19
\$\begingroup\$

Given an array of positive integers, output an array of all the elements that are greater than or equal to the adjacent ones. Most elements will have two adjacent elements; the first and last element are special cases, as they only have one adjacent element.

You may assume that the array contains at least two elements.

Test cases:

Input               | Output
[4,2,6,12,4,5,4,3]  | [4,12,5]
[1,2]               | [2]
[1,2,3,2,1]         | [3]
[3,2,1,2,3]         | [3,3]
[4,4]               | [4,4]
[2,4,4,4,1]         | [4,4,4]
[2,3,3,4]           | [3,4]
[4,3,3,4]           | [4,4]

This is , shortest code wins!

\$\endgroup\$
  • 1
    \$\begingroup\$ @PeterTaylor I think what's meant is "For the first or last element to be included in the output, ..." \$\endgroup\$ – xnor Jul 14 '17 at 18:50
  • \$\begingroup\$ @PeterTaylor xnor is correct. \$\endgroup\$ – Pavel Jul 14 '17 at 18:52
  • \$\begingroup\$ Related \$\endgroup\$ – Peter Taylor Jul 14 '17 at 18:56
  • \$\begingroup\$ Also related: Finding Local Extremes \$\endgroup\$ – Wrzlprmft Jul 15 '17 at 7:34
  • \$\begingroup\$ Can I propose [4,3,3,4] as a testcase. My solution didn't handle that one very well sadly. \$\endgroup\$ – JAD Jul 15 '17 at 10:49

19 Answers 19

5
\$\begingroup\$

Jelly,  13 12  11 bytes

0;;0»3\f"⁸Ẏ

A monadic link taking a list of positive integers and returning the filtered list containing only those which are greater than or equal to all their neighbours.

Try it online!


Previous 12 byter:

0;INżI>0ḄNMị

Previous 13 byter:

0;;0ṡ3M€ċ€2Tị

How?

0;;0»3\f"⁸Ẏ - Link: list of positive integers, A
0;          - a zero concatenated with A
  ;0        - concatenate a zero
     3\     - 3-wise reduce with:
    »       -   maximum (yields a list of the maximums in each overlapping window of 3)
         ⁸  - chain's left argument, A
        "   - zip with:
       f    -   filter keep (i.e. keep the maximal if it is [in] the [length 1 list 
            -                     of the] respective original element)
          Ẏ - flatten by one level
\$\endgroup\$
  • \$\begingroup\$ Well I think there may be a way to use 3-wise reduction but I have not worked it out. \$\endgroup\$ – Jonathan Allan Jul 14 '17 at 20:03
  • \$\begingroup\$ I was right - a 3-wise reduce with the maximum dyad, » - how about 10 though..? \$\endgroup\$ – Jonathan Allan Jul 14 '17 at 22:04
8
\$\begingroup\$

Python, 54 bytes

f=lambda l,*p:l and l[:p<=l[:1]>=l[1:2]]+f(l[1:],l[0])

Try it online!

I/O is with tuples rather than lists.


Python, 57 bytes

f=lambda l,p=0:l and l[:[p]<=l[:1]>=l[1:2]]+f(l[1:],l[0])

Try it online!

Alt 57:

f=lambda l,p=0:l and l[l<[max(p,*l[:2])]:1]+f(l[1:],l[0])
\$\endgroup\$
7
\$\begingroup\$

Mathematica 22 Bytes

Pick[#,MaxDetect@#,1]&
\$\endgroup\$
  • 1
    \$\begingroup\$ Incidentally, this would also work on higher dimension arrays. \$\endgroup\$ – Kelly Lowder Jul 14 '17 at 18:51
6
\$\begingroup\$

Haskell, 50 49 42 bytes

f l=[j|i:j:k:_<-scanr(:)[0]$0:l,k<=j,i<=j]

Try it online!

scanr(:)[0] makes a list of the tails of (0:l), each with a final 0, e.g. for l = [4,3,3,4]: [[0,4,3,3,4,0],[4,3,3,4,0],[3,3,4,0],[3,4,0],[4,0],[0]] which is pattern matched agains i:j:k:_ to extract all lists with at least 3 elements which are named i, j, and k. Keep j if its >= i and j.

Edit: Ørjan Johansen saved 7 bytes. Thanks!

\$\endgroup\$
  • 2
    \$\begingroup\$ i:j:k:_<-scanr(:)[0]$0:l is shorter. (Slightly adjusting the "standard" tails=scanr(:)[] trick.) \$\endgroup\$ – Ørjan Johansen Jul 14 '17 at 23:55
  • \$\begingroup\$ @ØrjanJohansen: oh, I've used that trick before myself, but somehow missed it here. Thanks a lot! \$\endgroup\$ – nimi Jul 15 '17 at 9:52
4
\$\begingroup\$

Dyalog APL, 31 30 28 22 21bytes

{⍵/⍨(⌈/=2⌷⊢)¨3,/∊0⍵0}

Try it online!

Explanation (I'm not good at explaining things):

0⍵0       - [0,input,0]   (it looks like a face!)
∊         - flatten
3,/       - split into overlapping sections of length 3.
(⌈/=2⌷⊢)¨ - Whether the middle element is the maximum (applied to every section)
⍵/⍨       - index
\$\endgroup\$
4
\$\begingroup\$

Haskell, 40 bytes

p%(h:t)=[h|h>=p,[h+1]>t]++h%t
_%e=e
(0%)

Try it online!

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

JavaScript (ES6), 40 bytes

a=>a.filter((e,i)=>!(e<a[i-1]|e<a[i+1]))
\$\endgroup\$
3
\$\begingroup\$

Python 3, 84 75* 71 bytes

lambda l,k=[0]:[j for x,j in enumerate(l)if(k+l+k)[x+2]<=j>=(k+l+k)[x]]

Try it online!


*@LeakyNun saved 9 bytes using a clever operator trick.

\$\endgroup\$
  • \$\begingroup\$ lambda l,k=[0]:[l[i]for i in range(len(l))if(k+l+k)[i+2]<=l[i]>=(k+l+k)[i]] \$\endgroup\$ – Leaky Nun Jul 14 '17 at 18:52
2
\$\begingroup\$

Jelly, 15 bytes

2ị⁼Ṁ
0;;0ṡ3Ç€Tị

Try it online!

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

05AB1E, 15  14  13 bytes

ü‹0¸«sĆÁü›+_Ï

Try it online!

Explanation

ü‹             # pairwise comparison for less than
  0¸«          # append 0
     s         # swap input to top of stack
      Ć        # enclose, append the head of the list
       Á       # rotate right
        ü›     # pairwise comparison for greater than
          +    # add the two boolean lists
           _   # logical negate
            Ï  # keep only elements of input that are true in the resulting list

Previous 15 byte solution

¬s¤)˜Œ3ùεZQ1è}Ï

Try it online!

Explanation

¬                # get head of input
 s¤              # get tail of input
   )˜            # wrap stack in flattened list
                 # produces the input list with the first and last element duplicated
     Œ3ù         # push sublists of length 3
        ε        # apply transformation on each triple
         ZQ      # ... check each element for equality to the max
          1è     # ... get the middle element
            }    # end transform
             Ï   # keep only elements of input that are true in the resulting list
\$\endgroup\$
2
\$\begingroup\$

R, 44 bytes

pryr::f(x[(x>=c(0,x)&x>=x[-1])[1:sum(x|1)]])

which evaluates to the function:

function (x) 
x[(x >= c(0, x) & x >= x[-1])[1:sum(x | 1)]]

Compares x to c(0,x), so with x shifted one position to the right. Also compares x to x[-1], so one position shifted to the left. These both are TRUE if there is a maximum there. & to take the AND of these booleans. Because of the wrapping nature of R's vectors when they are not the same length, we have to truncate the result at the length of x, which is found by taking sum(x|1). We then plug in the boolean vector, taking only the true indices of x and return that.

Note, because these logical operations are done with unequal length vectors, R will complain. A lot. But the correct output will be there amidst the warnings:

> pryr::f(x[(x>=c(0,x)&x>=x[-1])[1:sum(x|1)]])(c(4,2,6,12,4,5,4,3))
[1]  4 12  5
Warning messages:
1: In x >= c(0, x) :
  longer object length is not a multiple of shorter object length
2: In x >= x[-1] :
  longer object length is not a multiple of shorter object length
3: In x >= c(0, x) & x >= x[-1] :
  longer object length is not a multiple of shorter object length
\$\endgroup\$
2
\$\begingroup\$

R, 42 bytes

function(x)x[c(d<-diff(x),0)<=0&c(0,d)>=0]

Try it online!

2 bytes shorter than JAD's solution. diff computes successive differences; then keep only the entries greater than both neighbours.

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

Pyth, 20 bytes

-.b*YqYeSN.:++0Q03Q0

To be golfed...

Test suite.

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

R, 68 bytes

function(a)a[a==sapply(1:length(a),function(i)max(c(0,a,0)[i+0:2]))]

Try it online!

\$\endgroup\$
  • \$\begingroup\$ pryr::f(expression) is a shorter way to declare a function than function(a)expression. \$\endgroup\$ – JAD Jul 15 '17 at 10:28
  • \$\begingroup\$ Also, sum(a|1) is a shortcut for length(a). \$\endgroup\$ – JAD Jul 15 '17 at 10:50
  • \$\begingroup\$ See my solution for a shorter approach. \$\endgroup\$ – JAD Jul 15 '17 at 11:05
1
\$\begingroup\$

PHP, 67 bytes

for(;$g=$argv[+$i];$l=$g)$g<$l|$g<$argv[++$i]?:$r[]=$g;print_r($r);

Try it online!

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

Retina, 51 bytes

\d+
$*
^

M!`(?<=(^|1+) )(\1\d*)\b(?! 1\2)
^¶

%`1

Try it online

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

q, 39 bytes

{x where x = -1 _ next 3 mmax x,last x}
\$\endgroup\$
  • \$\begingroup\$ I've never heard of this language before. Do you know anywhere I can try or download it? \$\endgroup\$ – Pavel Jun 6 at 2:41
  • \$\begingroup\$ Sure, kx.com, docs: code.kx.com \$\endgroup\$ – skeevey Jun 6 at 7:03
1
\$\begingroup\$

Stax, 10 bytes

úâH◄(☼bM•Å

Run and debug it

It produces output as newline separated values on standard output.

Unpacked, ungolfed, and commented, it looks like this.

f       filter each value in input using the rest of the program; implicitly printing kept values
  x0|S  input pre- and post-pended with zero
  3B    split into batches of 3
  i@    get the i-th batch, where i is the iteration index
  |M=   is the current value equal to the max from the batch?

Run this one

Updated: Just found a 9-byte solution. Will update explanation later:

Stax, 9 bytes

▀▓ûa¥╓╧↨⌐

Run and debug it

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

Perl 5 -a, 37 bytes

map{$_>=$F[++$i]&$_>=$p&&say;$p=$_}@F

Try it online!

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.