don't count the second puddle because i don't ever "close" it by encountering it's level again. if i go off the edge, i should close it as best i can.

[acchow]

Ah, right. Now that I actually read your code, that's the real bug :p

The issue with this "state machine" approach is you don't have "lookahead".

In the two pass approach (approach from one side, then the other), you essentially have two machines, one implicitly serving as the "lookahead" for the other.

[igor47]

oops, sorry! i meant the "parlour tricks" part in response to this comment: https://news.ycombinator.com/item?id=6640033

not in response to the author!

[huhalu]

try this test case ([5,1,0,1],1)

[ledave123]

Try the list(reversed()) of the cases and this finds more bugs:

MISMATCH: [6, 7, 7, 4, 3, 2, 1, 5, 2] holds 10 (got 0) MISMATCH: [5, 3, 6, 7, 7, 4, 3, 2, 1, 5, 2] holds 12 (got 2)

Another mismatch: MISMATCH: [2, 0, 1] holds 1 (got 0) TRUE: [1, 0, 2] holds 1

[igor47]

thanks! i updated the gist (https://gist.github.com/igor47/7228586/revisions), but the solution is not nearly as simple as before.

[lelf]

Fails for [3,0,1,0,2]