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Challenges

Mediocre pop count

+3
−0

Given a sequence of letters, omit those with the highest or lowest pop count.

Terminology

The pop count or population count of a binary string is the number of 1s in it.

For this challenge, the pop count of a letter is the number of 1s in its ASCII character code in binary.

For example, the letter "Z" has ASCII character code 90, which is 1011010 in binary. There are 4 1s in this binary string, so the pop count of "Z" is 4.

Here is the pop count of every letter, in the format letter : pop count:

A : 2
B : 2
C : 3
D : 2
E : 3
F : 3
G : 4
H : 2
I : 3
J : 3
K : 4
L : 3
M : 4
N : 4
O : 5
P : 2
Q : 3
R : 3
S : 4
T : 3
U : 4
V : 4
W : 5
X : 3
Y : 4
Z : 4
a : 3
b : 3
c : 4
d : 3
e : 4
f : 4
g : 5
h : 3
i : 4
j : 4
k : 5
l : 4
m : 5
n : 5
o : 6
p : 3
q : 4
r : 4
s : 5
t : 4
u : 5
v : 5
w : 6
x : 4
y : 5
z : 5

Input

  • A sequence of letters
  • This may be a string or any ordered data structure of characters
  • The letters may contain a mixture of upper and lower case
  • The characters that count as letters for this challenge are A-Z and a-z, specifically ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz
  • The input may be empty (it may have length zero)

Output

  • A sequence of letters
  • This may be a string or any ordered data structure of characters. It does not need to match the input format (provided it is consistent between inputs)
    • For example, you may take input as an array of characters, and output as a string, provided this does not change for different inputs
  • All of the letters from the input are included in the output in the same order, except for those with the highest or lowest pop count, which are omitted
  • If all of the letters have either the highest or lowest pop count, the output is empty (a sequence of length zero)
  • Case must be maintained
    • a letter that is lower case in the input must be lower case in the output (if present)
    • a letter that is upper case in the input must be upper case in the output (if present)

Test cases

Test cases are in the format "input" : "output"

"" : ""
"a" : ""
"ab" : ""
"abc" : ""
"abcd" : ""
"abcde" : ""
"abcdef" : ""
"abcdefg" : "cef"
"A" : ""
"AB" : ""
"ABC" : ""
"ABCD" : ""
"ABCDE" : ""
"ABCDEF" : ""
"ABCDEFG" : "CEF"
"oAPwSoeHcoDretMBBoesoBsagHDoew" : "SecretMessage"
"SecretMessage" : "SecretMee"

Explanations are optional, but I'm more likely to upvote answers that have one.

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6 answers

+1
−0

Vyxal, 12 bytes

bṠ:₍gGvc†Tİ∑

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          İ  # Index into input
         T   # Indices where
      v      # For each
 Ṡ           # Sum of
b            # Binary digits
   ₍gG       # Does [min, max]
        †    # Not
       c     # contain
  :          # That value?
           ∑ # Concatenate into a single string
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+1
−0

APL (Dyalog APL), 61 bytes

{⍵/⍨~(⊢∊⌊/,⌈/)+⌿0 1↓2⊥⍣¯1⊢0,⎕UCS⍵}

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APL's style of filter works very well here, since we can check for the max and min elements here:

(⊢∊⌊/,⌈/)

and filter using the boolean mask later:

⍵/⍨~
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+1
−0

J, 46 char

Solution:

(]#~[:([:*./(<./,>./)~:"0 1])([:+/"1@#:a.&i.))

Test example:

   (]#~[:([:*./(<./,>./)~:"0 1])([:+/"1@#:a.&i.)) 'oAPwSoeHcoDretMBBoesoBsagHDoew'
SecretMessage

How it works:

a.&i. converts from text to ascii number
#: converts from ascii number to binary
+/"1 adds the 1s in the binary numbers
(<./,>./) finds greatest and lowest values in lists
 *./(<./,>./)~:"0 1 produces boolean indicating locations of non-greatest and non-lowest values
#~ selects from original text locations indicated by boolean
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+1
−0

Japt, 13 bytes

ÆüÈc¤ñÃé ¤¬øX

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ÆüÈc¤ñÃé ¤¬øX     :Implicit input of string U
Æ                 :Filter each character, X by
 ü                :  Group & sort U by
  È               :  Passing each character through the following function
   c              :    Charcode
    ¤             :    To binary string
     ñ            :    Sort
      Ã           :  End grouping
       é          :  Rotate right
         ¤        :  Slice off the first 2 elements
          ¬       :  Join
           øX     :  Contains X?
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+0
−0

Python, 89 bytes

lambda l,g=lambda c:ord(c).bit_count():[i for i in l if min(map(g,l))<g(i)<max(map(g,l))]

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+0
−0

JavaScript, 104 102 99 bytes

I/O as a character array.

a=>a.filter(x=>a<g(x)&g(x)<a[~-a.length],a=a.map(g=x=>[...Buffer(x)[0].toString(2)].sort()).sort())

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