Convert decimal to binary number, then count the consecutive 1's in the binary number.

Convert decimal to binary number, then count the consecutive 1's in the binary number.

	import java.util.*;
	public class HackerRank {
	public static void main(String[] args) {
	Scanner in = new Scanner(System.in);
	int input=0;
	int count=0,x=0,i=1,k=0;
	input=Integer.parseInt(in.nextLine().trim());
	String binary=Integer.toBinaryString(input);
	int [] arr=new int[binary.length()];
	do{
	if(String.valueOf(binary.substring(x,i)).contains("1")){
	count++;
	}else {
	i=x+1;
	arr[k]=count; //saving consecutive counts of 1 in an array.
	count=0; //resetting the value of counter
	k++;
	}
	x=i;
	i++;
	}while(i<=binary.length());
	if(k<1){
	arr[0]=count;
	System.out.println(arr[0]);
	}
	else{
	arr[k]=count; //adds the last consecutive 1's count after the zero.
	int max=arr[0]; //to print the maximum value of array.
	for(int f=0;f<arr.length;f++){
	if(arr[f]>max)
	max=arr[f];
	}
	System.out.println(max);
	in.close();
	}
	}
	}

view raw BinaryNumbers hosted with ❤ by GitHub

Input: 63

Output: 6
Explanation: 63 decimal format
binary format for 63 is 111111, so there are 6 consecutive 1's in the binary number.

another example: If input is 13, binary format is 1101, so the output is 2.

another example : input 439
output is 3.
Since binary format is 110110111. 3 consecutive 1's are found to be highest in it.

next example: input 65535
output is 16.
since binary format is 1111111111111111. 16 consecutive 1's are found to be highest in it.