[Codility] Binary Gap

# 문제

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. The number 32 has binary representation 100000 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. Given N = 32 the function should return 0, because N has binary representation '100000' and thus no binary gaps.

Write an efficient algorithm for the following assumptions:

N is an integer within the range [1..2,147,483,647].

 

# 풀이 (JAVA 8) 

import java.util.*;

class Solution {
    public int solution(int N) {
	        String binary = Integer.toBinaryString(N);
	        int last = 0;
	        HashSet<Integer> set = new HashSet<Integer>();
            
	        String[] zeros = binary.split("1"); 
             	int finish = zeros.length;
	        if (binary.charAt(binary.length() - 1) != '1'){finish = finish - 1;} // 마지막이 0일 때 
	        
	        for (int i = 1; i < finish; i++) {set.add(zeros[i].length());}
	        if (set.size() == 0) {return 0;}
	        return Collections.max(set);
    }
}

 

# 한 줄 느낌 

변수로 받은 int 를 2진법으로 변형한 뒤, 1과 1 사이의 0 의 개수를 카운트해서 가장 큰 수를 리턴하는 문제이다.

ex) 10001001 -> 3, 2 -> 3 

ex) 1111110 -> 0 -> 0 

 

간단한 문제로 큰 어려움 없이 풀 수 있었다. 나는 먼저 2진법으로 변환 후 String 타입의 binary로 변경했다.

String binary = Integer.toBinaryString(N);

 

그 뒤 1을 기준으로 split()을 통해 0들을 zeros 배열에 담았다.

String[] zeros = binary.split("1");

 

그리고 각 배열의 길이를 set에 담아주고 가장 큰 수를 return 하는 메소드를 만들었다. 

for (int i = 1; i < finish; i++) {set.add(zeros[i].length());}
if (set.size() == 0) {return 0;}
return Collections.max(set);