A binary gap within a positive integer N is any maximal sequence of consecutive zeros that is surrounded by ones at both ends in the binary representation of N.

For example, number 9 has binary representation 1001 and contains a binary gap of length 2. The number 529 has binary representation 1000010001 and contains two binary gaps: one of length 4 and one of length 3. The number 20 has binary representation 10100 and contains one binary gap of length 1. The number 15 has binary representation 1111 and has no binary gaps.

Write a function that, given a positive integer N, returns the length of its longest binary gap. The function should return 0 if N doesn’t contain a binary gap.

For example, given N = 1041 the function should return 5, because N has binary representation 10000010001 and so its longest binary gap is of length 5.

Assume that:

• N is an integer within the range [1..2,147,483,647].

Complexity:

• expected worst-case time complexity is O(log(N));
• expected worst-case space complexity is O(1).

So lets see how we can solve this. Actually it’s quite an easy thing to do. But what would be the fastest way to do it.

There are several options to solve this case, one way would be to convert it to a binary string and explode the string to a group, and you can just check which group has the most zeroes. Another way and fastest way would be to do bit checks on the number it self, this is what we gonna try.

``````func Solution(N int) int {

var max int = 0
var current int = 0
var startFound bool = false

for i := 31; i >= 0; i-- {
bit := ((N & (1 << uint(i))) > 0)
if startFound {
// we have found the start
if !bit {
current++
} else {
if max < current {
max = current
}
current = 0
}
} else {
startFound = bit
}
}

return max

}
``````

We can probably optimze the following code a bit more, but it should be sufficient for now, and I leave it up to you guys to see and decide how to do it.