Timeline for Is this word in standard order?
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
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| Feb 18, 2022 at 19:24 | history | edited | 12944qwerty | CC BY-SA 4.0 |
-4 bytes
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| Feb 18, 2022 at 18:57 | comment | added | pxeger |
50 bytes on the second one: lambda a,i=0:any(k^(i:=i+1)for k in{}.fromkeys(a)) (input \$ [1, 26] \$)
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| Jan 21, 2022 at 13:56 | history | edited | 12944qwerty | CC BY-SA 4.0 |
wow
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| Jan 21, 2022 at 13:52 | comment | added | 12944qwerty | @pxeger Wow.... i thought it couldn't get any smaller :O | |
| Jan 20, 2022 at 20:49 | comment | added | pxeger | down to 47 bytes by combining both answers | |
| Jan 20, 2022 at 20:42 | comment | added | pxeger |
Another -1 on @m90's version (outputs inverted): lambda a:0in map(int.__eq__,{}.fromkeys(a),range(len(a)))
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| Jan 20, 2022 at 20:11 | history | edited | 12944qwerty | CC BY-SA 4.0 |
deleted 60 characters in body
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| Jan 20, 2022 at 20:07 | comment | added | pxeger |
I just realised you can remove both sets of [] from your original answer, which makes it shorter overall: lambda a:any(i^j for i,j in enumerate({}.fromkeys(a)))
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| Jan 20, 2022 at 20:02 | history | edited | 12944qwerty | CC BY-SA 4.0 |
better solution
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| Jan 20, 2022 at 19:54 | comment | added | m90 |
Improvement to 58: lambda a:all(map(int.__eq__,{}.fromkeys(a),range(len(a))))
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| Jan 20, 2022 at 19:50 | comment | added | pxeger |
If you change i==j to i^j and change all to any, you can save a byte (output becomes inverted, but that's allowed)
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| Jan 20, 2022 at 19:48 | history | edited | 12944qwerty | CC BY-SA 4.0 |
added 34 characters in body
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| Jan 20, 2022 at 19:45 | comment | added | pxeger |
dict.fromkeys can be {}.fromkeys: see this tip of mine
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| Jan 20, 2022 at 19:19 | history | answered | 12944qwerty | CC BY-SA 4.0 |