Timeline for Count the lucky tickets within the given range
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Jul 23, 2018 at 12:20 | comment | added | Jo King♦ |
If you're keeping it as a function, you can move the n=0 part into the header, like def x(a,b,n=0)
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| Jul 22, 2018 at 14:03 | comment | added | Jonathan Allan | FYI single spaces can work as indents too, so your 135 without any other golfing is really 114. | |
| Jul 22, 2018 at 13:54 | comment | added | Jonathan Allan |
90 bytes using an unnamed function (declared as a lambda) and a sum for the counting - Try it online!. 86 is also possible by use of a clever golf by Asone Tuhid - lambda a,b:sum(sum(map(int,str(i)))==2*sum(map(int,str(i%1000)))for i in range(a,b+1))
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| Jul 22, 2018 at 13:38 | history | edited | user8276 | CC BY-SA 4.0 |
Implemented feedback from comments
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| Jul 21, 2018 at 22:15 | comment | added | Jo King♦ |
You can't take a and b as pre-declared variables. You either have to have a function or take them via input
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| Jul 21, 2018 at 17:08 | comment | added | Asone Tuhid |
I'd suggest replacing n=n+1 with n+=1 and moving it right after the if statement (if...:n+=1)
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| Jul 21, 2018 at 17:07 | comment | added | Asone Tuhid | Welcome to PPCG! Check out Tips for golfing in Python for tips and tricks, there's a similar thread for most languages if you're interested. Also, it's good practice to include a TIO link as demonstration. | |
| Jul 21, 2018 at 15:43 | comment | added | Wheat Wizard♦ | You don't need the indentation after the if btw. Also it will probably be cheaper to convert to string before you take the first or last 3 digits. | |
| Jul 21, 2018 at 15:36 | review | First posts | |||
| Jul 21, 2018 at 16:18 | |||||
| Jul 21, 2018 at 15:34 | history | answered | user8276 | CC BY-SA 4.0 |