# Solving Leetcode Interviews in Seconds with AI: Counting Bits


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "338" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> 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++)?  

	# Explanation
	Here's a breakdown of the approach, complexity, and code for efficiently counting bits:

*   **Dynamic Programming Principle:** The number of set bits in a number `i` can be derived from the number of set bits in `i // 2`. If `i` is even, `i` has the same number of set bits as `i // 2`. If `i` is odd, `i` has one more set bit than `i // 2`. This forms the basis of our DP solution.
*   **Iterative Calculation:** We iterate from 1 to `n`, building up the `ans` array based on the previously computed values.
*   **Bit Manipulation:** We use the right shift operator `>>` to efficiently perform integer division by 2 and the bitwise AND operator `&` to check if a number is odd.

*   **Time Complexity:** O(n), **Space Complexity:** O(n)

	
	# Code
	```python
	def countBits(n: int) -> list[int]:
    """
    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.
    """
    ans = [0] * (n + 1)
    for i in range(1, n + 1):
        ans[i] = ans[i >> 1] + (i & 1)
    return ans
	```