Introduction

In this blog post, we will explore how to solve the LeetCode problem "2220" 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

A bit flip of a number x is choosing a bit in the binary representation of x and flipping it from either 0 to 1 or 1 to 0. For example, for x = 7, the binary representation is 111 and we may choose any bit (including any leading zeros not shown) and flip it. We can flip the first bit from the right to get 110, flip the second bit from the right to get 101, flip the fifth bit from the right (a leading zero) to get 10111, etc. Given two integers start and goal, return the minimum number of bit flips to convert start to goal. Example 1: Input: start = 10, goal = 7 Output: 3 Explanation: The binary representation of 10 and 7 are 1010 and 0111 respectively. We can convert 10 to 7 in 3 steps: - Flip the first bit from the right: 1010 -> 1011. - Flip the third bit from the right: 1011 -> 1111. - Flip the fourth bit from the right: 1111 -> 0111. It can be shown we cannot convert 10 to 7 in less than 3 steps. Hence, we return 3. Example 2: Input: start = 3, goal = 4 Output: 3 Explanation: The binary representation of 3 and 4 are 011 and 100 respectively. We can convert 3 to 4 in 3 steps: - Flip the first bit from the right: 011 -> 010. - Flip the second bit from the right: 010 -> 000. - Flip the third bit from the right: 000 -> 100. It can be shown we cannot convert 3 to 4 in less than 3 steps. Hence, we return 3. Constraints: 0 <= start, goal <= 109 Note: This question is the same as 461: Hamming Distance.

Explanation

Here's the solution to determine the minimum bit flips to convert start to goal:

• XOR Operation: The core idea is to use the XOR (exclusive OR) operation between start and goal. The XOR operation results in a number where each bit is set to 1 if the corresponding bits in start and goal are different, and 0 if they are the same.

• Counting Set Bits: The number of set bits (bits equal to 1) in the XOR result directly corresponds to the minimum number of bit flips needed. Each set bit indicates a position where start and goal differ, and thus a bit flip is required.

• Bit Manipulation: Use bitwise operations to efficiently count the set bits in the XOR result.

• Runtime Complexity: O(1), Storage Complexity: O(1)

Code

    def minBitFlips(start: int, goal: int) -> int:
    """
    Calculates the minimum number of bit flips to convert start to goal.
    """
    xor_result = start ^ goal
    flips = 0
    while xor_result > 0:
        flips += xor_result & 1
        xor_result >>= 1
    return flips