How do you solve Substring Matching Pattern on LeetCode?

Substring Matching Pattern (LeetCode #3407) 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: The '*' can represent any sequence, so focus on the fixed parts of the pattern.

What is the time complexity of Substring Matching Pattern?

The optimal solution for Substring Matching Pattern runs in O(n) time and uses O(1) space. The time complexity is O(n) because we only check the start and end of the string once, making it linear in relation to the length of 's'.

What is the brute force solution for Substring Matching Pattern?

The brute force approach for Substring Matching Pattern is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks every possible substring of 's' to see if it matches the pattern 'p' with the '*' replaced by any sequence of characters. This means we will generate all substrings and check if they can fit the pattern. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Substring Matching Pattern?

Substring Matching Pattern uses the following concepts: String, String Matching. The recommended patterns to study are: String Matching, Two Pointers.

What are the interview tips for Substring Matching Pattern?

Clarify the problem statement; ensure you understand how '*' works. Think about edge cases, such as when 's' is shorter than the combined lengths of left and right parts. Practice breaking down the problem into smaller parts before coding.

What are common mistakes when solving Substring Matching Pattern?

Not checking the lengths of the left and right parts against 's'. Assuming '*' can only replace characters directly next to it.

#3407

Substring Matching Pattern

Easy

StringString MatchingString MatchingTwo 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 solution uses a single pass to check if the left and right parts of the pattern can be found in the string while allowing for any characters in between. This is more efficient than checking every substring.

⚙️

Algorithm

4 steps

1Step 1: Split the pattern 'p' into left and right parts around '*'.
2Step 2: Check if the left part exists at the start of 's'.
3Step 3: Check if the right part exists at the end of 's'.
4Step 4: If both parts match, return true; otherwise, return false.

solution.py3 lines

1def isMatch(s, p):
2    left, right = p.split('*')
3    return s.startswith(left) and s.endswith(right) and len(s) >= len(left) + len(right)

ℹ

Complexity note: The time complexity is O(n) because we only check the start and end of the string once, making it linear in relation to the length of 's'.

1The '*' can represent any sequence, so focus on the fixed parts of the pattern.
2The order of characters matters; ensure both left and right parts are checked.

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