31
\$\begingroup\$

A simple challenge for your Monday evening (well, or Tuesday morning in the other half of the world...)

You're given as input a nested, potentially ragged array of positive integers:

[1, [[2, 3, [[4], 5], 6, [7, 8]], 9, [10, [[[11]]]], 12, 13], 14]

Your task is to determine its depth, which is the greatest nesting-depth of any integer in the list. In this case, the depth of 11 is 6, which is largest.

You may assume that none of the arrays will be empty.

You may write a program or function, taking input via STDIN (or closest alternative), command-line argument or function argument and outputting the result via STDOUT (or closest alternative), function return value or function (out) parameter.

Input may be taken in any convenient list or string format that supports non-rectangular arrays (with nested arrays of different depths), as long as the actual information isn't preprocessed.

You must not use any built-ins related to the shape of arrays (including built-ins that solve this challenge, that get you the dimensions of a nested array). The only exception to this is getting the length of an array.

Standard rules apply.

Test Cases

[1]                                                               -> 1
[1, 2, 3]                                                         -> 1
[[1, 2, 3]]                                                       -> 2
[3, [3, [3], 3], 3]                                               -> 3
[[[[1], 2], [3, [4]]]]                                            -> 4
[1, [[3]], [5, 6], [[[[8]]]], 1]                                  -> 5
[1, [[2, 3, [[4], 5], 6, [7, 8]], 9, [10, [[[11]]]], 12, 13], 14] -> 6
[[[[[[[3]]]]]]]                                                   -> 7
\$\endgroup\$
  • 2
    \$\begingroup\$ After discussion in chat I've decided to allow length built-ins, because some languages require them to iterate over an array cleanly. \$\endgroup\$ – Martin Ender Feb 8 '16 at 23:09
  • 2
    \$\begingroup\$ Just for general education: is APL's built-in primitive for exactly this. \$\endgroup\$ – Adám Feb 9 '16 at 16:09
  • \$\begingroup\$ @MartinBüttner I've run into a tiny problem. I started doing this in java, unfourtunetly when testing inputs the commas are causing it to split the inputs into multiple command line arguments rather then one. Can I use the escape character \ in the inputs? EDIT: nevermind just tried it like that. That doesn't even work either. Darn can I not use CMD args? \$\endgroup\$ – Ashwin Gupta Feb 10 '16 at 5:21
  • \$\begingroup\$ @AshwinGupta can't you wrap the command line argument in quotes? You can also read input from STDIN or submit a function that takes an actual array object as a parameter. \$\endgroup\$ – Martin Ender Feb 10 '16 at 10:39
  • \$\begingroup\$ @MartinBüttner oh I didn't know that quotes thing I'll try it out. Currently just using Scanner. (System.in). I believe that is a form of STDIN? \$\endgroup\$ – Ashwin Gupta Feb 10 '16 at 22:02

37 Answers 37

20
\$\begingroup\$

K, 4 bytes

#,/\

In K, ,/ will join all the elements of a list. The common idiom ,// iterates to a fixed point, flattening an arbitrarily nested list completely. ,/\ will iterate to a fixed point in a similar way, but gather a list of intermediate results. By counting how many intermediate results we visit before reaching the fixed point (#), we get the answer we want: the maximum nesting depth.

"Count of join over fixed-point scan".

In action:

 (#,/\)'(,1
        1 2 3
        ,1 2 3
        (3;(3;,3;3);3)
        ,((,1;2);(3;,4)))
1 1 2 3 4
\$\endgroup\$
15
\$\begingroup\$

Retina, 10

  • Saved 1 byte thanks to @ӍѲꝆΛҐӍΛПҒЦꝆ
  • Saved 14 extra bytes thanks to @MartinBüttner
+`\w|}{

{

Here the input format is a bit contrived - _ characters are used for list separators, so an input would look like this {1_{{2_3_{{4}_5}_6_{7_8}}_9_{10_{{{11}}}}_12_13}_14}

  • Stage 1 - repeatedly remove }{ and all other \w characters. This has the effect of a) making all lists at all levels consist of only one element and b) removing all non-list-structural characters.
  • Stage 2 - count remaining {. This gives the deepest level of nesting.

Try it online.


If that's too much of a stretch, then the previous answer was:

Retina, 13

Assumes lists are contained in curly braces {}.

+`[^}{]|}{

{

Try it online.

\$\endgroup\$
  • 1
    \$\begingroup\$ Your code can be shortened to 13 bytes (11 if you stretch the input format a bit). Let me know if you want a hint. :) (I don't really want to post it myself, since it's virtually the same solution.) \$\endgroup\$ – Martin Ender Feb 9 '16 at 7:33
  • \$\begingroup\$ It's two things. a) You can save a byte or so by slightly tweaking the input format. b) You can save a lot of bytes regardless of that... can you find a shorter (and much simpler) solution if you try not to handle multiple test cases in a single run? \$\endgroup\$ – Martin Ender Feb 9 '16 at 18:19
  • \$\begingroup\$ I didn't even think of that. That's amount byte saved then. My change to the input format would have been even weaker. Regarding b) remember what Retina's very first and simplest mode of operation was? \$\endgroup\$ – Martin Ender Feb 9 '16 at 18:35
  • 1
    \$\begingroup\$ yep. My a) was referring to removing spaces from the input though. And you can then save two more bytes by using _ instead of , but that might be a bit of a stretch. \$\endgroup\$ – Martin Ender Feb 9 '16 at 18:37
  • \$\begingroup\$ @MartinBüttner Nice idea! Agreed - _ separators might be too contrived. So I'm leaving both versions in the answer \$\endgroup\$ – Digital Trauma Feb 9 '16 at 18:47
12
\$\begingroup\$

Python 2, 33 bytes

f=lambda l:l>{}and-~max(map(f,l))

Recursively defines the depth by saying the depth of a number is 0, and the depth of a list is one more than the maximum depth of its elements. Number vs list is checked by comparing to the empty dictionary {}, which falls above numbers but below lists on Python 2's arbitrary ordering of built-in types.

\$\endgroup\$
  • \$\begingroup\$ Length built-ins are now allowed if it helps. \$\endgroup\$ – Martin Ender Feb 8 '16 at 23:10
6
\$\begingroup\$

Pyth - 11 10 7 bytes

1 bytes saved thanks to @Dennis

4 bytes saved thanks to @Thomas Kwa

eU.usNQ

Try it online here.

Keeps on summing the array till it stops changing, which means its just a number, does this cumulatively to save all the intermediate results and gets length by making a urange with the same length as list and taking the last element.

\$\endgroup\$
  • \$\begingroup\$ m!!d can become &R1. \$\endgroup\$ – Dennis Feb 8 '16 at 22:21
  • \$\begingroup\$ @Dennis cool, that's smart \$\endgroup\$ – Maltysen Feb 8 '16 at 22:21
  • \$\begingroup\$ @ThomasKwa l isn't allowed in OP. \$\endgroup\$ – Maltysen Feb 8 '16 at 22:25
  • \$\begingroup\$ @ThomasKwa that is really smart, thanks! \$\endgroup\$ – Maltysen Feb 8 '16 at 22:31
  • \$\begingroup\$ Length built-ins are now allowed if it helps. \$\endgroup\$ – Martin Ender Feb 8 '16 at 23:10
6
\$\begingroup\$

Haskell, 43 bytes

'['#x=x-1
']'#x=x+1
_#x=x
maximum.scanr(#)0

Usage example: maximum.scanr(#)0 $ "[1, [[3]], [5, 6], [[[[8]]]], 1]" -> 5.

Haskell doesn't have mixed lists (Integer mixed with List of Integer), so I cannot exploit some list detection functions and I have to parse the string.

Im starting at the right with 0 and add 1 for every ], subtract 1 for every [ and keep the value otherwise. scanr keeps all intermediate results, so maximum can do it's work.

\$\endgroup\$
5
\$\begingroup\$

JavaScript (ES6), 35 bytes

f=a=>a[0]?Math.max(...a.map(f))+1:0

Explanation

Recursive function that returns the maximum depth of an array, or 0 if passed a number.

var solution =

f=a=>
  a[0]?                   // if a is an array
    Math.max(...a.map(f)) // return the maximum depth of each element in the array
    +1                    // add 1 to increase the depth
  :0                      // if a is a number, return 0

// Test cases
result.textContent =
`[1]                                                              -> 1
[1, 2, 3]                                                         -> 1
[[1, 2, 3]]                                                       -> 2
[3, [3, [3], 3], 3]                                               -> 3
[[[[1], 2], [3, [4]]]]                                            -> 4
[1, [[3]], [5, 6], [[[[8]]]], 1]                                  -> 5
[1, [[2, 3, [[4], 5], 6, [7, 8]], 9, [10, [[[11]]]], 12, 13], 14] -> 6
[[[[[[[3]]]]]]]                                                   -> 7`
.split`\n`.map(t=>(c=t.split`->`.map(p=>p.trim()),c[0]+" == "+c[1]+": "+(solution(eval(c[0]))==c[1]?"Passed":"Failed"))).join`\n`
<input type="text" id="input" value="[1, [[2, 3, [[4], 5], 6, [7, 8]], 9, [10, [[[11]]]], 12, 13], 14]" />
<button onclick="result.textContent=solution(eval(input.value))">Go</button>
<pre id="result"></pre>

\$\endgroup\$
  • \$\begingroup\$ Length built-ins are now allowed if it helps. \$\endgroup\$ – Martin Ender Feb 8 '16 at 23:10
4
\$\begingroup\$

MATL, 11 14 15 bytes

'}{'!=dYsX>

Curly braces are used in MATL for this type of arrays. Anyway, the input is taken and processed as a string, so square brackets could equally be used, modifying the two characters in the code.

Try it online!

          % implicitly take input as a string (row array of chars)
'}{'!     % 2x1 (column) char array with the two curly brace symbols
=         % 2-row array. First / second row contains 1 where '}' / '{' is found
d         % second row minus first row
Ys        % cumulative sum of the array
X>        % maximum of the array
          % implicitly display result
\$\endgroup\$
  • \$\begingroup\$ Length built-ins are now allowed if it helps. \$\endgroup\$ – Martin Ender Feb 8 '16 at 23:10
4
\$\begingroup\$

Octave, 29 bytes

@(a)max(cumsum(92-(a(a>90))))

Maps [ to 1 and ] to -1, then takes the max of the cumulative sum.

Input is a string of the form

S6 = '[1, [[3]], [5, 6], [[[[8]]]], 1]';

Sample run on ideone.

\$\endgroup\$
  • \$\begingroup\$ Should you use {, }? The Octave equivalent to the arrays in the OP are cell arrays, I think \$\endgroup\$ – Luis Mendo Feb 9 '16 at 10:42
  • \$\begingroup\$ @LuisMendo No, because that's 2 extra bytes :) Plus, since I never actually create the array, simply parse the input string, I don't think it matters. But you've reminded me to add the expected input to my answer. \$\endgroup\$ – beaker Feb 9 '16 at 16:20
  • \$\begingroup\$ True! Longer ASCII code \$\endgroup\$ – Luis Mendo Feb 9 '16 at 17:16
  • \$\begingroup\$ @LuisMendo Actually, 1 byte longer. That second comparison only needs to be greater than '9'. But you get the idea :D \$\endgroup\$ – beaker Feb 9 '16 at 17:20
4
\$\begingroup\$

Julia, 55 26 bytes

f(a)=0a!=0&&maximum(f,a)+1

This is a recursive function that accepts a one-dimensional array with contents of type Any and returns an integer. When passing an array to the function, prefix all brackets with Any, i.e. f(Any[1,Any[2,3]]).

The approach is pretty simple. For an input a, we multiply a by 0 and check whether the result is the scalar 0. If not, we know that a is an array, so we apply the function to each element of a, take the maximum and add 1.

Saved 29 bytes thanks to Dennis!

\$\endgroup\$
  • 2
    \$\begingroup\$ Dat golf. <filler> \$\endgroup\$ – El'endia Starman Feb 10 '16 at 7:27
3
\$\begingroup\$

Ruby, 53 bytes

i=0;p gets.chars.map{|c|i+=('] ['.index(c)||1)-1}.max

Input from STDIN, output to STDOUT.

i=0;                 initialize counter variable
p                    output to STDOUT...
gets                 get line of input
.chars               enumerator of each character in the input
.map{|c|             map each character to...
i+=                  increment i (and return the new value) by...
('] ['.index(c)||1)  returns 0 for ], 2 for [, 1 for anything else
-1                   now it's -1 for ], 1 for [, 0 for anything else
                     therefore: increment i on increase in nesting, decrement i
                       on decrease, do nothing otherwise
}.max                find the highest nesting level that we've seen
\$\endgroup\$
  • \$\begingroup\$ Length built-ins are now allowed if it helps. \$\endgroup\$ – Martin Ender Feb 8 '16 at 23:10
3
\$\begingroup\$

Jelly, 10 7 bytes

¬;/SпL

Try it online! or verify all test cases.

How it works

¬;/SпL  Main link. Input: A (list)

¬        Negate all integers in A. This replaces them with zeroes.
    п   Cumulative while loop.
   S       Condition: Compute the sum of all lists in A.
                      If the sum is an integer, it will be zero (hence falsy).
 ;/        Body:      Concatenate all lists in A.
      L  Count the number of iterations.

Update

While writing this answer, I noticed that Jelly behaves rather weirdly for ragged lists, because I calculated the depth of a list as the incremented minimum of depths of its items.

This has been addressed in the latest version, so the following code (6 bytes) would work now.

¬SSпL

This sums the rows of the array instead of concatenating them.

\$\endgroup\$
  • \$\begingroup\$ Presumably, ŒḊ is newer than the challenge? \$\endgroup\$ – caird coinheringaahing Oct 19 '17 at 16:28
  • \$\begingroup\$ You must not use any built-ins related to the shape of arrays (including built-ins that solve this challenge, that get you the dimensions of a nested array). \$\endgroup\$ – Dennis Oct 19 '17 at 16:39
3
\$\begingroup\$

Mathematica, 18 bytes

Max@#+1&//@(0#-1)&
\$\endgroup\$
  • \$\begingroup\$ Could you explain it, please? \$\endgroup\$ – skan Dec 20 '16 at 1:49
3
\$\begingroup\$

Mathematica, 27 20 bytes

Max[#0/@#]+1&[0#]-1&

Simple recursive function.

\$\endgroup\$
  • \$\begingroup\$ It's possible to void the If, saving 7 bytes. (Let me know if you want a hint.) \$\endgroup\$ – Martin Ender Feb 9 '16 at 7:39
  • \$\begingroup\$ @MartinBüttner I give up... A Replace-based solution is at least as long as this one... \$\endgroup\$ – LegionMammal978 Feb 9 '16 at 12:16
  • 1
    \$\begingroup\$ Mapping over an integer is a no-op: Max[#0/@#]+1&[0#]-1&. The -1 can also go inside the inner call like ...&[0#-1]&. \$\endgroup\$ – Martin Ender Feb 9 '16 at 12:30
3
\$\begingroup\$

PHP, 61 bytes

function d($a){return is_array($a)?1+max(array_map(d,$a)):0;}

recursive function that uses itself as a mapping function to replace each element with its depth.

\$\endgroup\$
  • \$\begingroup\$ I just noticed: The same thing in JS has only 35 bytes. Still pretty in php. \$\endgroup\$ – Titus Dec 20 '16 at 3:47
  • \$\begingroup\$ Nice, you beat me. But I updated mine and beat you back :) \$\endgroup\$ – aross Dec 20 '16 at 9:31
3
\$\begingroup\$

PHP, 84 72 64 63 60 bytes

Note: requires PHP 7 for the combined comparison operator. Also uses IBM-850 encoding

for(;$c=$argv[1][$i++];)$c>A&&$t=max($t,$z+=~ú<=>$c);echo$t;

Run like this:

php -r 'for(;$c=$argv[1][$i++];)$c>A&&$t=max($t,$z+=~ú<=>$c);echo$t;' "[1, [[3]], [5, 6], [[[[8]]]], 1]"
  • Saved 12 bytes by just counting braces of the string representation instead
  • Saved 8 bytes by simplifying string comparisons and using ordinal number of the char in case of [ and ]
  • Saved a byte by not casting $i to an int. String offsets are casted to an int implicitly
  • Saved 3 bytes by using combined comparison operator instead of ordinal number
\$\endgroup\$
  • \$\begingroup\$ Nice idea, great golfing! Check out mine. \$\endgroup\$ – Titus Dec 20 '16 at 2:42
2
\$\begingroup\$

C, 98 69 bytes

29 bytes off thanks @DigitalTrauma !!

r,m;f(char*s){for(r=m=0;*s;r-=*s++==93)r+=*s==91,m=r>m?r:m;return m;}

Takes an string as input and return the result as integer.

Live example in: http://ideone.com/IC23Bc

\$\endgroup\$
2
\$\begingroup\$

Python 3, 42 39 bytes

-3 bytes thanks to Sp3000

This is essentially a port of xnor's Python 2 solution:

f=lambda l:"A"<str(l)and-~max(map(f,l))

Unfortunately, [] > {} returns an unorderable types error, so that particular clever trick of xnor's cannot be used. In its place, -0123456789 are lower in ASCII value than A, which is lower than [], hence the string comparison works.

\$\endgroup\$
2
\$\begingroup\$

CJam (15 bytes)

q~{__e_-M*}h],(

Online demo

Dissection

q~      e# Read line and parse to array
{       e# Loop...
  _     e#   Leave a copy of the array on the stack to count it later
  _e_-  e#   Remove a flattened version of the array; this removes non-array elements from
        e#   the top-level array.
  M*    e#   Remove one level from each array directly in the top-level array
}h      e# ...until we get to an empty array
],(     e# Collect everything together, count and decrement to account for the extra []

For the same length but rather more in ugly hack territory,

q'[,-{:~_}h],2-
\$\endgroup\$
  • \$\begingroup\$ s/ugly/beautiful/ \$\endgroup\$ – Dennis Feb 10 '16 at 20:05
  • \$\begingroup\$ @Dennis, I was referring specifically to the use of '[,- to strip the string down to [], which relies on the contents being limited. The approach which flattens works regardless of the contents of the array. \$\endgroup\$ – Peter Taylor Feb 10 '16 at 22:05
  • \$\begingroup\$ The second one is prettier. The first one has two types of mismatched braces \$\endgroup\$ – Cyoce Oct 7 '16 at 0:21
2
\$\begingroup\$

Sed, 40 characters

(39 characters code + 1 character command line option.)

s/[^][]+//g
:;s/]\[//;t
s/]//g
s/\[/1/g

Input: string, output: unary number.

Sample run:

bash-4.3$ sed -r 's/[^][]+//g;:;s/]\[//;t;s/]//g;s/\[/1/g' <<< '[1, [[2, 3, [[4], 5], 6, [7, 8]], 9, [10, [[[11]]]], 12, 13], 14]'
111111

Sed, 33 characters

(32 characters code + 1 character command line option.)

If trailing spaces are allowed in the output.

s/[^][]+//g
:;s/]\[//;t
y/[]/1 /

Input: string, output: unary number.

Sample run:

bash-4.3$ sed -r 's/[^][]+//g;:;s/]\[//;t;y/[]/1 /' <<< '[1, [[2, 3, [[4], 5], 6, [7, 8]], 9, [10, [[[11]]]], 12, 13], 14]'
111111      
\$\endgroup\$
2
\$\begingroup\$

Hexagony, 61 bytes

Edit: Thanks @Martin Ender♦ for saving me 1 byte from the marvelous -1 trick!

|/'Z{>"-\..(.."/'&<'..{>-&,=+<$.{\$._{..><.Z=/...({=Z&"&@!-"

Try it online to verify test cases!

The images below are not modified but the flow is basically the same. Also note that this will return -1 if the input is not an array (i.e. without []).

I have lots of no-ops inside the Hexagon... I guess it can definitely be golfed more.

Explanation

In brief, it adds -1 when encounters a [ and adds 1 when encounters a ]. Finally it prints the max it has got.

Let's run along Test Case 5 to see its behaviour when it runs along the String [1, [[3]], [5, 6], [[[[8]]]], 1]:

It starts at the beginning and takes its input at the W corner:

Brackets

Since there is still input (not the null character \0 or EOL), it wraps to the top and starts the crimson path.

Here is what happens when from there till cute ><:

, reads [ into Buffer, and { and Z sets the constant Z to be 90. ' moves to Diff and - calculates the difference. For [ and ] the difference will be 1 and 3 respectively. For numbers and spaces and commas it'll be negative.

M1 M2

Then we run ( twice (once at the end of crimson path, one at the start after wrapping at the green path) to get -1 and 1 resp for [ and ]. Here we change the naming of Diff to Value. Add this Value to Depth. (I used Z& to ensure that it copies the right neighbor). Then we calculate lastMin - Depth and got a number on the Memory edge minLR.

Then we apply & (at the end of green path) to minLR: If the number is <=0, it copies the left value (i.e. lastMin - Depth <= 0 => lastMin <= Depth), otherwise it takes the right value.

We wraps to the horizontal blue path and we see Z& again which copies the minLR. Then we "& and made a copy of the calculated min. The brackets are assumed to be balanced, so the min must be <=0. After wrapping, the blue path go left and hit (, making the copy 1 less than the real min. Reusing the -, we created one more 1-off copy as a neighbor of Buffer:

M3

Note: copy is renamed as 1-off

When blue path hits \ and got a nice " and < catches it back to the main loop.

When the loop hits 1, , or or other numbers as input:

othersM4

The Diff will become negative and it got reflected back to the main loop for next input.

When everything has gone through the main loop, we reach EOL which makes Buffer -1 and it finally goes to the bottom edge:

M5

' moves the MP to the 1-off copy and ) increments it, and with ~ negation it got the correct Max Depth value which is printed with !

And the story ends with a @.

I guess I must have over complicating things a little bit. If I have had to only "move back" and "print" without incrementing and negation, I would have well saved 2 bytes without using the full Hexagon.

Great thanks to Timwi for Esoteric IDE and Hexagony Colorer!

\$\endgroup\$
  • \$\begingroup\$ You can save a byte by making use of the -1 from , by changing the last row to: @!-". (although I agree that it is probably possible to shave off a lot more or even fit this into side-length 4 with some restructuring). \$\endgroup\$ – Martin Ender Oct 12 '16 at 12:33
  • \$\begingroup\$ Haven't thought of making use of the -1! Will edit once I got my computer. If the temp is on left neighbor, I would have saved quite a few Z from using Z&. And there should be better ways to start the program with the implicit if. \$\endgroup\$ – Sunny Pun Oct 12 '16 at 15:51
2
\$\begingroup\$

brainfuck, 48 bytes

,[<++[>-<------]>++[+[<]>>[-]]+<,]-[<[>+<-]>>]<.

Formatted:

,
[
  <++[>-<------]>++
  [
    not close paren
    +
    [
      not open paren
      <
    ]
    >>[-]
  ]
  +<,
]
-[<[>+<-]>>]
<.

Takes input formatted like (1, ((3)), (5, 6), ((((8)))), 1) and outputs a byte value.

Try it online.

This stores the depth by memory location, moving the pointer right for ( and left for ) and ignoring other characters. Visited cells are marked with a 1 flag, so at the end of the main loop there will be depth + 1 flags to the right of the current cell. These are then added to print the final result.


A previous 69-byte solution using a different approach:

,
[
  >>++[<<->>------]<-<++
  [
    not close paren
    >++<+
    [
      not open paren
      >-<[-]
    ]
  ]
  <
  [
    [>+>]
    <[<-<]
    >
  ]
  >>[<+> >+<-]
  ,
]
<.

In this version, the depth and max depth are stored explicitly in cells.

\$\endgroup\$
1
\$\begingroup\$

Pyth, 15 13 bytes

-2 bytes by @Maltysen

eSm-F/Ld`Y._z

Counts the difference between the cumulative counts of [ and ], and takes the maximum. Y is the empty array, and its string representation (`) is conveniently [].

Try it here.

\$\endgroup\$
  • \$\begingroup\$ Length built-ins are now allowed if it helps. \$\endgroup\$ – Martin Ender Feb 8 '16 at 23:10
1
\$\begingroup\$

CJam, 19 22 23 bytes

0l{_91=\93=-+_}%:e>

Similar idea to my MATL answer.

Thanks to Peter Taylor for removing 3 bytes

Try it here

0                            push a 0
l                            read line as string
{            }%              map this block on the string
  _91=\93=-                  1 if it's an opening bracket, -1 if closing
           +_                cumulative sum
               :e>           fold maximum function
\$\endgroup\$
1
\$\begingroup\$

Perl 5, 34 bytes

32, plus two for -p

{s&]\[|[^][]&&g&&redo}$_=@a=/]/g

Stolen from Digital Trauma's Retina answer… which is 26% shorter than this. :-)

Or, equally:

{s&]\[|[^][]&&g&&redo}$_=y///c/2
\$\endgroup\$
  • \$\begingroup\$ @Cyoce, why? ] doesn't need escaping, except in brackets. \$\endgroup\$ – msh210 Oct 7 '16 at 6:11
  • \$\begingroup\$ @Cyoce, s&...&...&g is the substitution operator. See perldoc.perl.org/perlop.html \$\endgroup\$ – msh210 Oct 7 '16 at 9:40
1
\$\begingroup\$

Ruby, 51 characters

(Started as improvement suggestion for Doorknob's Ruby answer but ended differently. So I posted it as separate answer. Upvotes for the depth counting idea (?\\<=>$&, descending from '] ['.index(c)) should go to the original answer.)

m=i=0
gets.gsub(/\[|\]/){m=[m,i+=?\\<=>$&].max}
p m

Input: string, output: number.

Sample run:

bash-4.3$ ruby -e 'm=i=0;gets.gsub(/\[|\]/){m=[m,i+=?\\<=>$&].max};p m' <<< '[1, [[2, 3, [[4], 5], 6, [7, 8]], 9, [10, [[[11]]]], 12, 13], 14]'
6
\$\endgroup\$
1
\$\begingroup\$

Perl 6, 53 bytes

The closure:

{my ($m,$d);/[\[{$d++;$m=max $m,$d}|\]{$d--}|.]*/;$m}

Needs an argument, eg:

{my ($m,$d);/[\[{$d++;$m=max $m,$d}|\]{$d--}|.]*/;$m}("[[[3]][2]]")
3

Explanation:

{ my ($m,$d);                 # start closure, declare variables    

  /                           # start regex match

   [                          # start (non-capturing) group

     \[ {$d++;$m=max $m,$d} | # match [ and track depth; OR

     \] {$d--}              | # match ] and track depth; OR

     .                        # match one character

   ]*                         # repeat group

  /;                          # end regex

  $m                          # return max depth
}
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1
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Minkolang 0.15, 31 29 24 bytes

Overhauled my algorithm upon inspiration by Luis Mendo's CJam answer and saved 5 bytes!

od5&j$ZN.d"["=$r"]"=~++d

Try it here!

Explanation

Essentially, what this code does is keep a running total with +1 for each [ and -1 for each ], keeping track of the maximum value reached, outputting that maximum at the end. Looping is handled by the toroidal nature of Minkolang's codebox.

od           Take character from input and duplicate it (0 if input is empty)
  5&         Pop top of stack and skip the following five spaces if 0
    j$Z      Push the maximum value of the stack
       N.    Output as number and stop.

  d                  Duplicate top of stack for character tests
   "["=              +1 if the character is [
       $r            Swap top two items of stack
         "]"=~       -1 if the character is ]
              ++     Add twice
                d    Duplicate top of stack for the running total
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1
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Ruby, 41 characters

f=->a,d=1{a.map{|e|f[e,d+1]rescue d}.max}

Parameter: array, return: number.

Sample run:

2.1.5 :001 > f=->a,d=1{a.map{|e|f[e,d+1]rescue d}.max}
 => #<Proc:0x0000000214d258@(irb):1 (lambda)> 

2.1.5 :002 > f[[1, [[2, 3, [[4], 5], 6, [7, 8]], 9, [10, [[[11]]]], 12, 13], 14]]
 => 6 
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1
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Oracle SQL 11.2, 133 bytes

SELECT MAX(d)FROM(SELECT SUM(DECODE(SUBSTR(:1,LEVEL,1),'[',1,']',-1,0))OVER(ORDER BY LEVEL)d FROM DUAL CONNECT BY LEVEL<=LENGTH(:1));

Un-golfed

SELECT MAX(d)
FROM   (
         SELECT SUM(DECODE(SUBSTR(:1,LEVEL,1),'[',1,']',-1,0))OVER(ORDER BY LEVEL) d 
         FROM   DUAL 
         CONNECT BY LEVEL<=LENGTH(:1)
       );

The CONNECT BY creates one row per character in the input string.

The SUBSTR isolates the character corresponding to the row number.

The DECODE translates each '[' to 1, each ']' to -1 and every other character to 0.

The analytic SUM sums each 1, -1 and 0 from the preceding rows, including the current row;

The MAX sums is the depth.

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1
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Java 8, 95

This is a lambda expression for a ToIntFunction<String>. Input is taken as a String in the OP's examples format.

s->{int d=e=-1;for(String t:s.split("[")){d=++e>d?e:d;e-=t.split("]",-1).length()-1;}return d;}

fairly straightfoward. Split the string using [ as the delimiter. For each of them, increment the counter e and compare it with the counter d, keeping the larger of them in d. Then split the current iteration's string using ] as the delimiter this time and subtract the number of extra splits from e.

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