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
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