How do you solve Find Special Substring of Length K on LeetCode?

Find Special Substring of Length K (LeetCode #3456) 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: Look for consecutive characters efficiently.

What is the time complexity of Find Special Substring of Length K?

The optimal solution for Find Special Substring of Length K runs in O(n) time and uses O(1) space. The complexity is O(n) as we traverse the string only once, maintaining a count of consecutive characters.

What is the brute force solution for Find Special Substring of Length K?

The brute force approach for Find Special Substring of Length K is Brute Force, with O(n²) time complexity and O(1) space complexity. Check every possible substring of length k to see if it meets the criteria. This method is straightforward but inefficient for larger strings. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Find Special Substring of Length K?

Find Special Substring of Length K uses the following concepts: String. The recommended patterns to study are: Sliding Window, Two Pointers.

What are the interview tips for Find Special Substring of Length K?

Explain your thought process clearly. Consider edge cases like empty strings. Optimize your solution before coding.

What are common mistakes when solving Find Special Substring of Length K?

Not checking boundaries properly. Counting characters incorrectly.

#3456

Find Special Substring of Length K

Easy

StringSliding WindowTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(1)

Use a sliding window approach to efficiently check for valid substrings without redundant checks.

⚙️

Algorithm

3 steps

1Step 1: Traverse the string while counting consecutive characters.
2Step 2: When a sequence of k identical characters is found, check the characters before and after.
3Step 3: Return true if conditions are met, otherwise continue.

solution.py12 lines

1def has_special_substring(s, k):
2    n = len(s)
3    count = 1
4    for i in range(1, n):
5        if s[i] == s[i - 1]:
6            count += 1
7        else:
8            count = 1
9        if count == k:
10            if (i == k - 1 or s[i - k] != s[i]) and (i + 1 == n or s[i + 1] != s[i]):
11                return True
12    return False

ℹ

Complexity note: The complexity is O(n) as we traverse the string only once, maintaining a count of consecutive characters.

1Look for consecutive characters efficiently.
2Check boundaries to ensure distinct characters.

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