# Solving Leetcode Interviews in Seconds with AI: Number of Bit Changes to Make Two Integers Equal


	# 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](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
	> 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
	```python
	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
	```