This is a bit manipulation problem and the solution, although a bit verbose is really efficient beating 100% of all submission for Kotlin.

```
class Solution {
fun binaryGap(N: Int): Int {
if(Integer.bitCount(N) <= 1) return 0
// remove all zero LSB
var temp = N
while(true){
if(temp and 1 != 0) break
temp = temp shr 1
}
var ctr = 0
var max = Integer.MIN_VALUE
while(temp > 0){
temp = temp shr 1
ctr++
if(temp and 1 == 1) {
max = Math.max(max, ctr)
ctr = 0
}
}
return max
}
}
```

```
Runtime: 128 ms, faster than 100.00% of Kotlin online submissions for Binary Gap.
Memory Usage: 31 MB, less than 100.00% of Kotlin online submissions for Binary Gap.
```