# Extract Local Maxima [duplicate]

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]               | 
[1,2,3,2,1]         | 
[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!

• @PeterTaylor I think what's meant is "For the first or last element to be included in the output, ..."
– xnor
Jul 14 '17 at 18:50
• @PeterTaylor xnor is correct. Jul 14 '17 at 18:52
• Related Jul 14 '17 at 18:56
• Also related: Finding Local Extremes Jul 15 '17 at 7:34
• I've voted to close this as a duplicate of Finding Local Extremes as this is a subset of that challenge Dec 26 '20 at 23:27

# 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.

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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

• Well I think there may be a way to use 3-wise reduction but I have not worked it out. Jul 14 '17 at 20:03
• I was right - a 3-wise reduce with the maximum dyad, » - how about 10 though..? Jul 14 '17 at 22:04
• 10 bytes. Also 10 bytes Dec 26 '20 at 23:01
• @cairdcoinheringaahing indeed, I don't think either of those were possible at the time. Dec 27 '20 at 0:30

# Python, 54 bytes

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


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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)


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Alt 57:

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


## Mathematica 22 Bytes

Pick[#,MaxDetect@#,1]&

• Incidentally, this would also work on higher dimension arrays. Jul 14 '17 at 18:51

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

# Dyalog APL, 31302822 21bytes

{⍵/⍨(⌈/=2⌷⊢)¨3,/∊0⍵0}


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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


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


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## JavaScript (ES6), 40 bytes

a=>a.filter((e,i)=>!(e<a[i-1]|e<a[i+1]))


# Python 3, 84 75* 71 bytes

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


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

• lambda l,k=:[l[i]for i in range(len(l))if(k+l+k)[i+2]<=l[i]>=(k+l+k)[i]] Jul 14 '17 at 18:52

# Jelly, 15 bytes

2ị⁼Ṁ
0;;0ṡ3Ç€Tị


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# 05AB1E, 15 14  13 bytes

ü‹0¸«sĆÁü›+_Ï


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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


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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


# 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))
  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


# R, 42 bytes

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


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

# Pyth, 20 bytes

-.b*YqYeSN.:++0Q03Q0


To be golfed...

Test suite.

# R, 68 bytes

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


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

for(;$g=$argv[+$i];$l=$g)$g<$l|$g<$argv[++$i]?:$r[]=$g;print_r($r);  Try it online! # Retina, 51 bytes \d+$*
^

M!(?<=(^|1+) )(\1\d*)\b(?! 1\2)
^¶

%1


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# q, 39 bytes

{x where x = -1 _ next 3 mmax x,last x}

• I've never heard of this language before. Do you know anywhere I can try or download it? Jun 6 '19 at 2:41
• Sure, kx.com, docs: code.kx.com Jun 6 '19 at 7:03

# 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

# Perl 5-a, 37 bytes

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


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