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You can cut a lot of the boilerplate in your code by using existing functions (apologies if I get the argument orders wrong): def best_digit_choice_to_maxi

You can cut a lot of the boilerplate in your code by using existing functions (apologies if I get the argument orders wrong): def best_digit_choice_to_maximize_product(n, digits): numer = lambda d: reduce( sorted(d, reverse=True), 0, lambda a, e: a*10 + e) return max( itertools.combinations(digits, n), key = lambda e: numer(e) * numer(set(digits) - set(e)))

jmduke 12y ago
I don't mean this as a criticism -- as you are absolutely correct, your code does cut out a lot of the boilerplate and makes for a more compact function -- but I vastly prefer @cousin_it's function to yours in terms of readability. I think Python is a very pretty language but I always have a hard time reading Python code when it's heavy on the functional paradigms.
cousin_it 12y ago
My code runs in O(n log n) time and can deal with repeated digits. You converted it to something that takes exponential time and chokes on repeated digits. (BTW, I just realized that my code can be made O(n), by replacing the default sort with a counting sort :-))
- Strilanc 12y ago
  Hm, embarrassing, I just assumed you were doing the naive brute force. You're de-interleaving the digits with an order swap when the digits first start to differ.
  - cousin_it 12y ago
    Yeah, that's pretty much it.
thomasahle 12y ago
It's nice to have this version for testing the faster, more complicated one from OP. I can attest that they indeed produce the same answers. Two fixes: - If you use `(Counter(digits)-Counter(e)).elements()` in the last line, you can support repeated digits. - Reduce is `reduce(function, sequence[, initial]) -> value` so you should move the lambda to the first argument. All in all I think this is a very nice, succinct use of Python. The combinatorial parts of `itertools` are extreamly handy :)

jmduke 12y ago

I don't mean this as a criticism -- as you are absolutely correct, your code does cut out a lot of the boilerplate and makes for a more compact function -- but I vastly prefer @cousin_it's function to yours in terms of readability. I think Python is a very pretty language but I always have a hard time reading Python code when it's heavy on the functional paradigms.

cousin_it 12y ago

My code runs in O(n log n) time and can deal with repeated digits. You converted it to something that takes exponential time and chokes on repeated digits. (BTW, I just realized that my code can be made O(n), by replacing the default sort with a counting sort :-))

Strilanc 12y ago
Hm, embarrassing, I just assumed you were doing the naive brute force. You're de-interleaving the digits with an order swap when the digits first start to differ.
- cousin_it 12y ago
  Yeah, that's pretty much it.

thomasahle 12y ago

It's nice to have this version for testing the faster, more complicated one from OP. I can attest that they indeed produce the same answers. Two fixes: - If you use `(Counter(digits)-Counter(e)).elements()` in the last line, you can support repeated digits. - Reduce is `reduce(function, sequence[, initial]) -> value` so you should move the lambda to the first argument. All in all I think this is a very nice, succinct use of Python. The combinatorial parts of `itertools` are extreamly handy :)