# LeetCode 338. Counting Bits

## The Problem

Link to original problem on LeetCode.

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's**in the binary representation of i

## Examples

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++)?

## My Solution

The problem is very simple if you've already got a solution to 191. Number of 1 Bits. All you need to do is loop over the numbers from `0`

through `n`

, computer the Hamming Weight, and push that to an array to be returned.

`// To understand what this function is doing, see the`

// explanation linked in the preceding paragraph.

const hammingWeight = (n) => {

let count = 0;

while (n) {

n = n & (n - 1);

count++;

}

return count;

};

const countBits = (n) => {

const result = [];

for (let i = 0; i <= n; i++) {

result.push(hammingWeight(i));

}

return result;

};

Ok, so that's easy. Copy and paste! But can we do better? Yes, we can!

With the above example, we have to count the `1`

s from scratch for each value of `i`

. But we can actually use previous computations in our loop over `i`

to save us the trouble!

Think about the even numbers in binary. `10`

, `100`

, `110`

, `1000`

, etc. Notice that the final digit is always zero. If you shift the bits right by one, the count of `1`

s in the number does not change because you've only ever lopped off a `0`

.

The other thing to notice is that shifting each even number to the right by 1 bit is equivalent to dividing by `2`

. E.g., `1000 >> 1 === 100`

is the same as `8 / 2 === 4`

. So if we want to know how many `1`

s are in `8`

, we can go back to our previous count of how many `1`

s are in `4`

. This means that we can look back in our own results array for the value at the index of half the current number and find the count of `1`

s not including the final digit. To then account for the final digit, we use AND comparison to see if it is a `1`

or a `0`

. Putting it all together:

`const countBits = (n) => {`

const result = [0];

for (let i = 0; i <= n; i++) {

result[i] = result[i >> 1] + (i & 1);

}

return result;

};