monthconcatenation

Question

The goal is to output the number of the months given as input in a compact concatenated form which is still parsable if one knows the construction rules:

If either:

• January is followed by January, February, November or December; or
• November is followed by January or February

There must be a separator placed between.
Otherwise there should be no separator.

As such the output may be parsed. For example:

• March, April and September -> 349
• January, January and February -> 1-1-2
• January, January and December -> 1-1-12
• January, November and February -> 1-11-2
• November, January and February -> 11-1-2
• November and December -> 1112

Thus any run of ones is either a run of Novembers, a run of Novembers followed by a December, a run of Novembers followed by an October, or a January followed by an October. This may be parsed by looking to the right of such runs as the resulting string is read from left to right.

Input

A list of months numbers ([1-12]) in any format you want (list, JSON, separated by one or more characters, entered one by one by user, …).

the same month can be present more than once.

Output

The compact concatenated form described above. If a separator is needed, you can freely choose one.

Output examples for January, February and March (1-2-3):

• 1-23 (chosen for the test cases below)
• 1 23
• 1/23
• 1,23
• ...

Test cases

[1]       => 1
[1, 2]    => 1-2
[2, 1]    => 21
[12]      => 12
[1, 11]   => 1-11
[11, 1]   => 11-1
[1, 1, 1] => 1-1-1
[2, 11]   => 211
[11, 2]   => 11-2
[1, 2, 3] => 1-23
[11, 11]  => 1111
[1,1,11,1,12,1,11,11,12]     => 1-1-11-1-121-111112
[10,11,12,11,11,10,1,10,1,1] => 1011121111101101-1
[2,12,12,2,1,12,12,1,1,2,12,12,2,11,2,12,1] => 2121221-12121-1-21212211-2121

Rules

• As usual, this is code-golf, shortest answer in bytes wins.
• Standard loopholes are forbidden.

Answer 1

Python 2.7 96 bytes

Try it out here!

Huge thanks to Jonathan Allan for shortening the answer.

golfed:

b,o=input(),""
for a,n in zip(b,b[1:]):o+=`a`+"-"*(a%10==1and a+n in(2,3,12,13))
print o+`b[-1]`

Explanation:

runs through the input, checks if the current item is a special case and if so handles it, otherwise just concatenates the input, prints the output string.

ungolfed:

input = [2,12,12,2,1,12,12,1,1,2,12,12,2,11,2,12,1]
output = ""
for i in range(len(input) - 1):
    next = input[i + 1]
    item = input[i]
    if item == 1 and (next == 1 or next == 2 or next == 11 or next == 12) or item == 11 and (next == 1 or next == 2):
        output += str(item) + "-"
    else:
        output += str(item)

output += str(input[-1])
    
print output

Answer 2

Jelly, 20 bytes

%⁵=1a+e“£¤€Æ‘⁶ẋ⁸,⁹µ/

TryItOnline! or run all test cases

How?

%⁵=1a+e“£¤€Æ‘⁶ẋ⁸,⁹µ/ - Main link: the list
                   / - reduce
                  µ  - monadic chain separation
%⁵                   - left item mod 10
  =1                 - equal to 1? (Jan or Nov)
    a                - and
     +               - left item plus right item
      e              - is in
       “£¤€Æ‘        - jelly code page index list:  [2,3,12,13]
                           now True (1) for any of:
                               [1,1], [1,2], [1,11], [1,12], [11,1], or [11,2]
             ⁶       - literal space (the separator being used)
              ẋ      - repeated that many times
               ⁸     - left argument
                ,    - pair
                 ⁹   - right argument