Python 3, 43 bytes

lambda x:sorted(x,key='tibdfghklpqyj'.find)

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Sorts based on index in a sorted string. Inputs as a string and outputs as a list.

-14 bytes by replacing .index() with .find()

posted about 2 years ago

CC BY-SA 4.0

2y ago

dino‭

121 reputation 0 10 12 31

Vyxal, 12 bytes

µ«,←⋎„¶ɽ₌Ż«ḟ

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Outputs as a list of characters. Sort by index in (compressed) tibdfghklpqyj.

posted about 2 years ago

CC BY-SA 4.0

steffan153‭

121 reputation 0 8 12 6

Ruby, 53 51 bytes

->i{i.chars.sort_by{"tibdfghklpqyj".index(_1)||-1}}

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Works in Ruby 2.7 and Ruby 3.

Explanation

• ->i{...} is a short way to define a 1-argument lambda with parameter i
• .chars will turn a string into an array of characters
• .sort_by will do a normal sort, transforming each element for sorting purposes to whatever the block returns.
• _1 tells ruby that my block actually wanted a variable and that I'm using it at that location.
• a.index(b) returns the index of a in b. It returns nil when not found.
• a||-1 if a is falsy, make it -1 instead. (And falsy is much better defined than in JavaScript, this will only turn nil and false into -1)

Fun facts

In Ruby parentheses are optional, so I would have almost been able to do .index _1||-1. Unfortunately, the precedence of || is such that this now works on the argument (_1) rather than the output of the whole function. One could use or since it has different precedence than ||, but this doesn't save space because you now need an extra space. So all this yields us an equally long but possibly more convoluted:

->i{i.chars.sort_by{"tibdfghklpqyj".index _1 or-1}}

posted over 1 year ago

CC BY-SA 4.0

1y ago

Taeir‭

21 reputation 0 1 2 3

Japt, 23 16 15 13 bytes

Lexicographical sort using the custom alphabet tipbdghkflyqj. I/O as characters arrays, sorts in ascending order.

n`“pbdghkf§qj

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posted about 2 years ago

CC BY-SA 4.0

1y ago

Shaggy‭

2131 reputation 4 116 216 166

C (gcc), 152 bytes

r;e(c){r=c=='j'?5:strchr("bdfghklpqy",c)?4:c=='i'?3:c=='t'?2:1;}c;s;f(char*i,char*o){s?(e(*i),c=r,e(*o)):(qsort(i,strlen(i),s=1,f),puts(i));return r-c;}

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Output:

a
aa
jbita
kitc
now
jtus
jpyitzx

I didn't fine tune it much, but it's delightfully crazy :)

Explanation:

• The function f takes one input string and one output string as parameter. It returns int as per gcc's old C90 default behavior.
• Since in-place modification of the input should be fine according to Codidacts input/output rules, I actually never use the output but modify the input. (Would crash & burn in case of string literals though, but the challenge only said strings, so I skipped the copy from input to output buffer.)
• The global variable s gets zero-initialized and is used to mark the function's state. If not set, then it's the first call. If set, then the function has been called before. (This will have to be reset to zero externally, in order to execute multiple test cases in a row from main())
• s? checks if s is zero, if so execute qsort and then print.
• qsort uses the very same function f as its own callback to save a function. This abuses the fact that const void* and char* likely have the same representation. And the function returns int so it fulfils the requirements of a qsort callback. Curiously, qsort didn't raise a compiler error when passed the wrong type of function pointer (this doesn't sound conforming at all, gcc...).
• When qsort is called, the size of one item parameter also sets the state s to 1.
• f gets called by qsort many times and now the s? evaluates to 1. It calls the helper function e to evaluate both parameters (i and o are now the qsort callback parameters).
• e sets a global variable r to value 5,4,3,2 or 1 depending on how which set of characters the passed character belongs to.
• No benefit from typing out decimal values of these characters since they are all above ASCII value 100, might as well use 'a' character constants then.
• The result r from the first call to e is stored in c. Then after the next call r-c is returned as the result of qsort callback. This weighs 'j' as lowest, swap to c-r to get the other way around.
• The function actually always returns r-c but when s==0 nobody cares since the result is placed in the i buffer anyway.
• When qsort is done, we are back to the first execution of the function, so puts is called once before finishing.

posted about 2 years ago

CC BY-SA 4.0

2y ago by trichoplax‭

Lundin‭

1149 reputation 17 43 134 70