Inspired by these two SO questions (no doubt from the same class): print the elements in the subarray of maximum sum without adjacent elements java and Maximum sum of non adjacent elements of an array, to be printed.


Given a list of integers, output a subsequence consisting of non-adjacent elements that have the highest sum. Here some examples:

  • [1,2,3,-1,-3,2,5] would result in [1,3,5] (with a sum of 9) at the 0-based indices [0,2,6].
  • [4,5,4,3] would result in either [4,4] (with a sum of 8) at the 0-based indices [0,2] or [5,3] (also with a sum of 8) at the 0-based indices [1,3].
  • [5,5,10,100,10,5] would result in [5,100,5] (with a sum of 110) at either the 0-based indices [0,3,5] or [1,3,5].

What's most important about these examples above, the indices containing the elements are at least 2 apart from each other. If we look at the example [5,5,10,100,10,5] more in depth: we have the following potential subsequence containing non-adjacent items; with their indices below it; with their sums below that:

[[5],[10],[100],[10],[5],[5],[100,5],[10,5],[10,10],[5,5],[5,10],[5,100],[5,5],[5,10],[5,100],[5,10],[5,100,5],[5,100,5],[5,10,5],[5,10,10]]   // non-adjacent subsequences
[[5],[ 4],[  3],[ 2],[1],[0],[  3,5],[ 2,5],[ 2, 4],[1,5],[1, 4],[1,  3],[0,5],[0, 4],[0,  3],[0, 2],[1,  3,5],[0,  3,5],[0, 2,5],[0, 2, 4]]   // at these 0-based indices
[  5,  10,  100,  10,  5,  5,    105,    15,     20,   10,    15,    105,   10,    15,    105,    15,      110,      110,      20,       25]   // with these sums
                                                                                                            ^         ^                        // and these two maximums

Since the maximum sums are 110, we output [5,100,5] as result.

Challenge rules:

  • You are allowed to output key-value pairs of the index + value. So instead of [5,100,5] you can output [[0,5],[3,100],[5,5]] or [[1,5],[3,100],[5,5]] as result (or [[1,5],[4,100],[6,5]]/[[2,5],[4,100],[6,5]] when 1-based indexing is used instead of 0-based).
    • If you use key-value pairs, they can also be in reverse or random order, since it's clear which values are meant due to the paired index.
    • Outputting just the indices without values isn't allowed. It should either output the values, or the values/indices as key-value pairs (or two separated lists for 'keys' and 'values' of the same size if key-value pairs are not possible in your language of choice).
  • You are allowed to output all possible subsequences with the maximum sum instead of just one.
  • As you can see from the examples, the input-list can contain negative and duplicated values as well. You can assume the input-integers are within the range \$[-999,999]\$.
  • The output-list cannot be empty and must always contain at least one element (if a list would only contain negative values, a list containing the single lowest negative value would then be output as result - see last two test cases).
  • If there is one possible output but for multiple different indices, it's allowed to output both of them even though they might look duplicates. (i.e. the example above, may output [[5,100,5],[5,100,5]] for both possible index-combinations).

Test cases:

Input:                   Possible outputs:       At 0-based indices:     With sum:

[1,2,3,-1,-3,2,5]        [1,3,5]                 [0,2,6]                 9
[4,5,4,3]                [4,4]/[5,3]             [0,2]/[1,3]             8
[5,5,10,100,10,5]        [5,100,5]               [0,3,5]/[1,3,5]         110
[10]                     [10]                    [0]                     10
[1,1,1]                  [1,1]                   [0,2]                   2
[-3,7,4,-2,4]            [7,4]                   [1,4]                   11
[1,7,4,-2]               [7]                     [1]                     7
[1,2,-3,-4,5,6,-7]       [2,6]                   [1,5]                   8
[800,-31,0,0,421,726]    [800,726]/[800,0,726]   [0,5]/[0,3,5]/[0,2,5]   1526
[-1,7,8,-5,40,40]        [8,40]                  [2,4]/[2,5]             48
[-5,-18,-3,-1,-10]       [-1]                    [3]                     -1
[0,-3,-41,0,-99,-2,0]    [0]/[0,0]/[0,0,0]       [0]/[3]/[6]/[0,3]/
                                                  [0,6],[3,6]/[0,3,6]    0
  • 1
    \$\begingroup\$ powersetis a set of subsets isn't it? but it looks like you are returning a set of subsequences? [4,5,4,3] would result in either [4,4] where [4,4] is clearly not a set. \$\endgroup\$ Commented Apr 17, 2019 at 14:19
  • 1
    \$\begingroup\$ @Arnauld Yes, if the values are key-value pairs with their index it's clear which indexed values are meant in the input, so they can be in any order. Will also edit this into the challenge description. \$\endgroup\$ Commented Apr 17, 2019 at 16:47
  • 2
    \$\begingroup\$ Just to be sure: outputting the indices isn't an option, is it? \$\endgroup\$
    – Shaggy
    Commented Apr 17, 2019 at 22:19
  • 1
    \$\begingroup\$ @KevinCruijssen: That whole output in your link is the power set of [1, 2, 3, 4, 5]. The individual sublists aren't all power sets. "Power set" is not a correct term to describe what this challenge is asking for. \$\endgroup\$ Commented Apr 18, 2019 at 9:03
  • 2
    \$\begingroup\$ The classical term is "subsequence". This has the same problem of people thinking of contiguous subsequences, though. I would say "subset" if we were actually working with sets here, but these are definitely sequences - order matters and duplicates are allowed. \$\endgroup\$ Commented Apr 18, 2019 at 9:18

21 Answers 21


Husk, 11 bytes


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►Σ†!¹mü≈tṖŀ  Implicit input, say L=[4,5,3,4].
          ŀ  Indices: [1,2,3,4]
         Ṗ   Powerset: [[],[1],[2],[1,2],..,[1,2,3,4]]
        t    Tail (remove the empty list): [[1],[2],[1,2],..,[1,2,3,4]]
     m       For each,
      ü      de-duplicate by
       ≈     differing by at most 1.
             For example, [1,2,4] becomes [1,4].
  †          Deep map
   !¹        indexing into L: [[4],[5],[4],..,[5,4],[4,3]]
►            Maximum by
 Σ           sum: [5,4]
  • \$\begingroup\$ Just a comment to inform you someone outgolfed you, so I've switched the accepted answer. Sorry for the minor loss of rep as a result. :) \$\endgroup\$ Commented Oct 30, 2020 at 12:06

Haskell, 60 bytes

r%(h:t)=max(r%t)$(r++[h])%drop 1t
r%_=(sum r<$r,r)

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The helper function % recursively branches on choosing whether the include the first element and drop the second, or to skip the first element. It takes the maximum of all outcomes, which are tuples whose first element is the sum, and whose second element is the corresponding list which is extracted for the output.

To handle the rule that the empty list is disallowed even if it would have the smallest trick, we do a cute trick of writing sum r<$r rather than sum r.This makes a list whose elements all are sum r and whose length is that of r. That way, when we choose the maximum, we prioritize any list over an empty r, but otherwise comparisons depend on the first element which is sum r .


R, 136 125 bytes


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-6 bytes thanks to digEmAll, who incidentally also outgolfed me.

Returns the shortest subsequence as a list, breaking ties on lexicographically first by indices.

Brute-force generates all index subsequences, then Filters for those that are non-adjacent, i.e., where all(diff(x)>1). Then subsets [ into l using these indices, selecting [[ the first one where the sum is the max (which.max).

I'm pretty sure this is the first R answer I've ever written that uses Filter! sad, Filter is ungolfy, no wonder I've never used it...

  • \$\begingroup\$ 130 \$\endgroup\$
    – digEmAll
    Commented Apr 21, 2019 at 8:32

05AB1E, 14 bytes

Saved 1 byte thanks to Kevin Cruijssen


Try it online! or as a Test Suite


ā<               # push [0 ... len(input)-1]
  æ              # compute powerset
   ʒ    }        # filter, keep lists where:
      ≠W         # no element is 1 in the
     ¥           # deltas
    Ć            # of the list with the head appended
         è       # index into the input with each
          ΣO}    # sort by sum
             θ   # take the last element
  • \$\begingroup\$ You may not be happy, but it's still 4 bytes shorter than my initial solution. ;) And you can golf 1 more changing ¤ª to Ć. \$\endgroup\$ Commented Apr 17, 2019 at 16:40
  • \$\begingroup\$ @KevinCruijssen: Oh yeah! For some reason I had convinced myself I needed a repeat element at the end. Thanks! \$\endgroup\$
    – Emigna
    Commented Apr 17, 2019 at 17:00

Brachylog (v2), 14 bytes


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Function submission; input from the left, output from the right, as usual. Very slow; a five-element list is probably the maximum for testing on TIO.

 ~b              Prepend an arbitrary element to the input
   a             Take a prefix or suffix of the resulting list
    ~c           Ordered partition into contiguous sublists
      ∋₁         Take the second element
        ᵐ          of each sublist
{        }ᶠ      Find all possible ways to do this
           +ᵒ    Sort by sum
             t   Take the greatest

The results we get from prefixes aren't incorrect, but also aren't interesting; all possible results are generated via taking a suffix (which is possibly the list itself, but cannot be empty), but "suffix" is more verbose in Brachylog than "prefix or suffix", so I went with the version that's terser (and less efficient but still correct). The basic idea is that for each element we want in the output list, the partition into contiguous sublists needs to place that element and the element before into the same sublist (because the element is the second element of the sublist), so two consecutive elements can't appear in the result. On the other hand, it's fairly clear that any list without two consecutive elements can appear in the result. So once we have all possible candidate lists, we can just take the sums of all of them and see which one is largest.


Jelly, 8 bytes


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This is a function submission (it's almost usable as a full program, which is what the TIO link does, but the output format is a little weird; when used as a function, though, it just outputs the list of elements from the subsequence).


I found a much terser algorithm (loosely inspired by my previous Brachylog answer), saving 6 bytes over the Jelly and Brachylog answers and 3 bytes over the current leading answer.

ŒP         On the set of all subsequences of {the input},
  m 2        remove every second element of
   €         each subsequence,
      ÞṪ     then {return} the resulting subsequence with the maximal
     S       sum

The basic idea is that we don't look for the subsequence we want directly; rather, we find a subsequence with an additional element inserted in between each of the elements we want. This is always possible if the elements of the subsequence we want are not adjacent in the original list (because there'll always be at least one in between to insert), and never possible if the elements of the subsequence we want are adjacent in the original list. So all we have to do is look at all the subsequences of the input, then take alternating elements from them, to get a list of all the non-adjacent-item subsequences of the input.

This algorithm will return many of the non-adjacent-item subsequences more than once, but that doesn't matter, because we're just looking for a maximal element, and repeating elements of the set we're taking a maximum of has no influence on the maximum.


Jelly, 16 14 bytes


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Thanks to @EriktheOutgolfer for saving 2 bytes!


JavaScript (ES6),  138 132 130 129  126 bytes

Outputs key-value pairs.


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Step 1

We first compute the powerset of the input with \$[value, index]\$ pairs.

a.reduce((a, x, i) => // for each value x at position i:
  [                   //   update a[] to a new array consisting of:
    ...a,             //     all previous entries
    ...a.map(y =>     //     for each value y in a[]:
      [[x, i], ...y]  //       append [x, i], followed by all original entries
    )                 //     end of map()
  ],                  //   end of new array
  [[]]                //   start with a = [[]]
)                     // end of reduce()

Step 2

We then look for the maximum sum \$m\$ among these sets, discarding sets with at least two adjacent elements. The best set is stored in \$r\$.

.map(m =              // initialize m to a non-numeric value
  a =>                // for each entry a[] in the powerset:
  a.some(s = p =      //   initialize s and p to non numeric values
    ([v, i]) =>       //   for each value v and each index i in a[]:
    p - (             //     compute p - i
      s = ~~s + v,    //     add v to s
      p = i           //     update p to i
    ) < 2             //     if p - i is less than 2, yield true
  ) |                 //   end of some()
  s < m ||            //   unless some() was truthy or s is less than m,
  (r = a, m = s)      //   save a[] in r[] and update m to s
) && r                // end of map(); return r[]

Haskell, 81 80 bytes

f(a:b:c)=f(b:c)++map(a:)(f c)
f a=[]:map(:[])a

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f builds all valid subsequences by either skipping the next element (f(b:c)) or using it and skipping the next (map(a:)(f c)) and recursively work on the rest. For the result, build all subsequences (f), drop the empty subsequence (which occurs first in the list: tail), make pairs (<sum>,<subsequence>) (map((,)=<<sum)), find the maximum (pairs are compared in lexicographical order) -> maximum) and drop the sum (snd).

Edit: -1 byte thanks to @Lynn.

  • 1
    \$\begingroup\$ map(:[])a is a byte shorter than (pure<$>a) ^^ \$\endgroup\$
    – lynn
    Commented Apr 18, 2019 at 18:38

J, 47 bytes

>@{:@(<@#~/:+/@#~)1(-#~0=1 1+/@E."1-)[:i.&.#.=~

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-7 bytes thanks to FrownyFrog


J, 54 bytes

[:(>@{:@/:+/&>)]<@#~"1[:(#~0=1 1+/@E."1])[:#:@}[email protected]^#

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  • 1
    \$\begingroup\$ 47 \$\endgroup\$
    – FrownyFrog
    Commented May 10, 2019 at 7:54
  • \$\begingroup\$ Thanks you very much @FrownyFrog \$\endgroup\$
    – Jonah
    Commented May 11, 2019 at 1:37

T-SQL, 122 119 118 bytes

Input is a table variable.

This query picks all elements from the table variable, combining these with all non-adjacent elements with higher position values and show the text generated for the highest sum of these values.

WITH C(y,j,v)as(SELECT*,x*1FROM @
SELECT y+','+x,i,v+x

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Wolfram Language (Mathematica), 144 bytes


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Pyth, 19 bytes


Try it online here, or verify all the test cases at once here.

esDm@LQdtf!q#1.+TyUQ   Implicit: Q=eval(input())
                       Trailing Q inferred
                  UQ   Generate range [0-len(Q))
                 y     Take the powerset of the above
         f             Filter keep elements of the above, as T, using:
              .+T        Take differences of consecutive elements of T
           q#1           Keep those differences equal to 1
          !              Logical NOT - empty lists evaluate to true, populated ones to false
                       Result of the filter is those sets without consecutive numbers
        t              Drop the first element (empty set)
   m                   Map the remaining sets, as d, using:
     L d                 For each element of d...
    @ Q                  ... get the element in Q with that index
 sD                    Order the sets by their sum
e                      Take the last element, implicit print

Gaia, 24 bytes


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Ugh, E‡ does some weird stuff...according to the documentation, it should do something like "given length i set of lists X and length j set of indices Y, return X[i][Y[j]]", but instead returns [X[i][Y[j]] X[i][Y[-j]] where negative indexing represents the complement, so we have to do ev2% to extract only the ones we want.

e				| eval as a list l
 :				| dup
  w				| wrap as a list
   ;				| push l again
    ċ				| push [1..len(l)]
     z				| push all subsets of [1..len(l)] -- index powerset.
      ⟨      ⟩⁇			| filter this for:
       ọ			| deltas
        1>¦			| are greater than 1
           ẏ			| all (all deltas greater than 1)
	       ‼⁇		| filter for non-empty lists
		 E‡		| table extract elements. Given l and index set i, this pushes
				| [l[i] l[setdiff(1..l,i)]] for some reason
		   ev2%		| get the l[i] only by unlisting, reversing, and taking every other element
		       Σ⌠	| Get the one with the maximum sum
  • \$\begingroup\$ Out of curiosity, why does the output have two trailing ]] instead of one? \$\endgroup\$ Commented Apr 18, 2019 at 19:14
  • \$\begingroup\$ @KevinCruijssen Just another fun quirk of the interpreter; all lists are printed out like that, so [[1] [2]] gets printed [[1]] [2]]]] which makes it super hard to read/debug list output. \$\endgroup\$
    – Giuseppe
    Commented Apr 18, 2019 at 19:27
  • \$\begingroup\$ I think it's because of the expression re.sub(" ?$","]",result) in the interpreter which should instead be re.sub(" +$","]",result) but my python is super bad. \$\endgroup\$
    – Giuseppe
    Commented Apr 18, 2019 at 19:37

R, 108 107 bytes

function(v,M=-Inf){for(j in J<-seq(a=v))for(i in combn(J,j,,F))if(all(diff(i)>1)&sum(v[i])>sum(M))M=v[i]

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-1 thanks to @Giuseppe


Wolfram Language (Mathematica), 70 63 bytes


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High-level search

          Select[q=Rest@Subsets@#,                     ]        (*choose nonempty subsets of the input such that*)
                                  !FreeQ[q,          ]&         (*there exists a subset of the input which matches*)
                                           #~Riffle~_           (*this list, with an item inserted between adjacent elements*)
MaximalBy[                                              ,Tr,1]& (*and return one with the greatest total*)

,1 is required so as not to inadvertently return invalid sets (otherwise, for example, an input of {1,1,1} would result in an output of {{1,1},{1,1},{1,1}})


Haskell, 300 168 bytes

import Data.List
h(x:y)=fst$foldl(\a c->((fst a)&&(c-snd a>1),c))(1<2,x)y
z=snd.last.sortOn fst.map((,)=<<sum.snd).filter(h.fst).map unzip.subsequences.zip[0..]

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-132 bytes thanks to all the feedback from @nimi :)


Ungolfed (original)

import Data.List
import Data.Function

f :: [Int] -> [(Int, Int)] -- attach indices for later use
f [] = []
f xs = zip xs [0..length xs]

g :: [[(Int, Int)]] -> [([Int], [Int])] -- rearrange into list of tuples
g [] = []
g (x:xs) = (map fst x, map snd x) : g xs

h :: [Int] -> Bool -- predicate that checks if the indices are at least 2 apart from each other
h [] = False
h (x:xs) = fst $ foldl (\acc curr -> ((fst acc) && (curr - snd acc > 1), curr)) (True, x) xs
j :: [([Int], [Int])] -> [([Int], [Int])] -- remove sets that don't satisfy the condition
j xs = filter (\(elements, indices) -> h indices) xs

k :: [([Int], [Int])] -> [(Int, ([Int], [Int]))] -- calculate some of elements
k xs = map (\(elements, indices) -> (foldl1 (+) elements, (elements, indices))) xs

l :: [(Int, ([Int], [Int]))] -> ([Int], [Int]) -- grab max
l xs = snd $ last $ sortBy (compare `on` fst) xs

z -- put things together
  • 1
    \$\begingroup\$ Some tips: flip the element and its index within the pairs returned by f: f x=zip[0..length x]x, so f becomes f=zip[0..]. g is just g=map unzip. The function to filter with in j is h.fst (<- flipped pairs!). j=filter(h.fst). The foldl1+ from k is sum and with a pointfree pair making k=map((,)=<<sum.snd). sortBy(...) can be replaced by sortOn fst: l=snd.last.sortOn fst. Finally as you are using all functions only once, you can inline them into a single pointfree expression: z=snd.last.sortOn fst.map((,)=<<sum.snd).filter(h.fst).map unzip.subsequences.zip[0..] \$\endgroup\$
    – nimi
    Commented Apr 17, 2019 at 16:31
  • \$\begingroup\$ .... Try it online!. \$\endgroup\$
    – nimi
    Commented Apr 17, 2019 at 16:31
  • \$\begingroup\$ oh, and no need to import Data.Function anymore. \$\endgroup\$
    – nimi
    Commented Apr 17, 2019 at 16:32
  • \$\begingroup\$ That's great, thanks for the feedback :) \$\endgroup\$
    – bugs
    Commented Apr 17, 2019 at 17:05
  • \$\begingroup\$ Next h: we're looking for non-adjacent elements, i.e. the difference of adjacent indices must be >1. zipWith(-)=<<tail builds such a list of differences, but fails for the empty list, so we need an additional tail on the subsequences to get rid of it. Inline again. Try it online! \$\endgroup\$
    – nimi
    Commented Apr 17, 2019 at 17:27

Charcoal, 46 bytes


Try it online! Link is to verbose version of code. Explanation:


The variable u is predefined with an empty list. This is put in a list which is assigned to h. These variables act as accumulators. u contains the sublists that include the latest element of the input q while h contains the sublists that do not (and therefore are suitable for appending the next element of the input).


Loop over the elements of the input.


Save the list of sublists that contain the previous element.


Take all of the sublists that do not contain the previous element, append the current element, and save the result as the list of sublists that contain the current element. (I don't use Push here as I need to clone the list.)


Concatenate both previous sublists into the new list of sublists that do not contain the current element.


Concatenate the sublists one last time and remove the original empty list (which Charcoal can't sum anyway).


Compute the sums of all of the sublists.


Find an index of the greatest sum and output the corresponding sublist.


Japt -h, 22 bytes

Êo à k_mÄ øZÃm!gU

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Japt -h, 21 bytes

Ever have one of those challenges where you completely forget how to golf?!

ð¤à fÊk_än ø1îmgUÃñx

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ð¤à fÊk_än ø1îmgUÃñx     :Implicit input of array U
ð                         :Indices of elements that return true when
 ¤                        :  Converted to a base-2 string (to account for 0s)
  à                       :Combinations
    f                     :Filter by
     Ê                    :  Length (to remove the empty combination)
      k_                  :Remove elements that return true
        än                :  Deltas
           ø1             :  Contains 1
             Ã            :End remove
              ®           :Map
               m          :  Map
                gU        :    Index into U
                  Ã       :End map
                   ñ      :Sort by
                    x     :  Sum
                          :Implicit output of last element

Python 2, 63 70 65 bytes

f=lambda a:a and max([a[:1],a[:1]+f(a[2:]),f(a[1:])],key=sum)or a

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5 bytes thx to ArBo

  • \$\begingroup\$ Your test case [1, 7, 4, -2] [1, 4] 5 7 is getting the wrong answer. \$\endgroup\$
    – xnor
    Commented Apr 20, 2019 at 4:43
  • \$\begingroup\$ @xnor: fixed now. \$\endgroup\$
    – Chas Brown
    Commented Apr 20, 2019 at 9:44
  • \$\begingroup\$ 65 bytes \$\endgroup\$
    – ArBo
    Commented Apr 20, 2019 at 22:03

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