52
\$\begingroup\$

In the standard loopholes, the following is forbidden:

Claiming that your answer is written in "MyOwnLanguage", where the command x means "read a sequence of numbers, split them into groups of three, and print the last numbers of those groups where the second number is less than the first"

Here, we are going to do the exact same thing.

Task

Given a sequence of positive integers, whose length is divisible by 3, split them into groups of three, and print the last numbers of those groups where the second number is less than the first.

Testcases

Input               Output
[]                  []
[1,2,3,4,5,6,7,8,9] []
[2,1,3,5,4,6,8,7,9] [3,6,9]
[3,1,4,1,5,9,2,6,5] [4]
[100,99,123]        [123]
[123,123,456]       []
[456,123,789]       [789]

Scoring

This is . Shortest answer in bytes wins.

Standard loopholes apply, so remember not to have a built-in command x that does this task.

\$\endgroup\$
  • 31
    \$\begingroup\$ Hmmm... Now I'm really tempted to create MyOwnLanguage and add the x command... :P \$\endgroup\$ – DJMcMayhem May 9 '17 at 5:55
  • 6
    \$\begingroup\$ *remember not to have a built-in*‽ Well, if we already have it, we can use it, no? \$\endgroup\$ – Adám May 9 '17 at 6:28
  • 2
    \$\begingroup\$ @Adám According to the standard loopholes, you cannot have a language containing the built-in x specifically performing that function. \$\endgroup\$ – Leaky Nun May 9 '17 at 6:34
  • 33
    \$\begingroup\$ @LeakyNun Yes you can, you just cannot make such a language because of the challenge. If your language predates the challenge, it is acceptable. \$\endgroup\$ – Adám May 9 '17 at 6:38
  • 9
    \$\begingroup\$ If I call the builtin p, can I use it? \$\endgroup\$ – Mindwin May 9 '17 at 16:02

39 Answers 39

14
\$\begingroup\$

Octave, 32 bytes

@(L)L(x=3:3:end)(diff(L)(x-2)<0)

Try it online!

or

Verify test cases!

L3 = L(3:3:end)  %extract last elements of groups
d= diff(L)       % first difference of the list
y=d(1:3:end)     %extract first elements of each subgroup of the difference
idx = y<0        %check for negative numbers  
result = L3(idx)
\$\endgroup\$
13
\$\begingroup\$

Jelly, 9 8 bytes

>Ḋm3T×3ị

Try it online!

How it works

>Ḋm3T×3ị  Main link. Argument: A (array)

 Ḋ        Dequeue; yield A without its first element.
>         Compare the elements of A with the elements of the result.
  m3      Select each third element, starting with the first.
    T     Truth; get all indices of truthy elements.
     ×3   Multiply those indices by 3.
       ị  Unindex; retrieve the elements at the redulting indices.
\$\endgroup\$
12
\$\begingroup\$

Haskell, 30 29 bytes

x(a:b:c:l)=[c|b<a]++x l
x d=d

My first attempt at golfing Haskell, so I may have missed an optimization or two

-1 byte thanks to @JulianWolf

\$\endgroup\$
  • 4
    \$\begingroup\$ Nice answer! See codegolf.stackexchange.com/a/60884/66904 for a relevant tip; in particular, swapping the two definitions and writing the second (now first) as x d=d can save you a byte \$\endgroup\$ – Julian Wolf May 9 '17 at 16:21
  • \$\begingroup\$ Clever! I browsed that answer beforehand but must have missed the part where the definition reused the variable \$\endgroup\$ – user46863 May 9 '17 at 17:53
11
\$\begingroup\$

Mathematica, 37 bytes

Assuming this does satisfy the spec, ngenisis gets credit for this approach leading to a 1-byte saving!

BlockMap[If[#>#2,Print@#3]&@@#&,#,3]&

Pure function. BlockMap[...,#,3]& splits the input list into sublists of length 3 and then operates on each sublist with the function If[#>#2,Print@#3]&@@#&. The result is that each qualifying last number is printed. The function also returns a value (namely a list of Nulls a third as long as the input list), which seems to be allowed behavior.

Mathematica, 42 38 bytes

Thanks to Martin Ender for saving 4 bytes!

Cases[#~Partition~3,{a__,b_}/;a>0:>b]&

Pure function. #~Partition~3 does what you think. Cases[X,P:>Q] selects all the elements of X matching the pattern P, and returns the result of the transformation rule :>Q applied to each instance. Here, the pattern being matched is {a__,b_}/;a>0: b_ will match the last element of the list and a__ all the other elements (in this case, the first two); call them y and z for now. The sneaky a>0 then expands to y>z>0, which is the test we want to apply (valid because the spec says everything will be a positive integer). And the transformation rule is :>b, which simply replaces each matching ordered triple with its last element.

Original submission:

Last/@Select[#~Partition~3,#.{1,-1,0}>0&]&

Pure function; pretty much a straightforward implementation, other than #.{1,-1,0} which calculates the difference between the first and second elements of each 3-element sublist.

\$\endgroup\$
  • 3
    \$\begingroup\$ The dot product is neat, but #>#2&@@#& is shorter. But overall it's still shorter to use Cases instead of Select: Cases[#~Partition~3,{a__,b_}/;a>0:>b]& \$\endgroup\$ – Martin Ender May 9 '17 at 13:43
  • \$\begingroup\$ a>0:> has two kinds of magic in it! \$\endgroup\$ – Greg Martin May 9 '17 at 17:39
  • \$\begingroup\$ BlockMap is tantalizing here. \$\endgroup\$ – ngenisis May 10 '17 at 3:33
  • \$\begingroup\$ BlockMap[If[#>#2,#3,Nothing]&@@#&,#,3]& works and is only 39 bytes ... can we save a couple bytes? \$\endgroup\$ – Greg Martin May 10 '17 at 5:28
  • 1
    \$\begingroup\$ BlockMap[If[#>#2,Print@#3]&@@#&,#,3]& arguably satisfies the spec \$\endgroup\$ – ngenisis May 10 '17 at 16:26
8
\$\begingroup\$

Pyth, 10 bytes

eMf>FPTcQ3

Test suite

eMf>FPTcQ3
       cQ3    Chop the input into groups of size 3
  f           Filter on
     PT       All but the last element
   >F         Apply the greater than function
eM            Map to the last element
\$\endgroup\$
7
\$\begingroup\$

R, 35 bytes

(x=matrix(scan(),3))[3,x[2,]<x[1,]]
\$\endgroup\$
  • 3
    \$\begingroup\$ Is the leading whitespace necessary? \$\endgroup\$ – HyperNeutrino May 9 '17 at 18:24
  • 1
    \$\begingroup\$ nice elegant solution \$\endgroup\$ – MickyT May 9 '17 at 18:34
5
\$\begingroup\$

Brachylog (2), 14 bytes

~c{Ṫ}ᵐ{k>₁&t}ˢ

Try it online!

Brachylog rather struggles with this sort of problem. Note that this program has horrible computational complexity, as it brute-forces splitting the input into groups of 3 (having no "split into groups" builtin); it runs quickly with four groups but very slowly with five.

Explanation

~c{Ṫ}ᵐ{k>₁&t}ˢ
~c              Split into groups
  { }ᵐ          such that each group
   Ṫ            has three elements
      {     }ˢ  then on each element, skipping that element on error:
       k          with the list minus its last element
        >₁        assert that it's strictly decreasing
          &       and with the original list
           t      keep only its last element
\$\endgroup\$
  • \$\begingroup\$ Might be worth mentioning that l÷₃;?ḍ₍ is a faster alternative. \$\endgroup\$ – Leaky Nun May 9 '17 at 6:41
  • \$\begingroup\$ I had that in an earlier attempt (using / not ÷; they're equivalent here), but it's a byte longer so I discarded it while golfing it down. \$\endgroup\$ – user62131 May 9 '17 at 6:43
4
\$\begingroup\$

J, 14 bytes

_3&(>`[/\#]/\)

This evaluates to a monadic verb. Try it online!

Explanation

_3&(>`[/\#]/\)  Input is y.
_3&(    \    )  For each non-overlapping 3-element chunk of y,
    >`[/        check if first element is greater than second.
                Call the resulting array x.
_3&(        \)  For each non-overlapping 3-element chunk of y,
          ]/    take the last element.
         #      Keep those where the corresponding element of x is 1.
\$\endgroup\$
4
\$\begingroup\$

Alice, 12 11 bytes

Thanks to Leo for saving 1 byte.

I.h%I-rI~$O

Try it online!

Uses the code points of a string as the input list and outputs the character corresponding to the outputs that should be kept.

Explanation

I      Read x. Pushes -1 on EOF.
.h%    Compute x%(x+1). This terminates the program due to division by zero at EOF,
       but does nothing for non-negative x.
I      Read y.
-      Compute x-y. We only want to output z is this is positive.
r      Range. Pushes 0 1 ... n for positive n, and -n ... 1 0 for negative n
       (and simply 0 for n = 0). So this results in a positive number on top
       of the stack iff x-y is positive.
I      Read z.
~      Swap it with x-y > 0.
$O     Output z iff x-y > 0.
       Then the IP wraps to the beginning of the program to process the next triplet.
\$\endgroup\$
  • \$\begingroup\$ You can golf one byte using r instead of ex. TIO \$\endgroup\$ – Leo May 9 '17 at 21:36
  • \$\begingroup\$ @Leo that's brilliant, thank you! \$\endgroup\$ – Martin Ender May 10 '17 at 12:38
3
\$\begingroup\$

05AB1E, 8 bytes

Code:

3ôʒR`‹i,

Uses the 05AB1E encoding. Try it online!

\$\endgroup\$
3
\$\begingroup\$

dc, 30 bytes

[???sAz1[[lAps.]s.<.dx]s.<.]dx

I/O: one number per line.

\$\endgroup\$
3
\$\begingroup\$

Perl 5, 31 bytes

30 bytes of code + -p flag.

s/\d+ (\d+) (\d+)/$2if$1<$&/ge

Try it online!

Replaces each group of 3 numbers (\d+ (\d+) (\d+)) by the third ($2) if the second ($1) is less than the first ($&), and nothing otherwise.

\$\endgroup\$
3
\$\begingroup\$

CJam, 15 bytes

{3/{)\:>{;}|}%}

Anonymous block which expects argument on the stack, and leaves the result on the stack.

Try it online! (Runs all test cases)

Explanation

3/             e# Split the list into length-3 chunks.
  {            e# For each chunk:
   )           e#  Remove the last element.
    \:>        e#  Reduce the first 2 elements by greater than.
       {;}|    e#  If the first is not larger than the second, delete the third.
           }%  e# (end for)
\$\endgroup\$
3
\$\begingroup\$

Brain-Flak, 82 bytes

{([({}[{}()]<(())>)](<>)){({}())<>}{}{((<{}>))<>{}}{}<>{}{{}((<({}<>)<>>))}{}{}}<>

Try it online!

# Until the stack is empty (input is guaranteed to not contain 0)
{

  # Push 1 for greater than or equal to 0
  ([({}[{}()]<(())>)](<>)){({}())<>}{}{((<{}>))<>{}}{}<>{}
  #  ^------^  This part is Top - (Second + 1)

  # If the second number was less than the first...
  {{}

     # Get ready to push 2 zeros
     ((<

       # Move the next number to the other stack
       ({}<>)<>

     # Push those 2 zeros
     >))}

     # Pop 2 values.
     # This is either 2 zeros, or a 0 and a "last number" that shouldn't be printed
     {}{}

# End loop
}

# Switch to the stack where we stored the numbers to be printed
<>
\$\endgroup\$
3
\$\begingroup\$

Jelly, 10 bytes

s3µṪWx>/µ€

Try it online!

or

Verify test cases

-3 bytes thanks to @LeakyNun

Explanation

s3µṪWx>/µ€
s3         - split into groups of three
  µ     µ€ - on each group, do:
   ṪW      - return the third element as the only element of a list
     x     - repeat each element in that list the number of times
      >/   - corresponding to 1 if the second element of the group is greater than the first; 0 otherwise.
\$\endgroup\$
3
\$\begingroup\$

R, 37 bytes

Version with scan() which I do not like, but it makes it shorter.

x=scan();x[(i<--1:1)>0][x[!i]<x[i<0]]

Version with function() which is easier to test (41 byte)

f=function(x)x[(i<--1:1)>0][x[!i]<x[i<0]]

Thanks to the @Giuseppe! Nice idea to use recycling of index.

Test:

f(c())
f(c(1,2,3,4,5,6,7,8,9))
f(c(2,1,3,5,4,6,8,7,9))
f(c(3,1,4,1,5,9,2,6,5))
f(c(100,99,123))
f(c(123,123,456))
f(c(456,123,789))

Output:

> f(c())
NULL
> f(c(1,2,3,4,5,6,7,8,9))
numeric(0)
> f(c(2,1,3,5,4,6,8,7,9))
[1] 3 6 9
> f(c(3,1,4,1,5,9,2,6,5))
[1] 4
> f(c(100,99,123))
[1] 123
> f(c(123,123,456))
numeric(0)
> f(c(456,123,789))
[1] 789
\$\endgroup\$
  • \$\begingroup\$ you read x in from stdin by using x=scan() at the beginning instead of defining a function, You can also simply set i=c(1,2,0) since logical indices get recycled i.e., x=scan();i=c(1,2,0);x[!i][x[i>1]<x[i==1]] \$\endgroup\$ – Giuseppe May 9 '17 at 15:31
  • \$\begingroup\$ tio.run/nexus/… \$\endgroup\$ – Giuseppe May 9 '17 at 15:32
  • \$\begingroup\$ Thanks @Giuseppe! I do not like this x=scan() approach as it makes the input very cumbersome. And I can not make it repeatable then. \$\endgroup\$ – djhurio May 9 '17 at 19:34
  • 2
    \$\begingroup\$ Right, but the goal is to generate as short a code as possible. Unfortunately for both of us, somebody else found a better solution! \$\endgroup\$ – Giuseppe May 9 '17 at 19:39
  • \$\begingroup\$ Heh, I had as well idea to use matrix() but somehow I did not believed it will be possible to make so short. \$\endgroup\$ – djhurio May 9 '17 at 19:45
3
\$\begingroup\$

JavaScript (ES6), 46 44 42 41 39 bytes

a=>a.filter((_,y)=>y%3>1&a[y-1]<a[y-2])
  • Saved 2 bytes thanks to Neil.

Try It

Input a comma separated list of numbers, without any spaces.

f=
a=>a.filter((_,y)=>y%3>1&a[y-1]<a[y-2])
i.oninput=_=>o.innerText=JSON.stringify(f(i.value.split`,`.map(eval)))
console.log(JSON.stringify(f([])))                  // []
console.log(JSON.stringify(f([1,2,3,4,5,6,7,8,9]))) // []
console.log(JSON.stringify(f([2,1,3,5,4,6,8,7,9]))) // [3,6,9]
console.log(JSON.stringify(f([3,1,4,1,5,9,2,6,5]))) // [4]
console.log(JSON.stringify(f([100,99,123])))        // [123]
console.log(JSON.stringify(f([123,123,456])))       // []
console.log(JSON.stringify(f([456,123,789])))       // [789]
<input id=i><pre id=o>


Explanation

a=>              :Anonymous function taking the input array as an argument via parameter a
a.filter((_,y)=> :Filter the array by executing a callback function on each element,
                  with the index of the current element passed through parameter y.
                  If the function returns 0 for any element, remove it from the array.
y%3>1            :Check if the modulo of the current index is greater than 1.
                  (JS uses 0 indexing, therefore the index of the 3rd element is 2; 2%3=2)
&                :Bitwise AND.
a[y-1]<a[y-2]    :Check if the element at index y-1 in array a
                  is less than the element at index y-2
)                :End filtering method
\$\endgroup\$
  • 1
    \$\begingroup\$ Does y%3>1&a[y-1]<a[y-2] work? \$\endgroup\$ – Neil May 9 '17 at 9:34
  • \$\begingroup\$ Crossed out 44 is still 44 \$\endgroup\$ – Roman Gräf May 9 '17 at 19:15
  • \$\begingroup\$ What do you mean, @RomanGräf? \$\endgroup\$ – Shaggy May 9 '17 at 19:17
  • \$\begingroup\$ codegolf.meta.stackexchange.com/a/7427/56341 \$\endgroup\$ – Roman Gräf May 9 '17 at 19:21
  • \$\begingroup\$ A bug in "Arial,"Helvetica Neue",Helvetica,sans-serif" - well spotted @Roman \$\endgroup\$ – flurbius May 10 '17 at 18:07
2
\$\begingroup\$

Python 3, 43 42 bytes

1 byte thanks to xnor.

f=lambda a,b,c,*l:(b<a)*(c,)+(l and f(*l))

Try it online!

\$\endgroup\$
2
\$\begingroup\$

MATL, 10 bytes

IeI&Y)d0<)

The result is displayed as numbers separated by spaces.

Try it online!

Or verify all test cases. This displays a string representation of the output, so that an empty array is actually seen as []. Note that in MATL a number is the same as a singleton array, so [4] is shown as 4.

Explanation

Ie    % Implicit input. Reshape as a 3-row matrix (column-major order)
I&Y)  % Split into the third row and a submatrix with the other two rows
d     % Consecutive difference along each column of the submatrix
0<    % True for negative values
)     % Use as logical index into the original third row. Implicitly display
\$\endgroup\$
2
\$\begingroup\$

Röda, 15 bytes

{[_3]if[_2<_1]}

Röda is nearly as short as the golfing languages...

This takes three values from the stream, and pushes the third (_3) back, if the second (_2) is less than the first (_1).

The underscores are syntax sugar for for loops, so the program could be written as {{[a]if[b<c]}for a,b,c} or even {[a]for a,b,c if[b<c]}.

No TIO link, because it doesn't work on TIO for some reason (although works with the latest version of Röda that predates the challenge).

\$\endgroup\$
2
\$\begingroup\$

Java 7, 86 85 bytes

void c(int[]a){for(int i=-1;++i<a.length;)if(a[i++]>a[i++])System.out.println(a[i]);}

-1 byte thanks to @PunPun1000

Explanation:

Try it here.

void c(int[]a){                  // Method with integer-array parameter and no return
  for(int i=-1;++i<a.length;)    //  Loop over the array in steps of three at a time
    if(a[i++]>a[i++])            //   If the value of the current index is larger than the next:
      System.out.println(a[i]);  //    Print the value on the third index
                                 //  End of loop (implicit / single-line body)
}                                // End of method
\$\endgroup\$
  • \$\begingroup\$ @PunPun1000 Now you've only incremented the iteration by 2 instead of 3 though, and therefore giving incorrect results (like 3,9 for the test case 1,2,3,4,5,6,7,8,9 instead of 3,6,9). \$\endgroup\$ – Kevin Cruijssen May 9 '17 at 20:49
  • 1
    \$\begingroup\$ @Kevin_Cruijssen Oops you're right. You can still save a byte by using the increment operator though. You just have to start at -1 Try it online! \$\endgroup\$ – PunPun1000 May 9 '17 at 21:00
  • \$\begingroup\$ @PunPun1000 Ah, you're right, nice catch. Thanks! \$\endgroup\$ – Kevin Cruijssen May 9 '17 at 21:16
2
\$\begingroup\$

C#, 126 Bytes

using System.Linq;i=>Enumerable.Range(0,i.Length/3).Select(u=>3*u).Where(u=>i[u]>i[u+1]).Select(u=>i[u+2]);

If you want a whole program with the method it'd be 175 Bytes:

using System.Linq;namespace S{class P{static System.Collections.IEnumerable X(int[]i)=>Enumerable.Range(0,i.Length/3).Select(u=>3*u).Where(u=>i[u]>i[u+1]).Select(u=>i[u+2]);}}

Saved 7 Bytes with the help of TheLethalCoder

\$\endgroup\$
  • \$\begingroup\$ You can just print those... \$\endgroup\$ – Leaky Nun May 9 '17 at 6:52
  • \$\begingroup\$ @LeakyNun of course I could - but why should I? I asked wether it's neccessary, it's not, and it'd be more bytes, I guess. \$\endgroup\$ – MetaColon May 9 '17 at 6:57
  • \$\begingroup\$ (int[]i) can just be i no need for the type. \$\endgroup\$ – TheLethalCoder May 9 '17 at 15:24
  • \$\begingroup\$ @TheLethalCoder Updated it. \$\endgroup\$ – MetaColon May 9 '17 at 19:05
  • \$\begingroup\$ @MetaColon You don't need the braces around (i) either. \$\endgroup\$ – TheLethalCoder May 10 '17 at 7:49
2
\$\begingroup\$

Husk, 8 bytes

ṁΓȯΓ↑<C3

Try it online!

Explanation

This program is a bit involved, so bear with me.

ṁΓȯΓ↑<C3  Implicit input (list of integers).
      C3  Split into slices of length 3.
ṁ         Map over slices and concatenate results
 ΓȯΓ↑<    of this function, explained below.

The function ΓȯΓ↑< takes a list of length 3, x = [a,b,c]. The first Γ splits x into a and [b,c], and feeds them as arguments to the function ȯΓ↑<. This should be equivalent to ((Γ↑)<), but due to a bug/feature of the interpreter, it's actually equivalent to (Γ(↑<)), interpreted as a composition of Γ and ↑<. Now, a is fed to the latter function using partial application, the resulting function ↑<a is given to Γ, which deconstructs [b,c] into b and [c]. Then b is fed to ↑<a, resulting in a function that takes the first b<a elements from a list. This function is finally applied to [c]; the result is [c] if a>b, and [] otherwise. These lists are concatenated by to form the final result, which is printed implicitly.

Without the "feature", I would have 9 bytes:

ṁΓoΓo↑<C3
\$\endgroup\$
1
\$\begingroup\$

Python 2, 57 bytes

lambda i:[i[c+2]for c in range(0,len(i),3)if i[c+1]<i[c]]

Try it online!

\$\endgroup\$
1
\$\begingroup\$

CJam, 16 bytes

q~3/{~@@>S{;}?}%

The output is shown as numbers separated by spaces.

Try it online!

Explanation

q~               e# Read input list
  3/             e# List of sublists of length 3
   {         }%  e# Apply this to each sublist
    ~            e# Push sublist contents: 3 numbers
     @@          e# Rotate twice. This moves first two numbers to top
       >         e# Greater than?
        S{;}?    e# If so: push space (used as separator). Else: pop the third number
                 e# Implicitly display
\$\endgroup\$
1
\$\begingroup\$

PHP, 89 Bytes

<?print_r(array_filter($_GET,function($v,$k){return $k%3>1&&$_GET[$k-1]<$_GET[$k-2];},1));

Try it online!

\$\endgroup\$
1
\$\begingroup\$

JavaScript, 108 107 108 bytes

This is a valid JS anonymous (lambda) function. Add x= at the beginning and invoke like x([5,4,9,10,5,13]). Outputs as function return.

a=>(y=[],a.map((c,i)=>(i+1)%3?0:y.push(a.slice(i-2,i+1))),y.map(v=>v[1]<v[0]?v[2]:null).filter(c=>c|c==0))

The snippet takes in the input as a list of comma separated integers.

x=a=>(y=[],a.map((c,i)=>(i+1)%3?0:y.push(a.slice(i-2,i+1))),y.map(v=>v[1]<v[0]?v[2]:null).filter(c=>c|c==0))
martin.oninput = e => { dennis.innerHTML = x(martin.value.split`,`.map(c=>parseInt(c,10))) }
<input type=text id=martin><pre id=dennis>

\$\endgroup\$
  • \$\begingroup\$ What is the point at posting a longer solution, and using martin and dennis as id? \$\endgroup\$ – Leaky Nun May 9 '17 at 6:52
  • \$\begingroup\$ @LeakyNun Shaggy posted his solution while I was working on mine. But that was no reason for me to not post my solution. As for using the names as id's, I thought it'd be funny. \$\endgroup\$ – Arjun May 9 '17 at 7:03
  • \$\begingroup\$ This doesn't work for [5,4,9,10,5,13]. \$\endgroup\$ – Shaggy May 9 '17 at 10:50
  • \$\begingroup\$ @Shaggy That was a problem with the implementation of the test case snippet; nothing wrong with the solution. Actually, the value of the input element is always a string. So, splitting the string on , resulted it into being an array of strings rather than numbers! The solution is perfectly fine. Only the test case snippet was wrong. I have fixed that, now. Thanks for pointing that out! :) \$\endgroup\$ – Arjun May 9 '17 at 11:03
  • \$\begingroup\$ Oh, yeah, that explains the problem! Thank you, @Arjun. \$\endgroup\$ – Shaggy May 9 '17 at 11:04
1
\$\begingroup\$

Perl5.8.9, 73 60 bytes

while(@F){@b=splice@F,0,3;$b[1]<$b[0]&&print$b[2]}print"-"

(58+2 for the 'n' flag to read the whole file and a to autosplit). Assumes the input is lines of space separated numbers

Reduction thanks to Dada. Including the print at the end for visibility, that'd save 8 bytes if not.

\$\endgroup\$
  • \$\begingroup\$ Nice one! Have your well deserved +1! \$\endgroup\$ – Arjun May 9 '17 at 11:13
  • \$\begingroup\$ The output format being quite flexible, you don't really have to put that print"\n" at the end. Also, you can do $b[1]<$b[0]&&print"$b[2] "while@b=splice@a,0,3 to save 7 bytes. Finally, you can use -a flag instead of doing @a=split (it will do the same automatically and store the result in @F instead of @a); with Perl 5.8.9, you'll need -na while with recent Perls, -a is enough. That should get you to 47-48 bytes. \$\endgroup\$ – Dada May 9 '17 at 12:00
  • \$\begingroup\$ oh, i didn't know about -a. I still think I should do one output line per input line, the output is pretty incomprehensbile otherwise \$\endgroup\$ – Tom Tanner May 9 '17 at 14:03
1
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Clojure, 43 bytes

#(for[[a b c](partition 3 %):when(< b a)]c)

Boring :/

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1
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Bash, 61 bytes

a=($@)
for((t=0;t<$#;t+=3)){((a[t+1]<a[t]))&&echo ${a[t+2]};}

Try it online!

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