# Reverse odd runs

Reverse runs of odd numbers in a given list of 2 to 215 non-negative integers.

### Examples

0 1 → 0 1
1 3 → 3 1
1 2 3 → 1 2 3
1 3 2 → 3 1 2
10 7 9 6 8 9 → 10 9 7 6 8 9
23 12 32 23 25 27 → 23 12 32 27 25 23
123 123 345 0 1 9 → 345 123 123 0 9 1

• 1. I only understood the challenge after looking at the examples. I think runs of odd integers would be clearer than sequences. 2. I don't think setting an explicit upper limit is a good thing. If a language has only 8-bit integers, participating will be a lot harder. Jul 3, 2016 at 4:59
• Also, I'm not sure what further numeric computation refers to. Does it mean that I cannot return an immutable tuple or simply print the numbers? Jul 3, 2016 at 5:10
• @Dennis Updated as you suggested. It is to prevent input/output as string. Any suggestion for better wording?
Jul 3, 2016 at 5:14
• Why do you want to prevent string output? Jul 3, 2016 at 5:16
• Yes, looking at the other challenge, most of the answers rely on splitting on zeroes, whereas here you'd have to split on a condition, which most languages don't have a built-in for.
– xnor
Jul 3, 2016 at 7:14

# Python 2, 7568 63 bytes

5 bytes thanks to Dennis.

And I have outgolfed Dennis.

Credits to Byeonggon Lee for the core of the algorithm.

o=t=[]
for i in input():o+=~i%2*(t+[i]);t=i%2*([i]+t)
print o+t


Ideone it!

Old version: 75 bytes

• Tied, really. Also, I'm counting 81, not 75. I'm guessing you counted it with tabs, but the SE editor filled in spaces. Jul 3, 2016 at 6:02
• @DrGreenEggsandIronMan Your guess is correct. Tabs for readability. Either count the source or count the ideone one. Jul 3, 2016 at 6:05
• print doesn't need parens. Also, you only use a once, so there's no need for a variable. Jul 3, 2016 at 6:20

# APL, 21 20 bytes

{∊⌽¨⍵⊂⍨e⍲¯1↓0,e←2|⍵}


Explanation:

                  2|⍵ Select all the odd numbers
e←    Save that to e
0,      Append a 0
¯1↓        Delete the last element
e⍲           NAND it with the original list of odd numbers
⍵⊂⍨             Partition the list: (even)(even)(odd odd odd)(even)
⌽¨                 Reverse each partition
∊                    Flatten the list


Edit: Saved a ~ thanks to De Morgan's laws

• Hello, and welcome to PPCG! This is a good post. Jul 3, 2016 at 19:43

h%p|(l,r)<-span(odd.(h*))p=l++h:r
foldr(%)[]


Thanks to @xnor for recognizing a fold and saving two bytes.

• Nice method, especially the (h*)! You can save a byte on the base case by writing f x=x second to match the empty list, though it looks like a foldr is yet shorter: h%p|(l,r)<-span(odd.(h*))p=l++h:r;foldr(%)[]:
– xnor
Jul 5, 2016 at 0:32
• I knew it was just a foldr after all! Thank you.
– Lynn
Jul 5, 2016 at 7:02

# Python 2, 7975 73 bytes

def f(x):
i=j=0
for n in x+[0]:
if~n%2:x[i:j]=x[i:j][::-1];i=j+1
j+=1


This is a function that modifies its argument in place. Second indentation level is a tabulator.

Test it on Ideone.

• Where's the meta for this? Jul 3, 2016 at 6:02
• meta.codegolf.stackexchange.com/a/4942/12012 Jul 3, 2016 at 6:03
• Cant you use x[j:i:-1]?
– l4m2
Jun 4, 2020 at 17:52

# Jelly, 10 bytes

Ḃ¬ðœpUżx@F


### How it works

Ḃ¬ðœpUżx@F  Main link. Argument: A (array)

Ḃ           Bit; return the parity bit of each integer in A.
¬          Logical NOT; turn even integers into 1's, odds into 0's.
Left argument: B (array of Booleans). Right argument: A
œp       Partition; split A at 1's in B.
U      Upend; reverse each resulting chunk of odd numbers.
x@   Repeat (swapped); keep only numbers in A that correspond to a 1 in B.
ż     Zipwith; interleave the reversed runs of odd integers (result to the
left) and the flat array of even integers (result to the right).
F  Flatten the resulting array of pairs.


# Python 2, 78 75 bytes

def r(l):
def k(n):o=~n%2<<99;k.i+=o*2-1;return k.i-o
k.i=0;l.sort(key=k)


Super hacky :)

• what is k.i ? Jul 3, 2016 at 6:04
• @LeakyNun k.i=0 on the last line. It's just a variable.
– orlp
Jul 3, 2016 at 6:04
• I don't get it. Are k and k.i related? Jul 3, 2016 at 6:05
• @LeakyNun No. k.i is a persistent variable between calls of k. See it as a makeshift global without having to use the global keyword.
– orlp
Jul 3, 2016 at 6:06

## Python3, 96 bytes

Saved a lot of bytes thanks to Leaky Nun!

o=l=[]
for c in input().split():
if int(c)%2:l=[c]+l
else:o+=l+[c];l=[]
print(" ".join(o+l))

• 93 bytes
– l4m2
Jun 4, 2020 at 17:50

# C, 107 bytes

i;b[65536];f(){for(;i;)printf("%d ",b[--i]);}main(n){for(;~scanf("%d",&n);)n%2||f(),b[i++]=n,n%2||f();f();}

• I know this is an old answer but you can move the last statement in the loop into the for(;;) for -1 byte. As suggested in this tip
– c--
Jun 15 at 23:06

# Pyth, 14 bytes

s_McQshMBx0%R2

%R2Q   Take all elements of the input list modulo 2
x0       Get the indices of all 0s
hMB         Make a list of these indices and a list of these indices plus 1
s            Concatenate them
cQ             Chop the input list at all those positions
_M               Reverse all resulting sublists
s                 Concatenate them


Test cases

# MATL, 20 bytes

TiodgvYsG8XQ!"@gto?P


Input is a column array, using ; as separator.

Try it online!

### Explanation

Consider as an example the input array [1;2;3;5;7;4;6;7;9]. The first part of the code, Tiodgv, converts this array into [1;1;1;0;0;1;0;1;0], where 1 indicates a change of parity. (Specifically, the code obtains the parity of each entry of the input array, computes consecutive differences, converts nonzero values to 1, and prepends a 1.)

Then Ys computes the cumulative sum, giving [1;2;3;3;3;4;4;5;5]. Each of these numbers will be used as a label, based on which the elements of the input will be grouped. This is done by G8XQ!, which splits the input array into a cell array containing the groups. In this case it gives {[1] [2] [3;5;7] [4;6] [7;9]}.

The rest of the code iterates (") on the cell array. Each constituent numeric array is pushed with @g. to makes a copy and computes its parity. If (?) the result is truthy, i.e. the array contents are odd, the array is flipped (P).

The stack is implicitly displayed at the end. Each numeric vertical array is displayed, giving a list of numbers separated by newlines.

# Desmos, 167 bytes

k=mod(l,2)
K=l.length
a=[K...1]
L=a[k[a]=1]
g(q)=∑_{n=1}^{[1...q.length]}q[n]
A=g(join(1,sign(L-L[2...]-1)))
f(l)=l[\{k>0:\sort(L,A.\max+1-A)[g(k)],k\}+[1...K](1-k)]


Hooooly crap, this was pretty challenging to do in Desmos given its limited list functionalities, but after lots of puzzling and scratched ideas, I finally managed to pull together something that actually works.

Try It On Desmos!

Try It On Desmos! - Prettified

# Desmos, 94 93 bytes

g(M,j)=j-max([1...j]0^{mod(M,2)})
f(L)=L[[i+g(L[n...1],n+1-i)-g(L,i)fori=[1...n]]]
n=L.length


Try it on Desmos!

*-1 byte thanks to @AidenChow (remove space after for)

## How it works

g(M,j) gives the difference between j and the greatest index less than (or equal to) j of an even entry in M. For example, g([0,2,4,1,3,5,7,6,8],6) = 3 because the 6th element in the list is 5, and the last even entry before it is 4, which is 3 elements earlier.

f(L) is the solution to the challenge. It's of the form L[[h(i)for i=[1...n]]], which is a list indexing: h(i) gives the index of the value in L that should be at index i in the result list.

The main difficulty lies in h(i), defined as h(i) = i+g(L[n...1],n+1-i)-g(L,i). Note that g(M,j) = 0 whenever M[j] is even, so h(i)=i if L[i] is even (this corresponds to even entries not moving) since L[n...1][n+1-i]=L[i]=even

If L[i] is odd then we have a more complicated story: g(L,i) gives the distance to the element before the start of the odd run, and g(L[n...1],n+1-i) gives the distance to the element after the end of the odd run. Then g(L[n...1],n+1-i)-g(L,i) says how much closer the element is to the end of the run than the start of the run. For example, for L=[0,2,4,1,3,5,7,6,8] and i=6, this difference is -1, so L[6] has to move 1 element to the left, to where the 3 is currently.

In general regarding the difference g(L[n...1],n+1-i)-g(L,i):

• if the difference is positive, then the element is closer to the left than the right, and it needs to move right by the value of the difference
• if the difference is negative, then the element is closer to the right than the left, and it needs to move left by the negative of the difference
• if the difference is zero, then the element is at the center of an odd-length run, so it doesn't need to move at all.
• Wow, brilliant solution! You forgot to put {...} around mod(M,2) btw. Also I'm pretty sure you don't need a space between for and i. Jun 15 at 21:28
• @AidenChow Pastes in fine for me without the curly braces Jun 15 at 21:29
• Weird... it pastes like this for me. Jun 15 at 21:30
• Ah, I accidentally left wolfram2desmos enabled. Thanks for catching it Jun 15 at 21:32

# Clojure, 86 bytes

#(flatten(reduce(fn[a b](if(odd? b)(conj(pop a)(conj[b](last a)))(conj a b[])))[[]]%))


Here is the ungolfed version

#(flatten ; removes all empty vectors and flattens odd sequences
(reduce
(fn[a b]
(if(odd? b) ; if we encounter odd number in the seq
(conj(pop a)(conj[b](last a))) ; return all elements but last and the element we encountered plus the last element of current result
(conj a b[])) ; else just add the even number and the empty vector
)
[[]] ; starting vector, we need to have vector inside of vector if the sequence starts with odd number
%    ; anonymous function arg
)
)


Basically it goes through the input sequence and if it encounters even number it adds the number and the empty vector otherwise if it's an odd number it replaces the last element with this number plus what was in the last element.

For example for this seq 2 4 6 1 3 7 2 it goes like this:

• []<=2
• [2 []]<=4
• [2 [] 4 []]<=6
• [2 [] 4 [] 6 []]<=1
• [2 [] 4 [] 6 [1 []]]<=3
• [2 [] 4 [] 6 [3 [1 []]]]<=7
• [2 [] 4 [] 6 [7 [3 [1 []]]]]<=2
• [2 [] 4 [] 6 [7 [3 [1 []]]] 2 []]

And then flattening this vector gives the correct output. You can see it online here: https://ideone.com/d2LLEC

# J, 3331 30 bytes

[:;]<@(A.~2-@|{.);.1~1,2|2-/\]


## Usage

   f =: [:;]<@(A.~2-@|{.);.1~1,2|2-/\]
f 0 1
0 1
f 1 3
3 1
f 1 2 3
1 2 3
f 1 3 2
3 1 2
f 10 7 9 6 8 9
10 9 7 6 8 9
f 23 12 32 23 25 27
23 12 32 27 25 23
f 123 123 345 0 1 9
345 123 123 0 9 1


# Husk, 7 bytes

ṁ↔ġ¤&%2


Try it online!

## Explanation

ṁ↔ġ¤&%2  Implicit input, a list of integers.
ġ      Group by equality predicate:
¤ %2   Arguments modulo 2
&     are both truthy.
ṁ        Map and concatenate
↔       reversing.


# C#, 179178 177 bytes

s=>{var o=new List<int>();var l=new Stack<int>();foreach(var n in s.Split(' ').Select(int.Parse)){if(n%2>0)l.Push(n);else{o.AddRange(l);o.Add(n);l.Clear();}}return o.Concat(l);}


I use a C# lambda. You can try it on .NETFiddle.

The code less minify:

s => {
var o=new List<int>();var l=new Stack<int>();
foreach (var n in s.Split(' ').Select(int.Parse)) {
if (n%2>0)
l.Push(n);
else {
l.Clear();
}
}
return o.Concat(l);
};


Kudos to Byeonggon Lee for the original algorithm.

• You can drop the space at the foreach(var and change if(n%2==1) to if(n%2>0) to save 2 bytes (or actually 1, because your current answer is 179 bytes instead of 178). Feb 27, 2018 at 8:12
• @KevinCruijssen It was changed in the less minify section but not in the minify one. Also thank you for the foreach space! Feb 27, 2018 at 12:54

# JavaScript (ES6) 70 66 bytes

Edit 4 bytes saved thx @Neil

a=>[...a,[]].map(x=>x&1?o=[x,...o]:r=r.concat(o,x,o=[]),r=o=[])&&r

• :r=r.concat(o,x,o=[]), saves you a couple of bytes. I think you can then go on to save another two like this: a=>[...a,[]].map(x=>x&1?o=[x,...o]:r=r.concat(o,x,o=[]),r=o=[])&&r.
– Neil
Jul 3, 2016 at 11:26
• What is the meaning of ...o? Jul 3, 2016 at 16:20
• Jul 3, 2016 at 18:13
• @Neil the empty array used as the added element is a master stroke Jul 4, 2016 at 9:18

# Ruby, 51 bytes

->l{l.chunk(&:odd?).flat_map{|i,j|i ?j.reverse: j}}


Try it online!

Some slight variations:

->l{l.chunk(&:odd?).flat_map{|i,j|i&&j.reverse||j}}
->l{l.chunk(&:odd?).flat_map{|i,j|!i ?j:j.reverse}}
->l{l.chunk(&:even?).flat_map{|i,j|i ?j:j.reverse}}


### Ruby 2.7, 48 bytes

->l{l.chunk{_1%2}.flat_map{_1>0?_2.reverse: _2}}


Unsupported by TIO :(

# Pyth, 29 28 bytes

JYVQ=+J*%hN2+YN=Y*%N2+NY;+JY


Test suite.

Direct translation of my python answer (when has translating from python to pyth become a good idea?)

# TSQL 118 bytes

DECLARE @ TABLE(i int identity, v int)
INSERT @ values(123),(123),(345),(0),(1),(9)

SELECT v FROM(SELECT sum((v+1)%2)over(order by i)x,*FROM @)z
ORDER BY x,IIF(v%2=1,max(i)over(partition by x),i),i desc


Fiddle

# Stax, 15 10 bytesCP437

Çⁿ╜"}☻≥º╚(


Try it online!

Tied Jelly! So sad that packing only saved one byte.

Unpacked version with 11 bytes:

{|e_^*}/Frm


## Explanation

{|e_^*} is a block that maps all even numbers n to n+1, and all odd numbers n to 0.

{|e_^*}/Frm
{     }/       Group array by same value from block
|e            1 if the element is even, 0 if odd.
_^          Get another copy of the current element and increment by 1
*         Multiply them
F      For each group execute the rest of the program
r     Reverse the group
m    Print elements from the group, one element per line.


## Perl 5 with -p, 42 bytes

map{$_%2?$\=$_.$\:print$\.$_,$\=""}$_,<>}{


Try it online!

# J, 27 bytes

[:;]<@(|.^:{.);.1~1,2|2-/\]


Try it online!

A minor improvement over miles' J answer.

### How it works

[:;]<@(|.^:{.);.1~1,2|2-/\]  NB. Input: an integer vector V
1,2|2-/\]  NB. Pairwise difference modulo 2 and prepend 1
NB. giving starting positions for same-parity runs
]          ;.1~  NB. Cut V at ones of ^ as the starting point and
(|.^:{.)      NB. reverse each chunk (head of the chunk) times
NB.   (effectively, reverse odd chunks only)
<@              NB. and enclose it
[:;                 NB. Finally, flatten the result


# JavaScript (Node.js), 53 bytes

a=>a.map((c,k)=>o.splice(n=c&1?+n:k+1,0,c),o=n=[])&&o


Try it online!

# JavaScript (Node.js), 54 bytes

a=>a.map((c,k)=>o.splice(n=c&1?n:k+1,0,c),o=[],n=0)&&o


Try it online!

# R, 69 bytes

v=scan();ifelse(o<-v%%2,v[rep(2*cumsum(r<-rle(o)$l)-r+1,r)-seq(v)],v)  Try it online! Ungolfed, commented version: o=v%%2 # get the odd numbers in the list r=rle(o)$l          # find the run lengths of odd/even numbers

# reverse the indices of all the runs:
e=cumsum(r)         # the indices of the ends of each run
ie=rep(e,r)         # for each index in v, the index of the end of its run
is=rep(e-r+1,r)     # for each index in v, the index of the start of its run
irev=is+(ie-seq(v)) # reverses all runs

# the above 4 steps can be combined as:
irev=rep(2*cumsum(r)-r+1,r)-seq(v)

# so: use the reversed indices for odd numbers, and leave the even numbers unchanged
ifelse(o,v[irev],v)


# Clojure, 66 bytes

#(mapcat(fn[x](if(some odd? x)(reverse x)x))(partition-by odd? %))


Try it online!

# Factor + grouping.extras, 62 bytes

[ [ odd? ] group-by [ last2 swap [ reverse ] when ] map-flat ]


Attempt This Online!

# 05AB1E, 11 9 bytes

.γÈN>*}í˜


Explanation:

.γ    }    # Adjacent group the (implicit) input-list by:
È        #  Check if the value is even (1 if even; 0 if odd)
N>*     #  Multiply it by the 1-based index
# (adjacent odd values will be grouped together, and every even
# value is in its own separated group)
í   # After the adjacent group-by: reverse each inner list
˜  # Flatten the list of lists
# (after which the result is output implicitly)


# Vyxal, 10 bytes

λ&›₂¥*;ḊRf


Try it Online!

Port of 05AB1E.

Another 10-byter that I produced independently: ⁽₂Ḋƛh∷ßṘ;f