1980. Find Unique Binary String

#Problem

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.

https://leetcode.com/problems/find-unique-binary-string/

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.

#Test Cases

class Solution:
    def findDifferentBinaryString(self, nums: List[str]) -> str:

import pytest

from solution import Solution


def verify(res: str, nums: list[str]):
    assert res not in nums
    assert len(res) == len(nums[0])
    for d in res:
        assert d in '01'


@pytest.mark.parametrize('nums, expected', [
    (["01","10"], "11"),
    (["00","01"], "11"),
    (["111","011","001"], "101"),
])
@pytest.mark.parametrize('sol', [Solution()])
def test_solution(sol, nums, expected):
    res = sol.findDifferentBinaryString(nums)
    if res != expected:
        verify(res, nums)

#Thoughts

取 0 到 n（包含）的所有整数（共 n + 1 个），这些数字都可以表达成长度为 n 的二进制形式（左补零），且根据鸽笼原理，这 n + 1 个整数中，至少有一个不在 nums 中。

用一个集合记录 [0, n] 区间内所有整数，遍历 nums，对于 nums 中的每一个二进制串，转成整数后从集合中删除该数字。最后集合中至少会剩余一个数字。从集合剩余的数字中任取一个，格式化为长度为 n 的二进制字符串，返回即可。

时间复杂度 O(n²)，空间复杂度 O(n)。

#Code

class Solution:
    def findDifferentBinaryString(self, nums: list[str]) -> str:
        n = len(nums[0])
        candidates = set(range(n + 1))
        for num in nums:
            candidates.discard(int(num, 2))

        return format(candidates.pop(), f'0{n}b')