Question

How to get the number which repeated the fewest times ?

For example:

• from [1,2,3,4,5,6,6,2,3,4,6] return [1] because “1” is only repeated onces while others are repeated 2 or more times.
• from [1,1,1,2,3,3,4,4,5,6,6,2,3,4] return [2,6] because both “2” and “6” are only repeated twice instead of three or more times for other numbers.

Answer 1

This should work:

a.group_by{|e| a.count(e)}.min[1].uniq

---

ruby-1.9.2-p136 :040 > a =  [1,1,1,2,3,3,4,4,6,6,2,3,4]
ruby-1.9.2-p136 :041 > a.group_by{|e| a.count(e)}.min[1].uniq
 => [2, 6] 

ruby-1.9.2-p136 :044 > a =   [1,2,3,4,6,6,2,3,4,6]
ruby-1.9.2-p136 :045 > a.group_by{|e| a.count(e)}.min[1].uniq
 => [1]

Update: O(n) time

def least_frequent(a)
  counts = Hash.new(0)
  a.each{|e| counts[e] += 1}
  least =[nil, []]
  counts.each do |k,v|
    if least[0].nil?
      least[0] = v
      least[1] = k
    elsif v < least[0]
      least[0] = v
      least[1] = [k]
    elsif v == least[0]
      least[1] << k
    end
  end
  least[1]
end

Here are benchmarks(running this test 10,000 times) between the first and second method:

             user     system      total        real
first   10.950000   0.020000  10.970000 ( 10.973345)
better   0.510000   0.000000   0.510000 (  0.511417)

with an array set to:

a =  [1,1,1,2,3,3,4,4,6,6,2,3,4] * 10