# Time:  O(logn) = O(1) due to n is 32-bit number
# Space: O(1)

# Given a positive integer N, find and return the longest distance
# between two consecutive 1's in the binary representation of N.
#
# If there aren't two consecutive 1's, return 0.
#
# Example 1:
#
# Input: 22
# Output: 2
# Explanation: 
# 22 in binary is 0b10110.
# In the binary representation of 22, there are three ones,
# and two consecutive pairs of 1's.
# The first consecutive pair of 1's have distance 2.
# The second consecutive pair of 1's have distance 1.
# The answer is the largest of these two distances, which is 2.
# Example 2:
#
# Input: 5
# Output: 2
# Explanation: 
# 5 in binary is 0b101.
# Example 3:
#
# Input: 6
# Output: 1
# Explanation: 
# 6 in binary is 0b110.
# Example 4:
#
# Input: 8
# Output: 0
# Explanation: 
# 8 in binary is 0b1000.
# There aren't any consecutive pairs of 1's
# in the binary representation of 8, so we return 0.
#
# Note:
# - 1 <= N <= 10^9

class Solution(object):
    def binaryGap(self, N):
        """
        :type N: int
        :rtype: int
        """
        result = 0
        last = None
        for i in xrange(32):
            if (N >> i) & 1:
                if last is not None:
                    result = max(result, i-last)
                last = i
        return result

# Time:  O(logn) = O(1) due to n is a 32-bit number
# Space: O(1)

# Given a positive integer N, find and return the longest distance
# between two consecutive 1's in the binary representation of N.
#
# If there aren't two consecutive 1's, return 0.
#
# Example 1:
#
# Input: 22
# Output: 2
# Explanation: 
# 22 in binary is 0b10110.
# In the binary representation of 22, there are three ones,
# and two consecutive pairs of 1's.
# The first consecutive pair of 1's have distance 2.
# The second consecutive pair of 1's have distance 1.
# The answer is the largest of these two distances, which is 2.
# Example 2:
#
# Input: 5
# Output: 2
# Explanation: 
# 5 in binary is 0b101.
# Example 3:
#
# Input: 6
# Output: 1
# Explanation: 
# 6 in binary is 0b110.
# Example 4:
#
# Input: 8
# Output: 0
# Explanation: 
# 8 in binary is 0b1000.
# There aren't any consecutive pairs of 1's
# in the binary representation of 8, so we return 0.
#
# Note:
# - 1 <= N <= 10^9

class Solution(object):
    def binaryGap(self, N):
        """
        :type N: int
        :rtype: int
        """
        result = 0
        last = None
        for i in xrange(32):
            if (N >> i) & 1:
                if last is not None:
                    result = max(result, i-last)
                last = i
        return result

Update binary-gap.py