How to find and return a duplicate value in array

a = ["A", "B", "C", "B", "A"]
a.detect{ |e| a.count(e) > 1 }

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I know this isn't very elegant answer, but I love it. It's beautiful one liner code. And works perfectly fine unless you need to process huge data set.

Looking for faster solution? Here you go!

def find_one_using_hash_map(array)
  map = {}
  dup = nil
  array.each do |v|
    map[v] = (map[v] || 0 ) + 1

    if map[v] > 1
      dup = v
      break
    end
  end

  return dup
end

It's linear, O(n), but now needs to manage multiple lines-of-code, needs test cases, etc.

If you need an even faster solution, maybe try C instead.

And here is the gist comparing different solutions: https://gist.github.com/naveed-ahmad/8f0b926ffccf5fbd206a1cc58ce9743e

Simply find the first instance where the index of the object (counting from the left) does not equal the index of the object (counting from the right).

arr.detect {|e| arr.rindex(e) != arr.index(e) }

If there are no duplicates, the return value will be nil.

I believe this is the fastest solution posted in the thread so far, as well, since it doesn't rely on the creation of additional objects, and #index and #rindex are implemented in C. The big-O runtime is N^2 and thus slower than Sergio's, but the wall time could be much faster due to the the fact that the "slow" parts run in C.

You can do this in a few ways, with the first option being the fastest:

ary = ["A", "B", "C", "B", "A"]

ary.group_by{ |e| e }.select { |k, v| v.size > 1 }.map(&:first)

ary.sort.chunk{ |e| e }.select { |e, chunk| chunk.size > 1 }.map(&:first)

And a O(N^2) option (i.e. less efficient):

ary.select{ |e| ary.count(e) > 1 }.uniq