# Question

iven 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.

# Solution

Calculate this from 0 to num. For each number, we can find the number of 1s by removing its leading 1 bit.

```
class Solution {
public int[] countBits(int num) {
if(num == 0) {
return new int[]{0};
} else if (num == 1) {
return new int[]{0, 1};
}
int[] res = new int[num + 1];
res[0] = 0;
res[1] = 1;
int leadingBit = 2;
int i = 0;
while(leadingBit + i <= num) {
res[leadingBit + i] = res[i] + 1;
i++;
if(leadingBit == i) {
leadingBit *= 2;
i = 0;
}
}
return res;
}
}
```