How do you solve Find Unique Binary String on LeetCode?

Find Unique Binary String (LeetCode #1980) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Using the diagonal argument allows for guaranteed uniqueness without generating all combinations.

What is the brute force solution for Find Unique Binary String?

The brute force approach for Find Unique Binary String is Brute Force, with O(n²) time complexity and O(n) space complexity. The brute force approach involves generating all possible binary strings of length n and checking each one against the given list to see if it is unique. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Find Unique Binary String?

Find Unique Binary String uses the following concepts: Array, Hash Table, String, Backtracking. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Find Unique Binary String?

Always think about the constraints and how they affect your approach. Consider edge cases, such as the smallest and largest inputs. Practice explaining your thought process clearly, as communication is key in interviews.

What are common mistakes when solving Find Unique Binary String?

Failing to consider the uniqueness of the generated string properly. Not leveraging the properties of binary strings effectively.

#1980

Find Unique Binary String

Medium

ArrayHash TableStringBacktrackingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(n)	O(1)

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Intuition

Time O(n)Space O(1)

The optimal approach uses a clever technique called the 'diagonal argument'. By constructing a binary string based on the existing strings, we can ensure that the new string is unique without generating all possibilities.

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Algorithm

2 steps

1Step 1: Create a new binary string where the i-th bit is the opposite of the i-th bit of the i-th string in nums.
2Step 2: Return this newly constructed string as it cannot be in the original list.

solution.py6 lines

1# Full working Python code
2
3def findUniqueBinaryString(nums):
4    n = len(nums)
5    unique_string = ''.join('1' if nums[i][i] == '0' else '0' for i in range(n))
6    return unique_string

ℹ

Complexity note: The time complexity is O(n) because we iterate through the list once to construct the unique string. The space complexity is O(1) since we are using a fixed amount of space regardless of n.

1Using the diagonal argument allows for guaranteed uniqueness without generating all combinations.
2The brute force method is simple but inefficient for larger inputs.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.