What is the brute force solution for Find the Occurrence of First Almost Equal Substring?

The brute force approach for Find the Occurrence of First Almost Equal Substring is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible substring of `s` that has the same length as `pattern`. For each substring, we count how many characters differ from `pattern`. If the count is 1 or less, we have found an almost equal substring. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Find the Occurrence of First Almost Equal Substring?

Find the Occurrence of First Almost Equal Substring uses the following concepts: String, String Matching. The recommended patterns to study are: Sliding Window, Two Pointers.

What are the interview tips for Find the Occurrence of First Almost Equal Substring?

Always clarify the problem statement and constraints before jumping into coding. Discuss your thought process and potential approaches with the interviewer. Consider edge cases and test your solution with different input scenarios.

What are common mistakes when solving Find the Occurrence of First Almost Equal Substring?

Not correctly managing the character counts when sliding the window. Failing to account for edge cases where the substring is exactly equal.

#3303

Find the Occurrence of First Almost Equal Substring

Hard

StringString MatchingSliding 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)

The optimal approach uses a sliding window technique to maintain a count of character differences between the current substring of `s` and `pattern`. This allows us to efficiently check each substring without needing to re-evaluate characters we've already checked.

⚙️

Algorithm

4 steps

1Step 1: Initialize a variable to count character differences and a sliding window of size equal to the length of `pattern`.
2Step 2: Iterate through the first `m` characters of `s` to set up the initial difference count.
3Step 3: Slide the window across `s`, updating the difference count by removing the character that is sliding out and adding the character that is sliding in.
4Step 4: If the difference count is 1 or less at any point, return the starting index of that window.

solution.py16 lines

1def findAlmostEqualSubstring(s, pattern):
2    n, m = len(s), len(pattern)
3    diff_count = 0
4    for i in range(m):
5        if s[i] != pattern[i]:
6            diff_count += 1
7    if diff_count <= 1:
8        return 0
9    for i in range(m, n):
10        if s[i] != pattern[i - m]:
11            diff_count += 1
12        if s[i - m] != pattern[0]:
13            diff_count -= 1
14        if diff_count <= 1:
15            return i - m + 1
16    return -1

ℹ

Complexity note: The time complexity is O(n) because we only make a single pass through `s`, maintaining a count of differences without needing to re-evaluate the entire substring each time.

1Understanding the concept of character differences is crucial for solving substring problems.
2Using a sliding window approach can significantly reduce time complexity.

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