Skip to main content
Code Review

Return to Question

Question Protected by Jamal
added 65 characters in body; edited tags; edited title; edited tags
Source Link
200_success
200_success
  • 144.2k
  • 22
  • 188
  • 473

Advice to increase performance of printing Printing longest sequence of zeroes

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).

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).

Advice to increase performance of printing longest sequence of zeroes

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).

Printing longest sequence of zeroes

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).

Source Link
would_like_to_be_anon
would_like_to_be_anon
  • 473
  • 5
  • 10

Advice to increase performance of 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<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);
    }
}

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