class Solution(object):

def binaryGap(self, N):

"""

:type N: int

:rtype: int

"""

# Brute force method, naive iteration to find consecutive ones

# * One loop for 1st one

# * Second loop for 2nd one

# * We already traverse digits in between, so move forward the index to the second one for the next loop....

binary = bin(N)

#print binary

# Control vars

n = len(binary)

maxi = 0

# iteration until finding the 1st one

i = 0

while i < n:

if binary[i] == "1":

# iteration for 2nd one

j = i+1

while j < n:

# Break as soon as consecutive one is found and adjust index to the consecutive one

if binary[j] == "1":

maxi = max(maxi, j-i)

i = j-1 # Once you break after finding consecutive, while loop iterates i += 1, to handle that negate i before breaking

break

j += 1

i += 1

return maxi

"""

# Logic:

# * Convert number to a usable form of binary

# * Only the max is required, use 2 pointer method to return on first match --> does not work when consecutive ones need to be worked on.

binary = str(bin(N))[2:]

left = 0

right = len(binary)-1

while left <= right:

if binary[left] == "1":

break

left += 1

while right >= left:

if binary[right] == "1":

break

right -= 1

print right, left

return (right - left)

"""