Finding duplicate balls in a basket?

Question

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

Answer 1

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.

Answer 2

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"

Answer 3

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│
└───┘

Answer 4

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

Answer 5

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]

Answer 6

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

Answer 7

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

Answer 8

Mathematica 23

n contains the list.

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

Answer 9

Python 2.7, 67 chars

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

Answer 10

R - 18 chars

which(table(x)==1)

Answer 11

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=+/⍵∘.≡⍵}

Answer 12

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'