# Maximum box gap

You have a row of numbered boxes, with some filled with items and some empty. Your job is to output the number of boxes from the first to the last box, inclusive, minus the number of boxes of the largest gap between two filled boxes.

Rules

Your program should handle up to 1000 boxes. Score = size of program in bytes. -50 if you don't use a built-in sorting functionality. -150 if you program in assembly :D All cases will have at least one gap.

Input is a list of n-th boxes that are filled. Standard loop-holes aren't allowed.

Sample Input

9, 5, 4, 2


Functions with a list as input are fine.

Sample Output

5


This is the number of boxes between the 2nd and 9th box, inclusive, subtracted by the number of boxes of the gap between the 5th and 9th box (8 - 3 = 5)

• @MartinBüttner I've edited it so hopefully it's clearer.
– qwr
Oct 1, 2014 at 21:45
• @MartinBüttner Functions are okay.
– qwr
Oct 1, 2014 at 21:50
• Can I assume input to be like [9 5 4 2] ? Oct 1, 2014 at 21:58
• 1. Are the box numbers always in descending order? At least one answer seems to assume this. 2. Will the box numbers always be positive integers? Oct 2, 2014 at 2:41
• @Dennis No, and the boxes are numbered starting from 1.
– qwr
Oct 2, 2014 at 3:02

# CJam, -19-22 -23 (27 bytes - 50 bonus)

Vq~_:+,\&_(+]z{~-~:Xe>}/X+z


Try it online.

### Example run

\$ cjam box-gap.cjam <<< '[9 5 4 2]'; echo
5


### How it works

V                              " V := 0                                          ";
q~                            " Q := eval(input))                               ";
_:+,                        " S := range(sum(Q))                              ";
\&                      " T := S ∩ T          # Preserves the order of S. ";
_(+                   " U := T[1:] + [T]                             ";
]z                 " Z := zip(T, U)                                  ";
{       }/       " for (X,Y) ∊ Z:                                  ";
:-~:Me>         "   V := max(V, M := ~(X - Y))                    ";
M+z    " R := abs(V + M)                                 ";


## CJam, -8 (43 42 bytes - 50 bonus)

l~:A{e>}*A{e<}*-)A{_A\-f{-z}{e<}*}%{e>}*(-


Using tips by @Dennis!

Try it online here

Input is like:

[9 5 4 2]

• The bug with folding e< and e> is really annoying, isn't it? ;) Oct 1, 2014 at 21:59
• Yeah, 9 extra characters :( Oct 1, 2014 at 22:00
• You can use f{-z} instead of \f-{z}%. (I was going to suggest :z, but that doesn't seem to work either.) Oct 2, 2014 at 5:04
• Yeah, too many non : supported operations :( Oct 2, 2014 at 5:07
• Sadly. You don't need the \ before f. Oct 2, 2014 at 5:10

## Mathematica, 44 bytes

f=Max@#-Min@#-Max[RotateLeft[l=Sort@#]-l]+2&


Defines a function that takes one argument in form of a list {9, 5, 4, 2}.

• Wow, you're fast. What does purpose does RotateLeft have?
– qwr
Oct 1, 2014 at 21:51
• @qwr Well, it rotates the array to the left ^^. So {2,4,5,9} becomes {4,5,9,2}. Subtracting the original (sorted) list, gives me all gaps (plus one), i.e. {2,1,4,-7}. The last element is of course useless, but since it's always negative it doesn't matter for Max. Oct 1, 2014 at 21:53

## APL (Dyalog 14) (34 - 50 = -16)

{⎕ML←3⋄(1+V-⌊/⍵)-⌈/≢¨⊂⍨~⍵∊⍨⍳V←⌈/⍵}


This is a function that takes a list.

Or, for the same amount of characters:

⎕ML←3⋄(1+V-⌊/B)-⌈/≢¨⊂⍨~B∊⍨⍳V←⌈/B←⎕


This reads the list from the keyboard.

## Python 2 - 54 (117104-50)

x=input()
M=max(x)-min(x)+1
X=['1']*M
for i in x:X[i-min(x)]='0'
print M-len(max("".join(X).split('0')))


A program that pretends that the buckets are real, then awkwardly splits them into groups. Takes input like 9,5,4,2.