## Question

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:**

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

## Solution

**Result:** Accepted **Time:** 40 ms

Here should be some explanations.

```
/**
* Return an array of size *returnSize.
* Note: The returned array must be malloced, assume caller calls free().
*/
int* countBits(int num, int* returnSize) {
int * ary = malloc((num+2)*sizeof(int));
*returnSize = num + 1;
int ptr = 1;
int cnt = 2;
ary[0] = 0;
ary[1] = 1;
int limit = num / 2;
while(ptr <= limit)
{
ary[(ptr<<1)] = ary[ptr];
ary[(ptr<<1)|1] = ary[ptr]+1;
ptr++;
}
return ary;
}
```

**Complexity Analytics**

- Time Complexity:
- Space Complexity: