Wap to Count Number of Set Bits in Hexadecimal Number

class Solution { public int hex_bitcount(String S); }

that given a string S containing big-endian hexadecimal representation of a non-negative integer N returns the number of bits set to 1 in binary representation of N.

Assume that the length of the string does not exceed 100,000. Assume that the string contains only characters '0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'A', 'B', 'C', 'D', 'E', 'F'.

For example the function should return 5 when given S = "2F". The string "2F" represents the number 47. The binary representation of 47 is 101111 and it contains 5 bits set to 1.

Algorithm Used: Brian Kernighan’s Algorithm:
Subtraction of 1 from a number toggles all the bits (from right to left) till the rightmost set bit(including the righmost set bit). So if we subtract a number by 1 and do bitwise & with itself (n & (n-1)), we unset the righmost set bit. If we do n & (n-1) in a loop and count the no of times loop executes we get the set bit count.
Beauty of the this solution is number of times it loops is equal to the number of set bits in a given integer.

1 Initialize count: = 0
2 If integer n is not zero
(a) Do bitwise & with (n-1) and assign the value back to n
n: = n&(n-1)
(b) Increment count by 1
(c) go to step 2
3 Else return count


class Solution
{
int hex_bitcount ( String S )
{
int n=0,count=0;
if(S.length()<=100000)
{
n=Integer.parseInt(S,16);

while (n!=0)
{
n &= (n-1) ;
count++;
}
return count;
}
return -1;
}
public static void main(String a[])
{
Solution obj= new Solution();
System.out.println(obj.hex_bitcount("2F"));

}
}

Time Complexity O(logn)
Space Complexity O(1)
Run Here https://ideone.com/9vZ4V