How do you solve Camelcase Matching on LeetCode?

Camelcase Matching (LeetCode #1023) 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 pattern must be matched in order, but can have lowercase letters in between.

What is the time complexity of Camelcase Matching?

The optimal solution for Camelcase Matching runs in O(n) time and uses O(1) space. The optimal solution runs in O(n) time for each query since we only traverse the query string once, making it efficient.

What algorithm or data structure is used to solve Camelcase Matching?

Camelcase Matching uses the following concepts: Array, Two Pointers, String, Trie, String Matching. The recommended patterns to study are: Two Pointers, String Matching.

What are the interview tips for Camelcase Matching?

Always clarify the problem statement and constraints before jumping into coding. Think about edge cases, such as empty strings or patterns. Explain your thought process as you code to demonstrate your understanding.

What are common mistakes when solving Camelcase Matching?

Not handling cases where the pattern is longer than the query. Forgetting to check if all characters in the pattern are matched.

Camelcase Matching — LeetCode #1023 (Medium)

Tags: Array, Two Pointers, String, Trie, String Matching

Related patterns: Two Pointers, String Matching

Brute Force approach

Time complexity: O(n²). Space complexity: O(1).

The brute force approach checks each query against the pattern by trying to match characters one by one. If the characters in the pattern can be found in the same order in the query (with possible lowercase letters in between), we consider it a match.

The complexity is O(n²) because for each query (n), we may need to traverse the entire string (n) to check for matches.

Step 1: For each query, initialize two pointers: one for the query and one for the pattern.
Step 2: Iterate through the query characters, and for each character in the query, check if it matches the current character in the pattern.
Step 3: If it matches, move the pattern pointer forward. If it doesn't match, continue moving the query pointer.
Step 4: If you reach the end of the pattern, it means the query matches the pattern; otherwise, it doesn't.
Step 5: Store the result for each query.

For queries = ["FooBar", "FooBarTest", "FootBall", "FrameBuffer", "ForceFeedBack"], pattern = "FB": 1. For "FooBar": pIdx = 0, matches 'F' (pIdx = 1), matches 'B' (pIdx = 2) -> true 2. For "FooBarTest": pIdx = 0, matches 'F' (pIdx = 1), does not match 'B' -> false 3. For "FootBall": pIdx = 0, matches 'F' (pIdx = 1), matches 'B' (pIdx = 2) -> true 4. For "FrameBuffer": pIdx = 0, matches 'F' (pIdx = 1), matches 'B' (pIdx = 2) -> true 5. For "ForceFeedBack": pIdx = 0, matches 'F' (pIdx = 1), does not match 'B' -> false

Optimal Solution approach

Time complexity: O(n). Space complexity: O(1).

The optimal solution uses a two-pointer technique to efficiently match the pattern against each query. This approach reduces unnecessary checks and allows us to determine matches in linear time.

The optimal solution runs in O(n) time for each query since we only traverse the query string once, making it efficient.

Step 1: Initialize an empty result list.
Step 2: For each query, initialize two pointers: one for the query and one for the pattern.
Step 3: Traverse through the query and pattern simultaneously. If characters match, move both pointers; if they don't, only move the query pointer.
Step 4: If the pattern pointer reaches the end, it means the pattern is matched; otherwise, it doesn't.
Step 5: Store the result for each query.

Key Insights

The pattern must be matched in order, but can have lowercase letters in between.
Using two pointers allows for efficient matching without unnecessary checks.

Common Mistakes

Not handling cases where the pattern is longer than the query.
Forgetting to check if all characters in the pattern are matched.

Interview Tips

Always clarify the problem statement and constraints before jumping into coding.
Think about edge cases, such as empty strings or patterns.
Explain your thought process as you code to demonstrate your understanding.

#1023

Camelcase Matching

Medium

Array↗Two Pointers↗String↗Trie↗String Matching↗Two Pointers↗String Matching↗

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 two-pointer technique to efficiently match the pattern against each query. This approach reduces unnecessary checks and allows us to determine matches in linear time.

⚙️

Algorithm

5 steps

1Step 1: Initialize an empty result list.
2Step 2: For each query, initialize two pointers: one for the query and one for the pattern.
3Step 3: Traverse through the query and pattern simultaneously. If characters match, move both pointers; if they don't, only move the query pointer.
4Step 4: If the pattern pointer reaches the end, it means the pattern is matched; otherwise, it doesn't.
5Step 5: Store the result for each query.

solution.py8 lines

1def camelcaseMatching(queries, pattern):
2    def matches(query):
3        p_idx = 0
4        for char in query:
5            if p_idx < len(pattern) and char == pattern[p_idx]:
6                p_idx += 1
7        return p_idx == len(pattern)
8    return [matches(q) for q in queries]

ℹ

Complexity note: The optimal solution runs in O(n) time for each query since we only traverse the query string once, making it efficient.

1The pattern must be matched in order, but can have lowercase letters in between.
2Using two pointers allows for efficient matching without unnecessary checks.

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