# 01. Counting Bits

The problem can be found at the following link: [Question Link](https://leetcode.com/problems/counting-bits/?envType=daily-question\&envId=2023-09-01)

## My Approach

There are two methods to solve this problem

### 1. We can count number of bits by iterating every number till n and count each set bits in every number

## Time and Auxiliary Space Complexity

* **Time Complexity**: `O(nlogn)`
* **Auxiliary Space Complexity**: `O(1)`

## Code (C++)

```cpp
class Solution {
public:
    vector<int> countBits(int n) {
        vector<int> ans;
        for(int i=0;i<=n;i++){
            int x = i;
            int one =0;
            while(x){
                if(x&1) one++;
                x /=2;
            }
            ans.push_back(one);
        }
        return ans;
    }
};
```

### 2. Optimized approch

Let if we have X and Y in Such a way that, `X/2 = Y` then Number of set bits in X - Number of set bit in Y <= 1

```plaintext
eg let X = 7and Y = 3
then 7 / 2 = 3;

7 -> 1 1 1 number of set bit are 3
3 -> 0 1 1 number of set bit are 2

there difference is 3 - 2 <= 1

another eg
X = 12 and y = 6
then 12 / 2 = 6;

12 -> 1100 number of set bit are 2
6 -> 0110 number of set bit are 2

there difference is 2 - 2 <= 1
```

There can be 2 cases whether X is Odd or Even

1. if X is ODD

then the (LSB) Least Significant Bit will always be set and dividing by 2 means right shifting the number by 1 bit. so if last bit is set and right shift it by 1 bit than the last set bit will be lost. therfore the new number Y has 1 set bit less than in compare to X

2. if X is Even

then LSB will be equal to 0, therefore even we do right shift by1 bit then only this 0 bit will be lost and no set bit got lost so When we have X has Even,

no of set bit in X = no of set bit in Y and When X is ODD

no of set bit in X = 1 + no of set bit in Y

## Time and Auxiliary Space Complexity

* **Time Complexity**: `O(n)`
* **Auxiliary Space Complexity**: `O(1)`

## Code (C++)

```cpp
class Solution {
public:
    vector<int> countBits(int n) {
        vector<int> ans(n+1);
        ans[0] = 0;

        for(int i=1;i<=n;i++){
            ans[i] = ans[i/2] + i%2;
        }
        return ans;
    }
};
```

## Contribution and Support

For discussions, questions, or doubts related to this solution, please visit our [discussion section](https://leetcode.com/discuss/general-discussion). We welcome your input and aim to foster a collaborative learning environment.

If you find this solution helpful, consider supporting us by giving a `⭐ star` to the [rishabhv12/Daily-Leetcode-Solution](https://github.com/rishabhv12/Daily-Leetcode-Solution) repository.


---

# Agent Instructions: Querying This Documentation

If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question.

Perform an HTTP GET request on the current page URL with the `ask` query parameter:

```
GET https://code-render.gitbook.io/leetcode-editorials/09-september/01.counting-bits.md?ask=<question>
```

The question should be specific, self-contained, and written in natural language.
The response will contain a direct answer to the question and relevant excerpts and sources from the documentation.

Use this mechanism when the answer is not explicitly present in the current page, you need clarification or additional context, or you want to retrieve related documentation sections.