1980. Find Unique Binary String - Detailed Explanation

Problem Statement

Description:
You are given an array of unique binary strings, nums, where each string has the same length (say, n). Your task is to return any binary string of length n that is not present in the given array.

Since there are $(2^n$ ) possible binary strings of length n and only n strings are present in the input (with $(n < 2^n$ )), at least one binary string is guaranteed not to be in the list.

Examples:

Example 1:
- Input: nums = ["01", "10"]
- Output: "11"
- Explanation:
  - The array contains "01" and "10".
  - The string "11" (or "00") does not appear in the array, so "11" is a valid answer.
Example 2:
- Input: nums = ["00", "01"]
- Output: "10"
- Explanation:
  - The given binary strings are "00" and "01".
  - The string "10" (or "11") is not in the array, so "10" is a valid answer.
Example 3:
- Input: nums = ["1"]
- Output: "0"
- Explanation:
  - The single binary string is "1".
  - The string "0" is not present, making it a valid result.

Constraints:

The length of the input array is n.
Each binary string in the array has length n.
All strings in the array are unique.

Hints Before the Solution

Diagonalization Insight:
Think about how each binary string can be “differentiated” by looking at one character per string. If you look at the i-th character of the i-th string and flip it (change '0' to '1' and vice versa), the resulting string will differ from every string in the list at least at one index.
Avoid Brute Force When Possible:
Although you could generate all possible $(2^n$ ) binary strings and check which one is missing, this approach is not efficient. Instead, leverage the guarantee that the array has exactly n strings and use the diagonal method.

Approaches

Approach 1: Diagonalization (Optimal)

Idea:
For each index i (from 0 to n-1), look at the i-th character of the i-th string in the array and flip it. Construct a new binary string from these flipped characters.

Why It Works:

Assume the resulting string is ans.
For any string nums[i] in the array, the character at index i in ans is the opposite of the character at index i in nums[i].
Therefore, ans differs from each nums[i] in at least the i-th position, ensuring it cannot be in the input array.

Time Complexity:

O(n) iterations with O(1) work per iteration (accessing one character and flipping it), leading to O(n).
(Note: Since each string is of length n, if you consider reading the entire input, it is O(n²) overall—but the core construction is linear in the number of strings.)

Space Complexity:

O(n) for constructing the answer string.

Approach 2: Brute Force (Not Recommended)

Idea:
Generate all possible binary strings of length n (there are $(2^n$ ) of them) and check each one to see if it is absent from the input array.

Why It's Less Efficient:

For even moderate values of n, $(2^n$ ) grows exponentially.
Although the input size is small (since n strings of length n implies n is not large), the diagonal method is both elegant and optimal.

Code Implementations

Python Implementation

Python3

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Java Implementation

Java

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Complexity Analysis

Time Complexity:
- Diagonalization runs in O(n) time since it processes one character per each of the n strings.
- Considering the cost to access each character, if we treat each string's length as n, it is O(n) for the construction, and overall reading the input is O(n²).
Space Complexity:
- O(n) for the answer string (ignoring the input storage).

Step-by-Step Walkthrough with Visual Example

Consider the input: nums = ["01", "10"]

Let n = 2.
Iteration 1 (i = 0):
- Look at nums[0][0] which is '0'.
- Flip it to '1'.
- The result so far is "1".
Iteration 2 (i = 1):
- Look at nums[1][1] which is '0' (since "10", the second character is '0').
- Flip it to '1'.
- The result becomes "11".
Verification:
- "11" is not present in the input array ["01", "10"], so it is a valid answer.

Common Mistakes

Not Flipping the Character:
One might mistakenly copy the i-th character directly instead of flipping it, which could yield a string already present in the input.
Overcomplicating the Solution:
Trying to generate all $(2^n$ ) possible strings or using unnecessary nested loops when the diagonal method provides a simple O(n) solution.
Indexing Errors:
Be cautious with array indexing—ensure you access the correct i-th character of the i-th string.

Edge Cases

Single-Element Array:
If the array contains only one binary string (e.g., ["0"]), the method will flip the single character, producing "1", which is correct.
Uniform Pattern:
Even if all strings follow a certain pattern, the diagonalization always guarantees at least one differing bit per string.

Alternative Variation:
If the binary strings were not guaranteed to be of equal length, the problem would become more challenging as the diagonal method would not directly apply.
Related Problems for Further Practice:
- Missing Number (finding a missing element in a sequence)
- Construct Binary String (problems involving building strings under constraints)
- Find the Difference (determining differences between strings)