Reverse odd runs

Question

Inspiration.

Task

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

Answer 1

Python 2, 75 68 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

Answer 2

Python 2, 79 75 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.

Answer 3

APL, 21 20 bytes

{∊⌽¨⍵⊂⍨e⍲¯1↓0,e←2|⍵}

Try it || All test cases

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

Answer 4

Haskell, 46 44 bytes

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

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

Answer 5

Jelly, 10 bytes

Ḃ¬ðœpUżx@F

Try it online! or verify all test cases.

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.
  ð         Begin a new, dyadic link.
            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.

Answer 6

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))

Answer 7

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 :)

Answer 8

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();}

Answer 9

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.

Answer 10

J, 33 31 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

Answer 11

C#, 179 178 177 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 {
            o.AddRange(l);
            o.Add(n);
            l.Clear();
        }
    }
    return o.Concat(l);
};

Kudos to Byeonggon Lee for the original algorithm.

Answer 12

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

Answer 13

Stax, 15 10 bytes^CP437

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.

Answer 14

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.

Answer 15

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?)

Answer 16

JavaScript (ES6) 70 66

Edit 4 bytes saved thx @Neil

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

Answer 17

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

Answer 18

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

Answer 19

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}}

Answer 20

Perl 5 with -p, 42 bytes

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

Try it online!