1980. Find Unique Binary String
Description
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.
Solutions
Solution 1: Counting + Enumeration
Since the number of occurrences of '1' in a binary string of length $n$ can be $0, 1, 2, \cdots, n$ (there are $n + 1$ possibilities), we can certainly find a new binary string that has a different number of '1's from every string in nums
.
We can use an integer $mask$ to record the occurrence of '1' in all strings, i.e., the $i$-th bit of $mask$ is $1$ indicates that there is a string of length $n$ in which '1' appears $i$ times, otherwise it does not exist.
Then we start to enumerate the number of times '1' appears in a binary string of length $n$ from $0$. If the $i$-th bit of $mask$ is $0$, it means that there is no string of length $n$ in which '1' appears $i$ times. We can return this string as the answer.
The time complexity is $O(L)$, where $L$ is the total length of the strings in nums
. The space complexity is $O(1)$.
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