# Solving Leetcode Interviews in Seconds with AI: Find Unique Binary String


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1980" 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 array of strings nums containing n unique binary strings each of length n, return a binary string of length n that does not appear in nums. If there are multiple answers, you may return any of them.   Example 1:  Input: nums = ["01","10"] Output: "11" Explanation: "11" does not appear in nums. "00" would also be correct.  Example 2:  Input: nums = ["00","01"] Output: "11" Explanation: "11" does not appear in nums. "10" would also be correct.  Example 3:  Input: nums = ["111","011","001"] Output: "101" Explanation: "101" does not appear in nums. "000", "010", "100", and "110" would also be correct.    Constraints:  n == nums.length 1 <= n <= 16 nums[i].length == n nums[i] is either '0' or '1'. All the strings of nums are unique.  

	# Explanation
	Here's a breakdown of the solution:

*   **Diagonal Flipping (Cantor's Diagonal Argument):** Construct the missing string by examining the i-th character of the i-th string in the input array. Flip this character (0 becomes 1, and 1 becomes 0) to create the i-th character of the missing string. This guarantees the generated string will differ from every string in the input at least in one position, therefore ensuring it's unique.

*   **Iterative Construction:** Build the missing string character by character through a simple loop.

*   **Efficiency:** This approach avoids generating all possible binary strings, which would be exponential in `n`.

*   **Runtime & Storage Complexity:** O(n), O(n)

	
	# Code
	```python
	def findDifferentBinaryString(nums):
    """
    Finds a binary string of length n that does not appear in nums.

    Args:
        nums: A list of n unique binary strings of length n.

    Returns:
        A binary string of length n that is not in nums.
    """
    n = len(nums)
    result = ""
    for i in range(n):
        if nums[i][i] == '0':
            result += '1'
        else:
            result += '0'
    return result
	```