Printing longest sequence of zeroes

I am doing a coding exercise in codility and I came across this question:


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:


class Solution { public int solution(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.


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).

My code looks like this:

import java.util.*;

class Solution {
    public static int solution(int N) {
        return Optional.ofNullable(N)
        .map(Integer::toBinaryString)
        .filter(n -> n.length() > 1)
        .map(t -> {
            List counts = new ArrayList<>();
            int count = 0;
            for(int i = 0; i  0) {
                    counts.add(count);
                    count = 0;
                }
            }
            if(counts.size() > 0) {
                Collections.sort(counts);
                return counts.get(counts.size() - 1);
            }
            return 0;
        })
        .orElse(0);
    }
}

What else can I do to improve the performance of the aforementioned code? How do I determine the big-O complexity of this progra

Some comments:

• Converting to a binary string and then operating on the string is much slower and more laborious than operating on the bytes directly.

• There is no need to collect all counts and then take the largest as we can just calculate the largest as we go.

• Give your class and method appropriate names.

• Check the precondition that the input is positive in the beginning of the method.

Here is a streamlined but untested solution:

class BinaryGapCalculator {
    static int binaryGap(final int n) {
        if (n <= 0)
            throw new IllegalArgumentException("n must be positive; was " + n);
        final int start = Integer.lowestOneBit(n), end = Integer.highestOneBit(n);
        int curRun = 0, longestRun = 0;
        for (int bitmask = start; bitmask != end; bitmask <<= 1) {
            final int bit = n & bitmask;
            if (bit == 0)
                ++curRun;
            else {
                longestRun = Math.max(longestRun, curRun);
                curRun = 0;
            }
        }
        return Math.max(longestRun, curRun);
    }
}

Regarding big-O, all solutions are technically O(1) because there are finitely many inputs. But if we were to pretend that an int could store arbitrarily many bits, the optimal solution would be θ(b) where b ~ log(n) is the number of bits of n. This is because the algorithm I wrote is θ(b) and the optimum must be at least as efficient as my algorithm; also there is no tighter bound than θ(b) because asymptotically, at least half of the input has to be seen. So my algorithm is optimal (up to a constant).

StackExchange Code Review Q#122410, answer score: 6