O(nlogn) Algorithm - Find three evenly spaced ones within binary string

Question

I had this question on an Algorithms test yesterday, and I can\'t figure out the answer. It is driving me absolutely crazy, because it was worth about 40 points. I figure

Answer 1

I came up with something like this:

def IsSymetric(number):
    number = number.strip('0')

    if len(number) < 3:
        return False
    if len(number) % 2 == 0:
        return IsSymetric(number[1:]) or IsSymetric(number[0:len(number)-2])
    else:
        if number[len(number)//2] == '1':
            return True
        return IsSymetric(number[:(len(number)//2)]) or IsSymetric(number[len(number)//2+1:])
    return False

This is inspired by andycjw.

1. Truncate the zeros.
2. If even then test two substring 0 - (len-2) (skip last character) and from 1 - (len-1) (skip the first char)
3. If not even than if the middle char is one than we have success. Else divide the string in the midle without the midle element and check both parts.

As to the complexity this might be O(nlogn) as in each recursion we are dividing by two.

Hope it helps.