Given a positive integer n, find and return the longest distance between any two adjacent 1's in the binary representation of n. If there are no two adjacent 1's, return 0.
Two 1's are adjacent if there are only 0's separating them (possibly no 0's). The distance between two 1's is the absolute difference between their bit positions. For example, the two 1's in "1001" have a distance of 3.
 
Example 1:
Input: n = 22
Output: 2
Explanation: 22 in binary is "10110".
The first adjacent pair of 1's is "10110" with a distance of 2.
The second adjacent pair of 1's is "10110" with a distance of 1.
The answer is the largest of these two distances, which is 2.
Note that "10110" is not a valid pair since there is a 1 separating the two 1's underlined.

Example 2:
Input: n = 5
Output: 2
Explanation: 5 in binary is "101".

Example 3:
Input: n = 6
Output: 1
Explanation: 6 in binary is "110".

Example 4:
Input: n = 8
Output: 0
Explanation: 8 in binary is "1000".
There aren't any adjacent pairs of 1's in the binary representation of 8, so we return 0.

Example 5:
Input: n = 1
Output: 0

 
Constraints:

1 <= n <= 109

Source: leetcode | Difficulty: EASY
Tags: Bit manipulation

Starter code: class Solution:
    def binaryGap(self, n: int) -> int:
        

Solutions:
class Solution:

	def binaryGap(self, n: int) -> int:
		maxDist = 0
		currDist = 0
		while n:
			if n & 1 and currDist != 0:
				maxDist = max(maxDist, currDist)
				currDist = 1
			elif n & 1:
				currDist = 1
			elif not n & 1 and currDist != 0:
				currDist += 1
			n >>= 1
		return maxDist


class Solution:

	def binaryGap(self, n: int) -> int:
		bins = ''
		maxv = 0
		i = 0
		pre = None
		while n > 0:
			if n % 2 == 1:
				if pre is not None:
					maxv = max(maxv, i - pre)
				pre = i
			n //= 2
			i += 1
		return maxv


class Solution:

	def binaryGap(self, n: int) -> int:
		binary_string = bin(n)[2:]
		(left, right, current_distance, global_max) = (0, 1, 1, 0)
		print(binary_string)
		while right <= len(binary_string) - 1:
			print(left, right, current_distance, global_max)
			if binary_string[right] == '0':
				right += 1
				current_distance += 1
			else:
				global_max = max(global_max, current_distance)
				left = right
				right += 1
				current_distance = 1
		return global_max


class Solution:

	def binaryGap(self, N: int) -> int:
		bit_list = bin(N)[2:]
		stack = []
		dis = 0
		for b in bit_list:
			if not stack and b == '1':
				stack.append(b)
			elif stack and b == '0':
				stack.append(b)
			elif stack and b == '1':
				dis = max(dis, len(stack))
				stack = ['1']
		return dis


class Solution:

	def binaryGap(self, n: int) -> int:
		if n == 0:
			return 0
		while n & 1 == 0:
			n = n >> 1
		global_max = local_max = 0
		while n > 0:
			if n & 1 == 1:
				global_max = max(global_max, local_max)
				local_max = 1
			else:
				local_max += 1
			n = n >> 1
		return global_max