re: Daily Challenge #35 - Find the Outlier VIEW POST

FULL DISCUSSION
 

Bad solution! See comments below!

Ruby solution, I tried to do something different.

def findOutlier numbers
  sum = numbers.sum
  if sum.odd?
    numbers.find(&:odd?)
  else
    numbers.find(&:even?)
  end
end

puts findOutlier([2, 4, 0, 100, 4, 11, 2602, 36])
puts findOutlier([160, 3, 1719, 19, 11, 13, -21])
puts findOutlier([4, 8, 15, 16, 24, 42])
puts findOutlier([16, 6, 40, 66, 68, 28])
puts findOutlier([5, 9, 4, 155, 7])

When the list has no exception, the first element is returned; except when numbers looks like [3,3] (bug that you can fix by adding .uniq when computing the sum.)

 

I believe this would fail if the outlier is even and the rest are an odd number of odd numbers. The examples all have an even number of odd numbers. Can you check that?

 

It seems that I can’t make it right and as simple as it was. So I’ll stick with this solution:

def findOutlier numbers
  if numbers.first(3).count(&:even?) > 1
    numbers.find(&:odd?)
  else
    numbers.find(&:even?)
  end
end

puts findOutlier([2, 4, 0, 100, 4, 11, 2602, 36])
puts findOutlier([160, 3, 1719, 19, 11, 13, -21, 33])
puts findOutlier([160, 3, 1719, 19, 11, 13, -21])
puts findOutlier([4, 8, 15, 16, 24, 42])
puts findOutlier([16, 6, 40, 66, 68, 28])
puts findOutlier([3,3])
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