# Counting Bits

## Description

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.

## Solutions

### My solution

```cpp
class Solution {
public:
    vector<int> countBits(int num) {
        vector<int> ans(num + 1);
        for(int i = 1; i <= num; ++i){
            if(i % 2)
                ans[i] = ans[i - 1] + 1;
            else{
                int count = ans[((i - 1) ^ i) >> 1];
                ans[i] = ans[i - 1] - count + 1;
            }
        }
        return ans;
    }
};
```

### Optimal solution

```cpp
class Solution {
public:
    vector<int> countBits(int num) {
        vector<int> ans(num + 1);
        for(int i = 1; i <= num; ++i){
            if(i % 2)
                ans[i] = ans[i >> 1] + 1; // or ans[i - 1] + 1;
            else
                ans[i] = ans[i >> 1];
        }
        return ans;
    }
};
```
