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# 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.

```java
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;
    }
}
```