# 0338. Counting Bits

<https://leetcode.com/problems/counting-bits>

## Description

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`.

**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 <= 105`

**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++)?

## ac

```java
class Solution {
    public int[] countBits(int num) {
        // edge cases
        if (num < 0) throw new IllegalArgumentException("invalid input");

        // dp
        int[] dp = new int[num+1];
        dp[0] = 0;
        int base = 1;

        for (int i = 1; i <= num; i++) {
            if (i / base == 2) {
                dp[i] = 1;
                base = base << 1;
            } else {
                int idx = i % base;
                dp[i] = dp[idx] + 1;
            }
        }

        return dp;
    }
}
/*
power
f[i]
state: f[i] bit counts at i point
func: 1) if i/pow(2, power) == 2, f[i] = 1, power++;
        2) idx = i % pow(2, power), f[i] = f[idx] + 1;
init: f[0] = 0, power = 0
ans: f
*/
```

## ac2: bit manipulation

```java
class Solution {
    public int[] countBits(int num) {
        // edge case
        if (num < 0) return new int[0];

        int[] dp = new int[num+1];
        for (int i = 1; i < dp.length; i++) {
            dp[i] = dp[i>>1] + (i&1);
        }

        return dp;
    }
}
```

## ac3: bit manipulation

```java
class Solution {
    public int[] countBits(int num) {
        // edge case
        if (num < 0) return new int[0];

        int[] dp = new int[num+1];
        for (int i = 1; i < dp.length; i++) {
            dp[i] = dp[i&(i-1)] + 1;
        }

        return dp;
    }
}
```


---

# 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://jaywin.gitbook.io/leetcode/solutions/0338-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.