# Solving Leetcode Interviews in Seconds with AI: Convert to Base -2


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1017" 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 an integer n, return a binary string representing its representation in base -2. Note that the returned string should not have leading zeros unless the string is "0".   Example 1:  Input: n = 2 Output: "110" Explantion: (-2)2 + (-2)1 = 2  Example 2:  Input: n = 3 Output: "111" Explantion: (-2)2 + (-2)1 + (-2)0 = 3  Example 3:  Input: n = 4 Output: "100" Explantion: (-2)2 = 4    Constraints:  0 <= n <= 109  

	# Explanation
	Here's the breakdown of the solution:

*   **Iterative Conversion:** Repeatedly apply modulo and division operations, similar to base-2 conversion, but using -2 as the base.
*   **Handling Negative Remainders:** When the remainder is negative, adjust it to be positive by adding the absolute value of the base (2 in this case) and compensating by adding 1 to the quotient. This ensures correct handling in the negative base.
*   **Building the String:** Prepend the remainders (which are always 0 or 1 after the adjustments) to a string to build the base -2 representation.

*   **Runtime Complexity:** O(log n) & **Storage Complexity:** O(log n) - where n is the input integer.

	
	# Code
	```python
	def baseNeg2(n: int) -> str:
    """
    Converts an integer to its base -2 representation as a string.

    Args:
        n: The integer to convert.

    Returns:
        The base -2 representation of n as a string.
    """
    if n == 0:
        return "0"

    result = ""
    while n != 0:
        remainder = n % -2
        n //= -2

        if remainder < 0:
            remainder += 2
            n += 1

        result = str(remainder) + result

    return result
	```