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

This seemed liked a fun problem so I decided to try my hand at it.

I am making the assumption that 111000001 would find the first 3 ones and be successful. Essentially the number of zeroes following the 1 is the important thing, since 0111000 is the same as 111000 according to your definition. Once you find two cases of 1, the next 1 found completes the trilogy.

Here it is in Python:

def find_three(bstring):
    print bstring
    dict = {}
    lastone = -1
    zerocount = 0
    for i in range(len(bstring)):
        if bstring[i] == '1':
            print i, ': 1'
            if lastone != -1:
                if(zerocount in dict):
                    dict[zerocount].append(lastone)
                    if len(dict[zerocount]) == 2:
                        dict[zerocount].append(i)
                        return True, dict
                else:
                    dict[zerocount] = [lastone]
            lastone = i
            zerocount = 0
        else:
            zerocount = zerocount + 1
    #this is really just book keeping, as we have failed at this point
    if lastone != -1:
        if(zerocount in dict):
            dict[zerocount].append(lastone)
        else:
            dict[zerocount] = [lastone]
    return False, dict

This is a first try, so I'm sure this could be written in a cleaner manner. Please list the cases where this method fails down below.