Given a non negative integer number num. For every numbers i in the range 0 ≤ i ≤ num calculate the number of 1's in their binary representation and return them as an array.Example:For num = 5 you should return [0,1,1,2,1,2].Follow up:It is very easy to come up with a solution with run time O(n*sizeof(integer)). But can you do it in linear time O(n) /possibly in a single pass?Space complexity should be O(n).Can you do it like a boss? Do it without using any builtin function like __builtin_popcount in c++ or in any other language.

Hint:

1. You should make use of what you have produced already.
2. Divide the numbers in ranges like [2-3], [4-7], [8-15] and so on. And try to generate new range from previous.
3. Or does the odd/even status of the number help you in calculating the number of 1s?

1 public class Solution {2     public int[] countBits(int num) {3         int[] res = new int[num+1];4         for (int i=1; i<=num; i++) {5             res[i] = res[i>>1] + (i&1);6         }7         return res;8     }9 }