31
Sawako Ohno
Last seen 2 months ago
Member for 2 years, 12 months, 3 days
Difficulty Normal
Elegant solution, using the comparison of tuple; in tuple comparison if first element is same, then second element is compared.
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Thank you for your very clear and educative code ! The variable names clearly stand for what they mean and helped me understand the process.\
Can I ask why "field, dn, pop = q.popleft() " not "field, dn, pop = q.pop()" ?
I naively thought, if more filled version of "field" is added on the right, p
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Cool idea to use sum of quotient and remainder from divmod, instead of round or ceil.
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Learned about "partial" for the first time here.\
Also, I was not aware that \n expression can be reused across arguments:\
\1 in the second argument in re.sub() here referring to \w in the first argument.
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Like the combination of bool and re.search.
return bool(re.search(r"\d",password) and re.search(r"\D",password))
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When the number of digits after the decimal point is 1 it does not work. (Such as "$5,1")\
It should be like this to include those cases.
"\$(\d{1,3})(\.\d{3})*(,\d{1,2}(?!\d))?"
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In the case below, right move for "X" is (0,1)or(1,0)or(1,2)or(2,1) even 2 corners are still blank.\
(If you put "X" on (0,2), enemy puts "O" on (2,0) to prevent your win and it creates two-checks situation for "O", which means enemy's win.
x_and_o((
"0..",
".X.",
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14-liner: simple-przemyslaw.daniel
Learned to use complex number for two-dimension from this code.
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Cool use of divmod().
Also, cool use of class-producing function "make_broken(i)".
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Very smart in setting the range of y in (2len(M) - 1) and check if (y-x) in range.\
Mine was much longer as I produced 4 Diagonal patterns starting from each corner.
>[[M[y-x][x] for x in range(len(M)) if 0 <= y-x < len(M)] for y in range(2*len(M) - 1)]
Also smart here. Much more efficient to flip
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5-liner: easiest-przemyslaw.daniel
1
It seems that argument can be > 100.
In that case, it should be ['A+']*61 and min(percent,150) ?
Precondition: argument can be from 0 to 150
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I like the clearness of your code, but it seems that there are some edge cases in which you just cannot paint all countries with 4 colors with the strategy to start from country 0 and choose the smallest available color-id for the next country.
color_map(((10,10,10,10,10,2,2,2,2,),
(10,10,1
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