# Solving Leetcode Interviews in Seconds with AI: Minimize XOR


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2429" 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 two positive integers num1 and num2, find the positive integer x such that:  x has the same number of set bits as num2, and The value x XOR num1 is minimal.  Note that XOR is the bitwise XOR operation. Return the integer x. The test cases are generated such that x is uniquely determined. The number of set bits of an integer is the number of 1's in its binary representation.   Example 1:  Input: num1 = 3, num2 = 5 Output: 3 Explanation: The binary representations of num1 and num2 are 0011 and 0101, respectively. The integer 3 has the same number of set bits as num2, and the value 3 XOR 3 = 0 is minimal.  Example 2:  Input: num1 = 1, num2 = 12 Output: 3 Explanation: The binary representations of num1 and num2 are 0001 and 1100, respectively. The integer 3 has the same number of set bits as num2, and the value 3 XOR 1 = 2 is minimal.    Constraints:  1 <= num1, num2 <= 109  

	# Explanation
	Here's the breakdown of the solution:

*   **Count Set Bits:** Calculate the number of set bits (1s) in `num2`. This determines how many set bits `x` must have.
*   **Prioritize Matching Bits:** Iterate through the bits of `num1` from most significant to least significant. If a bit in `num1` is set, and we still need set bits in `x`, try to set the corresponding bit in `x` to 1 (matching `num1`) to minimize the XOR result. If a bit in `num1` is unset, try to set the corresponding bit in `x` to 0.
*   **Handle Remaining Set Bits:** After processing bits where `num1` has 1, and there are still set bits remaining to be placed in `x`, place the remaining set bits in `x` where `num1` had 0, from least significant bit up.

*   Runtime: O(log N), Storage: O(1)

	
	# Code
	```python
	def find_integer(num1: int, num2: int) -> int:
    """
    Given two positive integers num1 and num2, find the positive integer x such that:
    x has the same number of set bits as num2, and
    The value x XOR num1 is minimal.
    Note that XOR is the bitwise XOR operation.
    Return the integer x.
    The test cases are generated such that x is uniquely determined.
    The number of set bits of an integer is the number of 1's in its binary representation.

    For example:
    find_integer(3, 5) == 3
    find_integer(1, 12) == 3
    """

    set_bits_num2 = bin(num2).count('1')
    x = 0
    temp_num1 = num1
    bit_position = 0

    # Prioritize matching set bits in num1
    while temp_num1 > 0:
        if temp_num1 & 1 and set_bits_num2 > 0:
            x |= (1 << bit_position)
            set_bits_num2 -= 1
        bit_position += 1
        temp_num1 >>= 1

    # Handle remaining set bits (if any) by placing them in 0 bits of num1
    bit_position = 0
    temp_num1 = num1
    while set_bits_num2 > 0:
        if (num1 & (1 << bit_position)) == 0:
            x |= (1 << bit_position)
            set_bits_num2 -= 1
        bit_position += 1

    return x
	```