kurosawa4434 profile - python coding challenges

kurosawa4434

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Awesome Team Masayuki Hayashi

Last seen 40 minutes ago

Member for 8 years, 1 month, 24 days

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Tinus_Trotyl

Binair Interpretation-Tinus_Trotyl 3 1

It's elegant solution :) I think that using 'str.maketrans()' is more readable. Line 13: for word in code.translate(str.maketrans('.-', '01')).split(' ')).capitalize() More

Ylliw

1-liner-Ylliw 1 1

I have almost the same idea, and you are better :-) More

Merzix

Add turbo speed only for cases 12 < path < 21 and n == 5-Merzix 1 1

It is an interesting idea. But, How about the following case? assert g_key([[9, 9, 9, 9, 9], [9, 9, 9, 9, 9], [9, 9, 0, 0, 0], [9, 9, 0, 9, 9], [9, 9, 0, 9, 9]], 14) == 117 More

shankar294

First-shankar294 1

good solution! and I try more simple ( line : 2-12 ) words = [(ord(w[0])-96)*10 + int(w[1]) for w in list(pawns)] More

vit.aborigen

First-vit.aborigen 1

directions = [v for v in permutations([-2, -1, 1, 2], 2) if (abs(v[0]) + abs(v[1]) == 3)] Good :-D The Manhattan-distance of knight's movement is always 3 with 5x5 grid. More

Phil15

Short thanks to scipy.ndimage.label-Phil15 1 2

awesome solution. I used this solution to improve Visibility's `init.js`. More

HeNeArKr

Knapsack counter-HeNeArKr 1 1

Nice memoization. Line 30 equivalent the following: if not monster_ctr - lett_ctr: More

Tinus_Trotyl

First, concentric hexagon check-Tinus_Trotyl 1 1

I thought of another expression of Line 20-23: for center in {a+b for a in A[radius: -radius] for b in B[radius: -radius]} - obstacles: check = center More

kurosawa4434

First-kurosawa4434 1

Thank you very much for checking my code! I feel the code is redundant. But this is the most I can do now. >PEP8 violations (long lines) Thanks, I will check PEP8. More

HeNeArKr

Borrowing ideas-HeNeArKr 1 1

It is an interesting idea. But, How about the following case? assert g_key([[9, 9, 9, 9, 9], [9, 9, 9, 9, 9], [9, 9, 0, 0, 0], [9, 9, 0, 9, 9], [9, 9, 0, 9, 9]], 14) == 117 More

bsquare

Nice and documented solution with numpy-bsquare 1 1

good. Your result can be reduced. fix last code: return [int(n) for n in set(result) if result.count(n) % 2] More

ikaloveselina

First-ikaloveselina 1 1

nice solution. Python has a `defaultdict()` that has similar functionality to `append_to()`. from collections import defaultdict conn_hash = defaultdict(list) for a, b in lst: conn_hash[a].append(b) conn_hash[b].append(a) More

evsistr

sum(1 for d in DIGITS if...)-evsistr 1 1

Nice solution. another way of Line 16-17 : a = sum(lit[0] <= set(d) <= (lit[0] | broken[0]) for d in DIGITS) b = sum(lit[1] <= set(d) <= (lit[1] | broken[1]) for d in DIGITS) More