4
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Question

A basket is full n colored balls . Except one ball all the other colors are repeating. Find the one color that does not repeat,in O(n) space complexity.

What if each ball is repeated exactly twice?

you can consider it to be an array or a list or any container with a capacity of 100

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  • 6
    \$\begingroup\$ Unless you specify some kind of input and output format (or the flexibility of what you allow) people are going to make vastly different assumptions. \$\endgroup\$ – Peter Taylor Jul 9 '12 at 18:24
  • 2
    \$\begingroup\$ This question isn't suited for CodeGolf. You need to specify a winning criteria. Otherwise, try StackOverflow.com if you are needing help. \$\endgroup\$ – MrZander Jul 9 '12 at 21:15
  • \$\begingroup\$ Do you want us to find the duplicate balls as the title suggests? \$\endgroup\$ – Prince John Wesley Jul 10 '12 at 15:00
  • \$\begingroup\$ @PrinceJohnWesley Yes thats the whole idea!! \$\endgroup\$ – samairtimer Jul 10 '12 at 15:36
  • \$\begingroup\$ -1 no input format specified, no output format specified... How you output the repeated ball? Index? Is it one indexed or zero indexed? \$\endgroup\$ – Rohan Jhunjhunwala Jul 8 '16 at 22:31

12 Answers 12

10
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Python - 17 chars

Solely as an expression:

min(B,key=B.count)

Full, standalone program with input and output, 34 chars:

B=input()
print min(B,key=B.count)

edit: What gnibbler said.

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  • 1
    \$\begingroup\$ Python3 would be a little shorter - use input instead of raw_input and () around the print. Since the input format isn't specified, why not assign single letter codes to each colour? eg "rrgbb", then you can drop the split(). And use min instead of sorted FTW \$\endgroup\$ – gnibbler Jul 11 '12 at 23:23
  • \$\begingroup\$ Thanks! Yeah, I guess a string is within the bounds of 'any container' :D The question doesn't even mention input, so I think it's legit with 2.x's input() and have the user wrap quotes around it too. \$\endgroup\$ – daniero Jul 11 '12 at 23:51
6
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K, 17 15

&1=#:'=" "\:0:0

Takes input from stdin as a space separated list

k)&1=#:'=" "\:0:0
red red green blue blue
"green"
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  • \$\begingroup\$ I think you've got me beat this time - it takes me 15 characters just to get the separate strings. \$\endgroup\$ – Gareth Jul 11 '12 at 21:36
  • \$\begingroup\$ ...or maybe not - I've got the input down to 8 which at least gives me a fighting chance... :-) \$\endgroup\$ – Gareth Jul 11 '12 at 21:43
  • \$\begingroup\$ Given the way you successively golfed the last one I wouldn't be surprised to see you beat me again. My solution can't get any shorter, I think. \$\endgroup\$ – tmartin Jul 11 '12 at 21:57
  • \$\begingroup\$ No, I think 20 is my limit here. \$\endgroup\$ – Gareth Jul 11 '12 at 22:15
  • \$\begingroup\$ No need to wrap this in a function, &1=#:'=" "\:0:0 runs fine for 15. \$\endgroup\$ – skeevey Jul 12 '12 at 16:32
5
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J, 31 24 20 characters

{.(~./:#/.~);:1!:1[1

A lot prettier than the previous attempt. Takes input from keyboard.

Makes an assumption that wasn't necessary in the previous version - that there is exactly one colour with one ball. If there are 2 colours with one ball you'll get the first that appears in the list, if there are no colours with only one ball you'll get the colour with the least balls.

Example:

   {.(~./:#/.~);:1!:1[1
green green red blue blue
┌───┐
│red│
└───┘
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3
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SQL - 44

Find Unique Color

SELECT C FROM B GROUP BY C HAVING COUNT(C)=1

Find Duplicate Colors

SELECT C FROM B GROUP BY C HAVING COUNT(C)>1
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  • \$\begingroup\$ The right tool for the job! \$\endgroup\$ – Rohan Jhunjhunwala Jul 8 '16 at 22:32
2
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Python 2.7, 44 chars

Using the same sort of format as Keith Randall's answer of a function that takes an array and returns the element/colour that only occurs once.

R=lambda B:[a for a in B if B.count(a)<2][0]

Alternatively, if I need to read in colours from the command line and output the colour that is unique (which takes 60 chars):

B=raw_input().split();print[a for a in B if B.count(a)<2][0]
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  • \$\begingroup\$ This is O(1) space complexity - which is usually better than O(n), but technically isn't what the question asks for :) \$\endgroup\$ – gnibbler Jul 10 '12 at 23:32
  • 2
    \$\begingroup\$ I think you could use <2 rather than ==1 to save a char. \$\endgroup\$ – grc Jul 11 '12 at 1:24
  • \$\begingroup\$ @gnibbler: O(1) is a subset of O(n). \$\endgroup\$ – Keith Randall Jul 11 '12 at 3:00
  • \$\begingroup\$ @grc: Thanks, amended the solution. \$\endgroup\$ – JPvdMerwe Jul 11 '12 at 9:22
2
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Clojure - 44

#(some(fn[[c n]](if(= n 1)c))(frequencies%))

That's a lambda function. Example:

(#(some(fn[[c n]](if(= n 1)c))(frequencies%))
  [:red :red :red :blue :green :blue :yellow :yellow])
=> :green
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1
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Mathematica 23

n contains the list.

Cases[Tally@n,{x_,1}:>x]
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1
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GolfScript, 23

After a tremendous amount of help from w0lf, here's the source code.

' '/:x.&{x{1$=},,1=\;},

DEMO


28 characters with the error checking

? is displayed if there is not one unique color. For the sake of keeping the number of characters down to a minimum, the question mark was used (this of course can be changed to something more meaningful).

' '/:x.&{x{1$=},,1=\;},'?'or

DEMO


Commentary

' '/:x        # split up each element and assign it to 'x'
.&            # copy the last element in the stack and do a "setwise and"
{x            # start filtration process and invoke 'x'
{1$=}         # checks for an equality comparison
,,            # find the count of each unique character
1=            # compare value to '1'
\;            # leave only value that has count of '1' on the stack
},            # close filtration process and print element on stack
'?'or         # OPTIONAL: is printed if no element with a count of '1' is found
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0
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Python 2.7, 67 chars

from collections import*
R=lambda B:Counter(B).most_common()[-1][0]
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0
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R - 18 chars

which(table(x)==1)
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0
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APL (14)

B/⍨1=+/B∘.≡B←⎕

Takes input from the keyboard in the form of an APL list, so you need to put quotes around the colors 'like' 'this'. (Or use numbers.)

      B/⍨1=+/B∘.≡B←⎕
⎕:
      'green' 'green' 'red' 'blue' 'blue'
 red 

As a function (that takes a list, also 14 characters):

{⍵/⍨1=+/⍵∘.≡⍵}
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0
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Groovy - 22 chars

it.min{a->it.count(a)}

Attribution and test case:

lone={it.min{a->it.count(a)}}

l = ['red','green', 'red', 'blue', 'green']

assert lone(l) == 'blue'
assert lone(l) != 'red'
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