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Given an integer `n`

, return *an array *`ans`

* of length *`n + 1`

* such that for each *`i`

* *(`0 <= i <= n`

)*, *`ans[i]`

* is the number of *

`1`

`i`

.**Example 1:**

**Input:** n = 2
**Output:** [0,1,1]
**Explanation:**
0 --> 0
1 --> 1
2 --> 10

**Example 2:**

**Input:** n = 5
**Output:** [0,1,1,2,1,2]
**Explanation:**
0 --> 0
1 --> 1
2 --> 10
3 --> 11
4 --> 100
5 --> 101

**Constraints:**

`0 <= n <= 10`

^{5}

**Follow up:**

- It is very easy to come up with a solution with a runtime of
`O(n log n)`

. Can you do it in linear time`O(n)`

and possibly in a single pass? - Can you do it without using any built-in function (i.e., like
`__builtin_popcount`

in C++)?

```
class Solution {
public:
vector<int> countBits(int n) {
vector<int> ans;
// iterating fromt 0 to n
for(int i = 0; i<=n; i++)
{
// sum is initialised as 0
int sum = 0;
int num = i;
// while num not equals zero
while(num != 0)
{
// we have to count 1's in binary representation of i, therefore % 2
sum += num%2;
num = num/2;
}
// add sum to ans vector
ans.push_back(sum);
}
// return
return ans;
}
};
```

```
public int[] countBits(int num) {
int[] f = new int[num + 1];
for (int i=1; i<=num; i++) f[i] = f[i >> 1] + (i & 1);
return f;
}
```

```
def countBits(self, num: int) -> List[int]:
counter = [0]
for i in range(1, num+1):
counter.append(counter[i >> 1] + i % 2)
return counter
```

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