Introduction

In this blog post, we will explore how to solve the LeetCode problem "3226" 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

You are given two positive integers n and k. You can choose any bit in the binary representation of n that is equal to 1 and change it to 0. Return the number of changes needed to make n equal to k. If it is impossible, return -1. Example 1: Input: n = 13, k = 4 Output: 2 Explanation: Initially, the binary representations of n and k are n = (1101)2 and k = (0100)2. We can change the first and fourth bits of n. The resulting integer is n = (0100)2 = k. Example 2: Input: n = 21, k = 21 Output: 0 Explanation: n and k are already equal, so no changes are needed. Example 3: Input: n = 14, k = 13 Output: -1 Explanation: It is not possible to make n equal to k. Constraints: 1 <= n, k <= 106

Explanation

Here's the breakdown of the solution:

• Bitwise Comparison: Iterate through the bits of n and k from the least significant bit to the most significant bit.
• Counting Differences: If a bit is 1 in n and 0 in k, increment a counter (representing the number of changes needed). If a bit is 0 in n and 1 in k, it's impossible to transform n into k, so return -1.

• Early Exit: If n becomes equal to k during the bitwise comparison, return the current count.

• Runtime Complexity: O(log n) (where n is the larger of the two input integers).

• Storage Complexity: O(1).

Code

    def min_bit_flips(n: int, k: int) -> int:
    """
    Calculates the minimum number of bit flips needed to make n equal to k.

    Args:
        n: The initial integer.
        k: The target integer.

    Returns:
        The minimum number of bit flips needed, or -1 if it's impossible.
    """

    flips = 0
    for i in range(20):  # Iterate through bits (up to 10^6)
        bit_n = (n >> i) & 1
        bit_k = (k >> i) & 1

        if bit_n == 1 and bit_k == 0:
            flips += 1
        elif bit_n == 0 and bit_k == 1:
            return -1

        if n == k:
            return flips

    if n != k: #This condition is relevant if numbers require more than 20 bits
        return -1

    return flips