Find longest sequence of zeros in binary representation of an integer. (BinaryGap)

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].

PLEASE SOLVE ON YOUR OWN FIRST BEFORE LOOKING INTO THE SOLUTION

Solution 1 :

1) first try to find if there is 1 present in binary representation
2) if present then set binaryGapFoundFlag = true to ensure that there is a chance of a binary gap
3) if binaryGapFoundFlag = true then count the number of zeroes until we get 1 again
4) if we get 1 again then previous binary gap has ended and new binarygap has started from that point. check if current count of zeroes = maxlength (we are using thjs to find the binary gap with max count). Also currlength of binary gap to zero because previous binary gap ended and new binary gap has started.

Solution 2 :

I think we can also do this by regex. feel free to share if possible by regex in comment section below

CODE

class Solution {
    public static void main(String[] args) {
      System.out.println(new Solution().solution(677));
    }
   
    public int solution(int N) {
        if(N<0 || N>2147483647){
            return 0;
        }
        String s = Integer.toBinaryString(N);
        int maxLength=0;
        int currLength=0;
        boolean binaryGapFoundFlag = false;
       
        for(int i=0;i<s.length();i++){
           if('1' == s.charAt(i)){
               binaryGapFoundFlag = true;
           }
           if(binaryGapFoundFlag && '0' == s.charAt(i)){
               currLength++;
           }
           if(currLength > 0 && '1' == s.charAt(i)){
               maxLength = currLength>maxLength ? currLength : maxLength;
               binaryGapFoundFlag=true;
               currLength=0;
           }
        }
        return maxLength;
    }
}

OUTPUT : 

2