34
Last seen 2 years ago
Member for 11 years, 5 months, 27 days
Difficulty Normal
Well, your solution is O(n) but need unnecessary treatment, I think.
At least, why do you make 2 Counters?
May I suggest this following, very similar to yours but with less memory used:
def checkio(data):
from collections import Counter
count = Counter(data)
return [n for n i
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Changing the beginning of the year is a nice idea.
Maybe extracting a function to compute days would be nicer:
def days_diff(date1, date2):
def days(date):
y, m, d = date
correction = m <= 2
y -= correction; m += 12*correction - 3
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To my mind, this solution is not really creative.
Moreover, it presents several flaws:
* Use python unpacking with * operator instead of enumerating indexes: _date(*date1)_
* As pedrobueno mentions, *timedelta.days* is useful here.
* Why 2 _return_ statements? Last one is useless (as extra variabl
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Python 3 annotations, very nice!
Just to mention it to future readers ;-) :
Switch lines 11 and 12 and you have a solution that runs 20x faster, rather than this one which runs in more than 1s on my slow (very slow) computer!
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I definitely like the representation you have made, keeping stack and moves in same list; even better with the use of `slice`.
However, I can't tell I'm fond of this list/''.join step on tuples. If you wanted to introduce some performance with the use of `deque`, I have the feeling that looping two
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Again a wrong solution on left-down diagonal:
Tested with:
checkio([[1, 5, 4, 4, 3],
[2, 2, 4, 1, 5],
[1, 4, 3, 5, 4],
[4, 3, 3, 2, 5]])
Correction is in line 20:
if (x+4 <= l_x and y+1 >= 4 and
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And certainly possible to go down by changing type test.
Edit: Well, Veky's idea breaks down character count: 'd*0==0'
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Simple approach (not as ugly as you said!)
Just to give you a more pythonic implementation:
def checkio(number):
pig = newpigs = 1
fooddif, food = 0, number
while food:
oldfood, food = food, max(food - pig, 0)
fooddif = max(fooddif, ol
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Easier: the most frequent day is always the first day of the year.
Second day of the year has the same frequency than the first one when its year is leap.
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Such exotic lib with use of recent RecursionError, I bet this solution is Veky's!
Nice union-find algorithm, although handling cycles by exceeding max depth recursion limit seems a bit inefficient (and wrong on very big graph - 1000 edges, isn't it? - but I don't hope such big tests at CheckIO :-P)
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Nice solution (quite the same on my own, except I reasoned on k-1).
Just a bit surprised by last line ; why not simply ;-) :
return max(t, number - s)
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This is effectively a DFS. You can search for Dijkstra's algorithm if you want to learn more about.
Just a few things:
lines 8-14: you can factorize these lines for 'clearer' code
pair = drones.split("-")
for index, drone in enumerate(pair):
if node == drone and pair[index - 1] no
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This code searches through _text_ 26 times - it is not what I call a "Speedy" solution.
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Smart way to "limit" your grid but here you should use 'infinite' grid by memoizing live cells, truly easier.
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Better than first one, really more readable!
But still a niptick: imagine you have 2 live cells (0,0) and (100, 100).
Are you really gonna loop through 10,000 (-2) dead cells for nothing?
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I'm glad you chose to code such a mathematical masterpiece.
I didn't have the courage to write it down !
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for golfing, here is your solution in 65 characters ;)
check_pangram=lambda t:len(set(filter(str.islower,t.lower())))>25
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Strings are also iterable, so you can use _for_ loop and _in_ operator on them directly.
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'd*0==0'... Amazing, really.
By the way, 1 more character to delete: space between 0 and 'and' ;)
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