Hello and welcome back to one day one answer series. In this chapter, we will need to return the position of either an odd or even number from a list of numbers following the below condition.

If there is only one odd number in a list of odd-even numbers then return the position of that odd number or else return the position of the even number. The solution for this question is very straight forward, after seeing the below solution do leave your own solution below this post.

def iq_test(numbers): num_dict = dict() num_dict['even'] = 0 num_dict['odd'] = 0 number_list = numbers.split(' ') index = 1 even_index = 0 odd_index = 0 for num in number_list: if(int(num) % 2 == 0): num_dict['even'] += 1 even_index = index else: num_dict['odd'] += 1 odd_index = index index += 1 if(num_dict['even'] > num_dict['odd']): return odd_index elif(num_dict['odd'] > num_dict['even']): return even_index

As you can see we do not need to worry about the position which will be returned by the program because there is only a single odd number or a single even number so there will be only one position that program needs to worry about.

## Leave a Reply