Find Unique Binary String Leetcode Problem 1980 [Python Solution]

Find Unique Binary String problem, a LeetCode problem. We will provide a Python solution for this problem.

If you are a beginner looking to understand how to approach this problem, you’re in the right place.

Problem Overview

The problem statement is as follows: 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.

To clarify, you are given an array of binary strings, and your task is to find a binary string of the same length that is not present in the input array.

You can return any valid solution if multiple answers exist.

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.

Understanding the Constraints

Before diving into the solution, let’s analyze the problem constraints:

n is equal to nums.length, which means the number of strings in nums is equal to the length of each string.
The length of each string in nums is n.
Each string in nums consists of only ‘0’ and ‘1’.
All the strings in nums are unique.

Now that we have a clear understanding of the problem, let’s discuss how we can approach it.

Solution Approach

The problem can be solved using a backtracking approach.

The idea is to generate binary strings of length n until we find one that is not present in the input array nums.

Once we find such a string, we can return it as the solution.

Here’s a step-by-step explanation of the solution approach:

Create a set, strSet, to store the binary strings from the nums array.

This will allow us to quickly check if a string exists in the input.

Implement a backtracking function that takes two parameters: i (the current index) and cur (the current string being generated).

We’ll start with i = 0 and an initial cur string of all zeros.

In the backtracking function, we’ll have a base case.

If i is equal to the length of nums, we’ve successfully generated a binary string of length n.

At this point, we’ll check if this string exists in the strSet.

If it does, we continue generating the next possible string.

If it doesn’t, we return this string as the solution.

We’ll perform two recursive calls in the backtracking function:

First, we’ll update cur[i] to ‘0’ and make a recursive call with i + 1.

This represents the decision to choose ‘0’ at the current position.

Then, we’ll update cur[i] to ‘1’ and make a recursive call with i + 1.

This represents the decision to choose ‘1’ at the current position.

As soon as we find a valid binary string that is not in the strSet, we return it, effectively terminating the backtracking.
We’ll call the backtracking function with initial values i = 0 and cur as an array of ‘0’s.

The result of this function will be our answer.

Let’s now provide the Python solution for this problem.

Find Unique Binary String Python Code Solution

View on GitHub ➹

class Solution:
    def findDifferentBinaryString(self, nums: List[str]) -> str:
        # Convert input nums into a set for quick lookup
        strSet = set(nums)

        def backtrack(i, cur):
            # Base case: Check if we've generated a binary string of length n
            if i == len(nums):
                res = "".join(cur)
                return res if res not in strSet else None

            # Recursive call with '0' at the current position
            res = backtrack(i + 1, cur)
            if res:
                return res

            # Recursive call with '1' at the current position
            cur[i] = "1"
            res = backtrack(i + 1, cur)
            if res:
                return res

        # Start the backtracking with an initial string of all '0's
        return backtrack(0, ["0" for _ in range(len(nums))])

The code above defines a Python class Solution with a method findDifferentBinaryString that takes a list of binary strings as input and returns the desired unique binary string.

Time and Space Complexity

Let’s analyze the time and space complexity of our solution:

Time Complexity: In the worst case, the backtracking function will be called 2^n times (where n is the length of the binary strings) before finding a valid solution.

This is because we explore all possible binary strings of length n.

Additionally, checking if a string exists in the set has a time complexity of O(n), and we do this at most 2^n times.

Therefore, the overall time complexity of the solution is O(n^2).

Space Complexity: The space complexity is determined by the space used for the set strSet and the recursive call stack.

The set strSet contains at most n binary strings, so its space complexity is O(n).

The recursive call stack can have a maximum depth of n, resulting in O(n) space complexity.

Thus, the total space complexity is O(n).

Reasoning Behind Our Approach

The problem of finding a unique binary string can be efficiently solved using a backtracking approach.

We start by generating binary strings of length n, making choices of ‘0’ or ‘1’ at each position.

By keeping track of the generated strings in a set, we can quickly check if a generated string already exists in the input.

This allows us to avoid generating unnecessary strings and find a unique binary string efficiently.

Conclusion

In this blog post, we tackled the Find Unique Binary String problem from LeetCode.

We provided a Python solution using a backtracking approach to find a binary string of length n that is not present in the given array of binary strings.

The time complexity of our solution is O(n^2), and the space complexity is O(n).

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If you want to explore the problem further or test the solution, you can find the original question on LeetCode here.

Happy coding!