## DEV Community

Abhishek Chaudhary

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# Binary Gap

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 = 8
Output: 0
Explanation: 8 in binary is "1000".
There are not any adjacent pairs of 1's in the binary representation of 8, so we return 0.

Example 3:

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

Constraints:

• `1 <= n <= 109`

SOLUTION:

``````class Solution:
def binaryGap(self, n: int) -> int:
curr = "{:b}".format(n)
n = len(curr)
mdist = 0
prev = -1
for i in range(n):
if curr[i] == "1":
if prev >= 0:
mdist = max(mdist, i - prev)
prev = i
return mdist
``````