2107 Find Unique Binary String

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**.

class Solution:
    def findDifferentBinaryString(self, nums: List[str]) -> str:
        arr = []
        s = set()
        v = 0
        for k in nums:
            arr.append(int(k, 2))
            s.add(int(k, 2))
        #print(arr)
        for k in range(0, pow(2,len(nums[0]))):
            if k not in s:
                v = k
                break
        x = format(v, 'b')
        #print(x)
        if len(nums[0]) > len(str(x)):
            return ("0" * (len(nums[0]) - len(str(x)))) + str(x)
        return str(x)