I hate broccoli!

Question

My mom really wants me to eat broccoli, but I hate it and never want to eat it. Mom wants to trick me so she cuts it into small pieces, puts it in a salad and mixes it. Help me find out if the salad contains broccoli!

Input:

Array of mixed letters of all the ingredients. For a single broccoli ingredient it could be:

[c,o,r,b,l,c,i,o]

Output:

Boolean or other convienient output allowing to distinguish if a salad contains broccoli or not.

Rules:

• all possible ingredients are onion, broccoli, celery and beans
• input array is always formed of mixing valid ingredients (meaning that it can be always decoded back to the original ingredients array)
• input array cannot be decoded into more than one valid ingredient combination
• each ingredient can occur 0 or more times in the salad

Sample test cases:

true stands for a salad with broccoli

[c,o,r,b,l,c,i,o] //true
[o,b,n,o,i,i,e,l,n,a,o,r,n,s,b,o,c,c] //true
[l,c,b,r,r,o,r,c,i,i,c,o,c,b,o,c,c,b,i,o,l,o,l,o] //true

[] //false
[o,n,i,o,n] //false
[b,n,s,i,a,o,n,b,a,s,e,e,n,o,n] //false
[b,r,o,c,c,o,l,i,e,y,e,r,l,a,e,e,n,s,y] //false
[e,n,s,o,e,n,i,o,a,o,o,e,n,n,e,l,n,l,e,o,e,r,c,l,o,y,i,r,y,r,y,b,e,l,n,n,c,c,r,c,e,y,i,e] //false

This is code-golf so the shortest code in bytes wins!

Answer 1

Python 2, 34 32 bytes

-2 bytes thanks to @dingledooper

lambda a:cmp(*map(a.count,'bs'))

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Returns 1 for true, 0 for false

Answer 2

J, 14 bytes

[:=/1#.'cy'=/]

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Idea of invariant based on letter counts from mathjunkie's clever answer -- be sure to upvote him.

• 0 for contains broccoli
• 1 for not

Tests if the number cs is equal to the number of ys. This will only be true when no broccoli is present.

Answer 3

perl -ple, 14 bytes

$_=y-y---y-c--

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Prints 0 if no broccoli is present, something else if there is.

Answer 4

MATLAB/Octave, 26 bytes

@(b)b&&diff(sum(b'=='cl'))

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Credit to @mathjunkie's answer for the idea. Interestingly comparing c and l counts also works because celery has one of each, but broccoli has a different number.

MATLAB likes to expand == into 2D arrays if the inputs are vectors in differing orientations. However this doesn't work with empty arrays. Fortunately empty is false in MATLAB and non-empty with any non-zero elements is true, so we can simply use the input as a logical true-false to capture that case.

Answer 5

05AB1E, 7 6 bytes

„bsS¢Ë

-1 byte by reversing the output: will output 0 if it contains broccoli and 1 if not.

Try it online or verify all test cases.

6 bytes alternative which outputs 1 if it contains broccoli and 0 if not:

AS¢üÊн

Try it online or verify all test cases.

Explanation:

„bs     # Push string "bs"
   S    # Convert it to a character-list: ["b","s"]
    ¢   # Count each in the (implicit) input
     Ë  # Check if the counts are the same for both letters (1 if truthy; 0 if falsey)
        # (after which it is output implicitly as result)

A       # Push the lowercase alphabet: "abcdefghijklmnopqrstuvwxyz"
 S      # Convert it to a character-list: ["a","b","c",...,"z"]
  ¢     # Count each in the (implicit) input-list
   ü    # For each overlapping pair of counts:
    Ê   #  Check that they are not equal to one another (1 if truthy; 0 if falsey)
     н  # Pop and only leave the first check (for letters 'a' and 'b')
        # (after which it is output implicitly as result)

If the amount of b and s/a are the same, it means the input only contains beans and no broccoli.

Answer 6

APL (Dyalog Classic), ¹⁹ 12 bytes

>/+/↑'cl'=⊂⍞

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Note that you have to add ⎕← to the beginning for it to output. Originally created with ngn/apl.

>/+/↑'cl'=⊂⍞
            ⍞ ⍝ get input
     'cl'=⊂   ⍝ make two packed vectors: characters equal to c and l
  +/↑         ⍝ turn into two row matrix and add (to get number of 'c's and 'l's
>/            ⍝ Are there more 'c's than 'l's?

returns broccoli: 1, no broccoli: 0

Answer 7

JavaScript (ES6),  33  32 bytes

Saved 1 byte by not forcing a Boolean value, as suggested by @SteveBennett

Takes a string and returns undefined (falsy) for a broccoli-free salad (yummy!) or a string (truthy) otherwise (yuck!).

s=>s.split`b`[s.split`s`.length]

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Answer 8

Charcoal, 7 bytes

›№θb№θs

Try it online! Link is to verbose version of code. Outputs a Charcoal boolean; - for broccoli, nothing for none. Explanation:

 №θb    Number of `b`s in input
›       Exceeeds
    №θs Number of `s`s in input
        Implicitly print

Answer 9

JavaScript (SpiderMonkey), 32 bytes

s=>s.map(c=>++this[c],r=y=0)|r>y

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JavaScript (Node.js), 37 bytes

s=>s.map(c=>n+=~~{r:1,y:-1}[c],n=0)|n

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Answer 10

K (oK), 11 bytes

Solution:

</+/"sb"=/:

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Explanation:

Shamelessly stolen "more s' than b's" logic from mathjunkie:

</+/"sb"=/: / the solution
    "sb"=/: / "sb" equal (=) each-right (/:)
  +/        / sum
</          / s less than b? (aka no brocoli?)

Answer 11

Zsh, 22 characters

((${#1//a}==${#1//b}))

Expects the ingredients as a single command line parameter.
Sets the exit code to 1 for broccoli alert and 0 for no broccoli.

(Just to keep the exit code semantics of 0 = Ok / non-0 = error. Using < for comparison would save 1 character but reverse the result's encoding.)

Sample run:

manatwork@manatwork ~ % . ./broccoli-detector.zsh obnoiielnaornsbocc
manatwork@manatwork ~ % echo $?
1
manatwork@manatwork ~ % . ./broccoli-detector.zsh broccolieyerlaeensy
manatwork@manatwork ~ % echo $?                                      
0

Try it online! / Try all test cases online!

Answer 12

Retina 0.8.2, 18 bytes

+`[^bs]|bs|sb

1`b

Try it online! Link includes test cases. Explanation:

+`[^bs]|bs|sb

Delete everything that's not a b or an s, plus any remaining bs or sb pairs that result from bean salad.

1`b

Check whether there is any broccoli.

Answer 13

Jelly,  7  6 bytes

ċⱮ⁾abE

A monadic Link accepting a list of characters which yields 0 if the salad contains broccoli and 1 if not.

Try it online! Or see the test-suite.

How?

Tests whether the number of 'b's (in both 'broccoli' & 'beans') is equal to the number of 'a's (in only 'beans').

ċⱮ⁾abE - Link: list
 Ɱ     - map across...
  ⁾ab  - ...what: list of characters = ['a', 'b']
ċ      - ...applying: count
     E - all equal?

Answer 14

Java (JDK), 41 bytes

s->s.reduce(0,(a,c)->c<98?a-1:c<99?a+1:a)

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Uses a chars()-Stream as input, returns any strictly positive value if it contains brocoli.

Credits

• -5 bytes thanks to Elgirhath by comparing b to a instead of b to s, and by allowing any positive value instead of a strict boolean.

Answer 15

asm2bf, 73 bytes

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Takes input on stdin as a string; outputs either ASCII(0x00) or ASCII(0x01).

@l
in r1
ceqr1,.b
cadr2,1
ceqr1,.s
cadr3,1
jnzr1,1
gt r2,r3
outr2

Commented code:

; Essentially, our game plan is to compare amount of `b' and `s' occurences.
; Start a new label
@loop
; Read a character from stdin, put it in r1.
  in r1
; If r1 = 'b' (note the way of expressing a character constant), set
; the condition flag.
  ceq r1,.b
; If the condition flag is set, add 1 to r2
  cadd r2,1
; If r1 = 's', set the condition flag. Otherwise, clear it.
  ceq r1,.s
; If the condition flag is set, add 1 to r3
  cadd r3,1
; If r1 is bigger than zero, we didn't hit eof yet, so continue reading.
; Note: The golfed version is based on a certain property beyond explaining.
; It's related to the way how labels work under the hood.
  jnz r1,%l
; The loop finished.
; Compare r2 and r3 (the accumulators).
gt r2,r3
; Display the result
out r2

Answer 16

Japt, 6 bytes

Takes input as a string with reversed output.

èÕ¶Uèc

Try it - includes all test cases, footer negates the output for easier testing

Answer 17

C# (Visual C# Interactive Compiler), 39 bytes

i=>i.Count(x=>x==98)>i.Count(x=>x=='s')

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Answer 18

C (gcc), ^{57 \$\cdots\$ 53} 52 bytes

Saved 3 bytes thanks to l4m2!!!

Saved a byte thanks to Arnauld!!!

a;f(char*s){for(a=0;*s;++s)a+=*s-99?*s>120:-1;a=!a;}

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Takes a string as input and returns a \$1\$ for edible and a \$0\$ for grossing me out!

Answer 19

PHP, 44 42 bytes

-2 bytes thanks to @Elgirhath

$b=count_chars($a,1);echo+($b[98]>$b[97]);

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Answer 20

Javascript, 46 characters

Shortening Yaraslav's answer:

s=>(q=z=>s.filter(e=>e==z).length)('b')>q('s')

Javascript, 62 characters

x=>([,a,b]=/(a*)(b*)/.exec(x.sort().join``),b.length>a.length)

Maybe I missed something? Just counts if there are more b than a.

Answer 21

Google Sheets, 34 33 bytes

^{Saved 1 byte thanks to Steve Bennett}

=CountIf(A:A,"b")>CountIf(A:A,"s"

Input is in column A. When you exit the cell, Sheets will automatically add the trailing parentheses.

There's nothing really clever about this solution besides implementing the concept from others.

Answer 22

JavaScript (Node.js), 62 bytes

s=>s.filter(e=>e=='b').map(Date)>s.filter(e=>e=='s').map(Date)

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JavaScript (Node.js), 56 bytes

s=>s.filter(e=>e=='b').length>s.filter(e=>e=='s').length

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