Problem Informations

• Problem Index: 1980
• Problem Link: Find Unique Binary String
• Topics: Array, Hash Table, String, Backtracking

Problem 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:

• $n == \text{nums.length}$
• $1 \leq n \leq 16$
• $\text{nums}[i].\text{length} == n$
• $\text{nums}[i]$ is either '0' or '1'.
• All the strings of nums are unique.

Intuition

The problem requires finding a binary string of length $n$ that is not present in the given list of binary strings, where each string is of length $n$ and all strings are unique. A straightforward way to ensure this is to leverage the properties of the strings and construct a new string that intentionally deviates from each of the input strings in a systematic manner. Specifically, considering the constraints and structure of the problem, a diagonal traversal approach can be effectively utilized.

Approach

The algorithm is centered around constructing a new binary string by flipping a specific character in each of the input strings. This method ensures the newly created string deviates from all the input strings at least in one position, making it impossible for this newly formed string to be present in the input list.

1. Initialize the Result String: Begin with an empty result string that will ultimately hold the binary sequence not present in nums.

2. Iterate Over Input Strings: Traverse each binary string at index i in the list nums. The list has $n$ strings, and each string itself is of length $n$.

3. Construct the Different String:

   • At each index i, focus on the $i^{th}$ character of the string nums[i].
   • Flip the $i^{th}$ character of nums[i]. Specifically, if this character is ‘0’, append ‘1’ to the result string, and if it is ‘1’, append ‘0’.
   • This manipulation ensures that the constructed string will differ from the input string nums[i] at least at the $i^{th}$ position.

4. Return the Result String: Once the iteration and manipulation complete, return the constructed string. This string is guaranteed not to be in nums as it differs from each nums[i] at the $i^{th}$ position.

5. Guarantee of Uniqueness: Since all input strings in nums are unique and n == \text{nums.length}, this approach reliably produces a string not present in nums, satisfying the problem’s requirements. This method takes advantage of the fact that flipping each diagonal element ensures the newly formed string will not match any of the provided ones. This approach is both efficient given the constraints and simple in its logic.

Code

C++

class Solution {
public:
    string findDifferentBinaryString(vector<string>& nums) {
        string result = "";

        // Iterate through each string in the array
        for (int i = 0; i < nums.size(); i++) {
            // Construct a new binary string by flipping the ith character 
            // of the ith string in nums (if it's '0', append '1' to result; 
            // if it's '1', append '0')
            result += nums[i][i] == '0' ? '1' : '0';
        }

        // Return the constructed binary string which does not appear in nums
        return result;
    }
};

Python

class Solution:
    def findDifferentBinaryString(self, nums):
        result = ""

        # Iterate through each string in the array
        for i in range(len(nums)):
            # Construct a new binary string by flipping the ith character 
            # of the ith string in nums (if it's '0', append '1' to result; 
            # if it's '1', append '0')
            result += '1' if nums[i][i] == '0' else '0'

        # Return the constructed binary string which does not appear in nums
        return result

Complexity

• Time complexity: $O(n)$
  The algorithm iterates through each string in the nums array exactly once, constructing a new binary string by flipping the $i$th character of the $i$th string. Each operation of flipping a bit is $O(1)$, and since this is done $n$ times, the overall time complexity is $O(n)$.

• Space complexity: $O(n)$
  The space complexity is determined by the string result that is being constructed. This string has a length of $n$, as each iteration appends one character to it. Consequently, the additional space required is proportional to the length of the output string, resulting in a space complexity of $O(n)$. This complexity does not include the input space, as it is generally not considered in space complexity calculations.