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)
        .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) {
                    count = 0;
            if(counts.size() > 0) {
                return counts.get(counts.size() - 1);
            return 0;

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


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)
            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).

  • \$\begingroup\$ ...and complexity is \$O(n)\$ \$\endgroup\$ – Andriy Kryvtsun Mar 9 '16 at 20:14
  • \$\begingroup\$ For an arbitrary large int, if we pretend that instructions - including popcount - execute in constant time, there is a logarithmic solution. \$\endgroup\$ – vnp Mar 9 '16 at 21:51
  • 1
    \$\begingroup\$ @vnp Constant popcount sounds unrealistic because in some sense it has to scan the whole number (but note that I counted my shift and mask as constant, which is technically incorrect, but I am assuming there would be a way to select a bit in constant time). Regardless, how would this logarithmic algorithm work? \$\endgroup\$ – Reinstate Monica Mar 9 '16 at 22:01
  • 1
    \$\begingroup\$ Sorry, got carried away. Of course there is no logarithmic solution. What I had in mind is that popcount((x ^ (x - 1)) - 1 gives the length of a run of zeroes from least significant in a constant time. Then shift right by the run length, negate, get rid of run of (used to be) ones and negate again to reach next run of zeroes. Surely still linear. \$\endgroup\$ – vnp Mar 9 '16 at 22:27
  • 1
    \$\begingroup\$ Or change the way you use start and end in the loop. You can shift position instead of incrementing it, and use it directly as a bitmask. \$\endgroup\$ – One Lyner Feb 5 '19 at 9:02

Here is the different way of implementation


  1. Convert the integer value to Binary format.
  2. Iterate from first index to last index.
  3. check whether the value is 0, if yes, increment the current iteration value, if not calculate the max_zero value by comparing current iteration with existing max_zero value


public static int binaryGap(final int n) {
    String binRep = Integer.toBinaryString(n);
    int currentItr = 0;
    int maxZeroSeq = 0;
    for (int index = 0; index <binRep.length(); index ++) {
        if (binRep.charAt(index) == '0') {
        } else {
            maxZeroSeq = Math.max(maxZeroSeq, currentItr);
            currentItr = 0;
    if (currentItr!=0) {
        if (currentItr > maxZeroSeq) {   
            return currentItr;
    return Math.max(maxZeroSeq, currentItr);
  • \$\begingroup\$ Try with N=28. It will return 2 which is wrong. Algorithm above returns 0 what is correct because gap should be surrounded with 1 from both sides \$\endgroup\$ – avalon Feb 5 '20 at 13:32

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