Find unique binary string

class Solution:
    def findDifferentBinaryString(self, nums: List[str]) -> str:
        n = len(nums)
        # 预处理对应整数的哈希集合
        vals = {int(num, 2) for num in nums}
        # 寻找第一个不在哈希集合中的整数
        val = 0
        while val in vals:
            val += 1
        # 将整数转化为二进制字符串返回
        res = "{:b}".format(val)
        return '0' * (n - len(res)) + res

Find Unique Binary String

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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.