# 338. 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: 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.

Credits:Special thanks to @ syedee for adding this problem and creating all test cases.

## Solution <a href="#solution" id="solution"></a>

```java
public class Solution {
      public int[] countBits(int num) {
        if (num < 0) return new int[0];

        int[] result = new int[num + 1];
        result[0] = 0;

        int offset = 1;
        for (int i = 1; i <= num; i++) {
            if (offset * 2 == i) {
                offset *= 2;
            }

            result[i] = result[i-offset] + 1;
        }

        return result;
    }
}
```


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