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 canβt make it right and as simple as it was. So Iβll 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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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 canβt make it right and as simple as it was. So Iβll stick with this solution: