Counting Bits

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 1:

Input: 2
Output: [0,1,1]

Example 2:

Input: 5
Output: [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.

class Solution {
    public int[] countBits(int num) {
        int result[] = new int[num + 1];
        for (int i = 0; i <= num; i++)
            // Every number has same bits as number>>1 (number/2), combined with a 0 or 1
            result[i] = result[i >> 1] + (i & 1);
        return result;
    }
}