Find Unique Binary String

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.
SOLUTION:
class Solution:
    def findDifferentBinaryString(self, nums: List[str]) -> str:
        n = len(nums)
        nums = set(nums)
        for num in nums:
            for i in range(n):
                curr = num[:i] + str(1 - int(num[i])) + num[i + 1:]
                if curr not in nums:
                    return curr

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