## DEV Community

Abhishek Chaudhary

Posted on

# 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
``````