868 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

Approach 1:

Using a list to store index position of 1's

Convert the number to the binary format . Then iterate over the binary string. If the value is 1 then add the index position to the list. After traversing the entire string. Calculate the required max distance using the list created.

Time : O(logn)

Space: O(logn)

Approach 2:

By storing previous position where a 1 is found

While on iteration on the binary string , we will store the indexes , where a ‘1' is found and when current position also is found to have a 1 , we will be calculating the distance, by the formula (prev -current), if this current distance is greater than maximum distance found so far, then we will have to update the maximum distance variable by the new distance.

Time : O(logn)

Space: O(1)

Approach 3:

The same solution above can be implemented in a different way by using it operation , rather than creating a binary string

class Solution:
    def binaryGap(self, n: int) -> int:
        prev,i,max_=-1,0,0
        while(n>0):
            if(n&1==1):
                if prev>=0:
                    max_=max(max_,i-prev)
                prev=i
            n>>=1 
            i+=1    
        return(max_)

Time : O(logn)

Space: O(1)