Abinary 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. e. g.: the number 20 has binary representation 10100 and contains one binary gap of length 1. the number 15 has binary representation and has no binary gaps. write a function: class solution { public int binary_gap(int n); } 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.

This is the best way:

class Solution {    public static int solution(int N) {        return Optional.ofNullable(N)        .map(Integer::toBinaryString)        .filter(n -> n.length() > 1)        .map(t -> {            List<Integer> counts = new ArrayList<>();            int count = 0;            for(int i = 0; i < t.length(); i++)            {                if(t.charAt(i) == '0') {                    count += 1;                } else if(count > 0) {                    counts.add(count);                    count = 0;                }            }            if(counts.size() > 0) {                Collections.sort(counts);                return counts.get(counts.size() - 1);            }            return 0;        })        .orElse(0);    }}

It can sometimes be called a process chart.