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# Counting Bits

## [Counting Bits](https://leetcode.com/problems/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.

## Solutions

### 🧠 Cpp

```cpp
class Solution
{
public:
    vector<int> countBits(int num)
    {
        vector<int> res(num+1);

        for(int i = 0; i <= num; ++i)
            res[i] = __builtin_popcount(i);

        return res;
    }
};
```