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Question:

This question already has an answer here:

I have an array like [1,1,1,2,4,6,3,3] and I would like to get the list of repeated elements, in this case [1,3]. I wrote this:

`my_array.select{|obj|my_array.count(obj)>1}.uniq `

But it is tragically inefficient (o(nÂ²)). Do you have a better idea? If possible concise.

Thanks

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Solution:1

Inspired by Ilya Haykinson's answer:

`def repeated(array) counts = Hash.new(0) array.each{|val|counts[val]+=1} counts.reject{|val,count|count==1}.keys end `

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Solution:2

Using Ruby's Set library:

`require 'set' ary = [1,1,1,2,4,6,3,3] dups = Set.new test_set = Set.new ary.each {|val| dups.add(val) unless test_set.add?(val)} dups.to_a # [1, 3] `

I believe this should be O(n), because Set#add and Set#add? are constant-time operations, as far as I know.

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Solution:3

How about something like this? It will run in O(n).

`a = [1,1,1,2,4,6,3,3] b = {} a.each { |v| if b.has_key? v then b[v] = b[v]+1 else b[v]=1 end } b.reject { |k,v| if v > 1 then false else true end }.keys `

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Solution:4

A O(n) solution (change `<< x`

to `+ [x]`

and `update`

to `merge`

to make it purely functional):

`rs = xs.inject([[], {}]) do |(out, seen), x| [(seen[x] == 1 ? (out << x) : out), seen.update(x => (seen[x] || 0)+1)] end[0] `

A much simpler yet less space-efficient approach:

`rs = xs.group_by { |x| x }.select { |y, ys| ys.size > 1 }.keys `

The same idea avoiding the intermediate hash using a "list-comprehension":

`rs = xs.group_by { |x| x }.map { |y, ys| y if ys.size > 1 }.compact `

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Solution:5

Using `inject`

`[1,1,1,2,4,6,3,3].inject({}){ |ele, n| ele[n] = nil; ele }.keys # => [1, 2, 4, 6, 3] `

### EXPLANATION:

`ele`

hash it's initialled to `{}`

, each iteration a key with the number `n`

and `nil`

value is added to the `ele`

hash. At the end `ele`

is returned as:

`{1=>nil, 2=>nil, 4=>nil, 6=>nil, 3=>nil} `

We only want the keys, so `.keys`

ends the job.

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Solution:6

Some ideas: you'd have to figure out the correct library data structures:

1Sort the array O(nlogn), then run through the array

2Create a set, search for the current array element in the set and if not found, insert and proceed for all the elements -- O(nlogn) again.

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Solution:7

I was thinking of counting how many times a unique element appears in array. It may be really inefficient just like the original suggestion but it was fun looking at the problem. I didn't do any benchmarks on larger arrays so this is just an excercise.

`a = [1,1,1,2,4,6,3,3] dupes = [] a.uniq.each do |u| c = a.find_all {|e| e == u}.size dupes << [u, c] unless c == 1 end puts dupes.inspect # dupes = [[1, 3], [3, 2]] # 1 appears 3 times # 3 appears twice # to extract just the elment a bit cleaner dupes = a.uniq.select do |u| a.find_all {|e| e == u}.size != 1 end puts dupes.inspect # returns [1,3] `

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Solution:8

This will work if the duplicated entries are always consecutive, as in your example; otherwise you would have to sort first. each_cons examines a rolling window of the specified size.

`require 'set' my_array = [1,1,1,2,4,6,3,3] dups = Set.new my_array.each_cons(2) {|a,b| dups.add(a) if (a == b)} p dups.to_a `

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