Ruby - indexOfOddManOut.txt

determine which is the ‘odd man out’ between even/odd string of numbers.
return the index+1 of the odd man out.


- - - - - my solution - - - - -
def iq_test(numbers)
    evenCount = 0
    oddCount = 0
    
    arr = numbers.split(' ')
    numArr = arr.map { |element| element.to_i }
    
    numArr.map { |number| number.even? ? evenCount+=1 : oddCount+=1 }

    if evenCount == 1
        numArr.map { |number| if number.even?
                return numArr.index(number)+1
            end}
    else
        numArr.map { |number| if number.odd?
            return numArr.index(number)+1
        end
        }
    end
    
end

iq_test("2 4 7 8 10")


- - - alt - - -

def iq_test(numbers)
  evenCount = 0
  even = nil
  oddCount = 0
  odd = nil
  numbers.split(' ').each_with_index { |v, i|		<- - - takes ‘value’ and ‘index’ in one fell swoop
    if v.to_i.odd?
      oddCount += 1
      odd = i+1
    else
      evenCount += 1
      even = i+1
    end
  }
  evenCount == 1 ? even : odd
end


- - - nicer still - - -

def iq_test(numbers)
  nn = numbers.split.map{|n| n.to_i%2}.join
  par = nn[0]
  par != nn[1] ? par != nn[2] ? 1 : 2 : nn.index("10"[par.to_i])+1
end



this is much sexier. it is destructive, but if you only need the index, then it doesn’t matter... but it ‘%2’s all the values.
then compares [1,1,1,1,1,0,1,1,1] for non equality. looks for the index.