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Tested Python solution for LeetCode 1980 with 20 pytest cases. Generate a practice environment with lcpy.
lcpy gen -n 1980 # by problem number
lcpy gen -s find_unique_binary_string # by problem name
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.
Input: nums = ["01","10"]
Output: "11"
Explanation: "11" does not appear in nums. "00" would also be correct.
Input: nums = ["00","01"]
Output: "11"
Explanation: "11" does not appear in nums. "10" would also be correct.
Input: nums = ["111","011","001"]
Output: "101"
Explanation: "101" does not appear in nums. "000", "010", "100", and "110" would also be correct.
n == nums.length1 <= n <= 16nums[i].length == nnums[i] is either '0' or '1'.nums are unique.class Solution:
# Time: O(n) single pass over the diagonal
# Space: O(n) for the result string
def find_unique_binary_string(self, nums: list[str]) -> str:
# Cantor diagonal: flipping nums[i][i] differs from nums[i] at position i,
# so the result differs from every string in nums.
return "".join("1" if s[i] == "0" else "0" for i, s in enumerate(nums))
| Time | Space |
|---|---|
| O(n) single pass over the diagonal | O(n) for the result string |