What is the time complexity of Check If a String Contains All Binary Codes of Size K?

The optimal solution for Check If a String Contains All Binary Codes of Size K runs in O(n) time and uses O(n) space. The time complexity is O(n) because we only traverse the string once, and the space complexity is O(n) due to the storage of unique substrings in the set.

What is the brute force solution for Check If a String Contains All Binary Codes of Size K?

The brute force approach for Check If a String Contains All Binary Codes of Size K is Brute Force, with O(n²) time complexity and O(n) space complexity. The brute force approach checks every possible substring of length k in the given binary string. If we find all possible binary codes of length k, we return true; otherwise, false. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Check If a String Contains All Binary Codes of Size K?

Check If a String Contains All Binary Codes of Size K uses the following concepts: Hash Table, String, Bit Manipulation, Rolling Hash, Hash Function. The recommended patterns to study are: Hash Map, Sliding Window.

What are the interview tips for Check If a String Contains All Binary Codes of Size K?

Always clarify the problem constraints and edge cases with the interviewer. Explain your thought process while coding to demonstrate your understanding. Consider both time and space complexity in your solutions.

What are common mistakes when solving Check If a String Contains All Binary Codes of Size K?

Not considering edge cases where the string length is less than k. Failing to check if the size of the set matches 2^k before returning.

#1461

Check If a String Contains All Binary Codes of Size K

Medium

Hash TableStringBit ManipulationRolling HashHash FunctionHash MapSliding Window

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(n)

The optimal solution uses a sliding window approach to efficiently track the unique substrings of length k. This avoids the overhead of repeatedly checking for duplicates.

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Algorithm

4 steps

1Step 1: Create a set to store unique binary codes of length k.
2Step 2: Loop through the string with a sliding window of size k, adding each substring to the set.
3Step 3: If the size of the set equals 2^k at any point, return true.
4Step 4: If the loop completes and the size is not equal to 2^k, return false.

solution.py7 lines

1def hasAllCodes(s, k):
2    seen = set()
3    for i in range(len(s) - k + 1):
4        seen.add(s[i:i + k])
5        if len(seen) == 2 ** k:
6            return True
7    return len(seen) == 2 ** k

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Complexity note: The time complexity is O(n) because we only traverse the string once, and the space complexity is O(n) due to the storage of unique substrings in the set.

1The number of distinct binary codes of length k is always 2^k.
2Using a set allows for efficient tracking of unique substrings.

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