interview-practice.rb

=begin
Given an array, find the int that appears an odd number of times.

There will always be only one integer that appears an odd number of times.
967 ms
=end
def find_it(seq)
  #your code here
  counts = Hash.new(0)
  
  seq.each { |num| counts[num] += 1 }
  
  counts.each_key { |key| return key if counts[key] % 2 != 0 }
end

#814ms
def find_it(seq)
  seq.find { |n| seq.count(n).odd? }
end