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, 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

    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