Comments on: Another Quick Coding Challenge

By: OJ

OJ — Fri, 25 Jul 2008 03:23:18 +0000

So that generates the set of numbers right? It doesn’t pick the biggest difference and the number of elements?

the Haskell equivalent would probably be:

[ n | n <- [1..1000], length (show n) == length (dup (show n)) ]

OJ — Fri, 25 Jul 2008 03:23:00 +0000

So that generates the set of numbers right? It doesn’t pick the biggest difference and the number of elements?

the Haskell equivalent would probably be:

[ n | n <- [1..1000], length (show n) == length (dup (show n)) ]

Bill Mill — Fri, 25 Jul 2008 02:20:35 +0000

My answer in python was the same as commenter “DeWitt” on that site – I was disappointed when he chimed in before me:

(i for i in xrange(start, end) if len(str(i)) == len(set(str(i))))

Bill Mill — Fri, 25 Jul 2008 02:20:00 +0000

My answer in python was the same as commenter “DeWitt” on that site – I was disappointed when he chimed in before me:

(i for i in xrange(start, end) if len(str(i)) == len(set(str(i))))