20
Ruby, 128 .. 51 49 bytes
->w{%w(o ~).map{|c|w.scan(/_*._*(.)/).count [c]}}
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No error checking.
How:
Ignore all underscores
All mice going towards the piper have the head in an odd-numbered (0-based) position.
All mice heading away from the piper have the tail in an odd-numbered position.
Take every 2nd character (skipping underscores), ...
16
sed, 12
s/.(.)/&\1/g
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13
Python, 45 43 38 27 bytes
For inputs up until 239999:
lambda n:n/100%100<60>n%100
You can try it online! Thanks @Jitse and @Scurpulose for saving me several bytes ;)
For inputs above 239999 go with 36 bytes:
lambda n:n/100%100<60>n%100<60>n/4e3
12
brainfuck, 13 or 11 bytes
>,[.,[..,>]<]
Assumes EOF->0. :(
You can try it online!
(Assuming input also continues to give zeroes indefinitely after the first EOF, we can do it in 11 bytes:
,[.,[..>],]
though this eats memory.)
11
Jelly, 3 bytes
ḤÐe
A full program accepting a string which prints the result
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How?
Utilises the fact that the implementation of the double atom, Ḥ, is multiplication by two, and that strings are lists of Python characters. In Python when a character is multiplied by two it becomes a Python string of length two (e.g. 'x'*2=='xx'). Lastly ...
10
Python, 38 bytes
f=lambda s:s and s[:2]+s[1:2]+f(s[2:])
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Takes the first two characters, then the second character, then removes the first two characters and recurses.
39 bytes
f=lambda s,c=2:s and s[:c]+f(s[1:],3-c)
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Flips between taking the first two or first one character.
10
x86-16 machine code, IBM PC DOS, 14 bytes
Binary:
00000000: b401 cd21 f7d9 78f8 b40e cd10 ebf2 ...!..x.......
Unassembled:
START:
B4 01 MOV AH, 1 ; DOS read char from STDIN
CD 21 INT 21H ; put char in AL
F7 D9 NEG CX ; toggle CX sign
78 F8 JS START ; if negative, loop to next char
B4 0E MOV AH, 0EH ...
10
Python 3.8, 75 68 97 63 bytes
lambda s,n:max(l:=[s[j:j+n]for j in range(len(s))],key=l.count)
You can try it online! Increased byte count because, as @xnor kindly pointed out, Python's str.count doesn't count overlapping substrings... But then @xnor's reformulation of what I was doing allowed to slice a third of the bytes!
How:
l:=[s[j:j+n]for j in range(...
8
J, 13 bytes
[:;_2|.\_3<\]
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Example input: 'flybuys'
_3<\] Box every group of 3
┌───┬───┬─┐
│fly│buy│s│
└───┴───┴─┘
_2|. Split that into groups of 2, and reverse each one:
┌───┬───┐
│buy│fly│
├───┼───┤
│s │ │
└───┴───┘
[:; Raze the result:
buyflys
8
05AB1E, 13 11 10 bytes
'_KāÈÏaTS¢
Port of @GB's Ruby answer, so make sure to upvote him!!
-1 byte thanks to @Grimmy by taking the input as character-list.
Try it online or verify all test cases.
Explanation:
'_K '# Remove all "_" from the (implicit) input-list
ā # Push a list in the range [1, length] (without popping the list)
È ...
8
Perl 6, 33 25 bytes
-7 bytes thanks to Kevin Cruijssen
60>*.polymod(100,100).max
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7
JavaScript (Node.js), 608 597 586 551 bytes
Takes input as (source)(input), where input is an array containing single characters and/or integers.
s=>I=>eval([..."@CDEJKLMNQTUWYZ^_bkmqw"].reduce((s,c)=>(a=s.split(c)).join(a.pop()),"(R=X=>(H={},o=P=[],S=[],z=x=p=i=0,gUs[p]?~(j=` \n`.indexOf(s[p++]N?j:gK:3,hUgK<2?Mn*2+j):V=n,GUx=...
7
Haskell, 33 24 bytes
By noticing that obviously f [] = [] also fits the pattern f s = s and removing one extra space, we get to
f(a:b:s)=a:b:b:f s
f s=s
from my starting point of
f []=[]
f (a:b:s)=a:b:b:f s
f s=s
Now my answer matches xnor's Haskell answer.
You can try it online!.
7
Perl 6, 24 bytes
{S:g/(...)(..?.?)/$1$0/}
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Simple regex substitution that swaps every pair of three and up to three.
7
JavaScript (Node.js), 80 66 65 58 57 bytes
Returns [towards, away].
s=>Buffer(s).map(c=>c%3?i^=~c:a[c&1]+=++i&1,a=[i=0,0])&&a
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How?
An efficient way to determine whether the character is a mouse body part is to take the ASCII code modulo \$3\$.
Quite conveniently, we can also distinguish between P and _ ...
7
Ruby, 98 95 94 bytes
Really simple string interpolation. I don't think there's all that much to optimize here that will save a lot of bytes.
Takes the byte count as a parameter in order to save 3 bytes, as per Kevin Cruijssen's suggestion.
->n,b,c,l{"# #{n}, #{b} bytes
#{c.gsub /^/,' '*4}
You can [try it online][tio]!
[tio]: "+l}
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7
05AB1E, 5 bytes
ŒIù.M
Try it online or verify the smaller test cases or verify the larger test case (which is a bit slow).
Explanation:
Π# Push all substrings of the (implicit) input-string
Iù # Only keep substrings of a length equal to the second input-integer
.M # Only keep the most frequent item of the remaining substrings
# (...
7
JavaScript (ES9), 88 85 82 bytes
Takes input as (string)(n).
s=>g=n=>n?s.replace(eval(`/\\b${s.match(/(?<=^| )./g).join``}/g`),`(${g(n-1)})`):s
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Commented
s => // s = string
g = n => // g is a recursive function taking the counter n
n ? // if n is not ...
6
J, 8 bytes
#~1 2$~#
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1 2$~# Shape $~ the list 1 2 (repeating it cyclically as needed) until it matches the length # of the input.
#~ Use this 1 2 1 2... "mask" to Copy #~ the implicit input elementwise. That is, make 1 copy of the first char, 2 copies of the 2nd char, 1 copy of the third char, etc.
6
Haskell, 24 bytes
f(a:b:t)=a:b:b:f t
f s=s
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6
Octave / MATLAB, 16 bytes
@(s)s(1:2/3:end)
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Explanation
Non-integer values are automatically rounded (with a warning) in colon index expressions.
6
APL (Dyalog), 41 38 bytes
-3 bytes thanks to Bubbler
2⌽'mehttp://',∊'.',¨⍨⎕A⍴¨⍨⌊1+62×.7*⍒⎕A
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Outputs the URL with the letters capitalised. Uses the 0 indexed formula \$ \lfloor 1 + 62 \times 0.7^{25-x} \rfloor \$, since \$ (\frac{10}{7})^{x-25} = ((\frac{7}{10})^{-1})^{x-25} = (\frac{7}{10})^{25-x}\$
Explanation:
2⌽ ...
6
05AB1E, 25 24 23 21 bytes
žXAS.7Ƶ-t∞-m>×….me¬ýJ
-1 byte thanks to @Neil's analysis that *(10/7)** is the same as /.7**.
-3 bytes thanks to @Grimmy using a different formula and ingenious use of ý!
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Explanation:
The formula used to get the correct amount of characters of the alphabet, where \$n\$ is the 1-based index of the alphabetic ...
6
Python 2, 47 bytes
Port of G B's Ruby answer.
lambda s:map(s.replace('_','')[::2].count,'~o')
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6
05AB1E (legacy), 56 53 52 51 50 bytes
“# ÿ, ÿ¡Ï“I4ú»"[tio]"D“You€© [try€•€Ø]ÿ!“s¶·ý„: IJ
Inputs in the order name, byte-count, [code-lines], link, where the [code-lines] is a list of lines.
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50 bytes alternative (credit to @Grimmy):
“:# ÿ, ÿ¡Ï“I4ú»“You€© [try€•€Ø][tio]!“Â6£R¨¶·ýÀ$ú«
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Explanation:
“# ÿ, ÿ¡Ï“ # Push ...
6
APL (Dyalog Extended), 33 32 bytes
-1 byte thanks to Bubbler.
Anonymous infix lambda. Takes count as left argument and string as right argument.
{('\b',⊃¨⍵⊆⍨≠⍵)⎕R(1⌽')(',⍵)⍣⍺⊢⍵}
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{…} "dfn"; number is ⍺ and text is ⍵:
⊢⍵ on the text:
⍣⍺ repeat given number of times
(…)⎕R(…) PCRE Replace:
')(',⍵ the text prefixed by ")("
1⌽ ...
6
Gema, 8 characters
<J>="$0"
Copy of Adám's QuadR solution, just adjusted the syntax and the explanation:
Replace
<J> one or more lowercase letters
with
"$0" the quoted match.
Sample run:
bash-5.0$ gema '<J>="$0"' <<< '[abc,def,ghi,jkl]'
["abc","def","ghi","jkl"]
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6
APL (Dyalog Extended), 19 bytesSBCS
-10 bytes thanks to Kevin Cruijssen.
Anonymous tacit prefix function. Takes argument as integer.
⍱59<100∘⊤
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100∘⊤ convert To base-100
59< are they, each, greater than 59?
⍱ are none of them true?
5
Python 2, 34 bytes (xnor's improvement)
lambda s:`[c*11for c in s]`[7::10]
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Python 2, 36 bytes
lambda s:`sum(zip(s,s,s),())`[7::10]
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Slices characters 7, 17, 27… out of a string like this:
('a', 'a', 'a', 'b', 'b', 'b', 'c', 'c', 'c', 'd', 'd', 'd', 'e', 'e', 'e')
...
5
Gema, 13 characters
<u3><u3>=$2$1
Sample run:
bash-5.0$ echo -n 'flybuys' | gema '<u3><u3>=$2$1'
buyflys
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Only top voted, non community-wiki answers of a minimum length are eligible
