What is the time complexity of Number of Matching Subsequences?

The optimal solution for Number of Matching Subsequences runs in O(n + m), where n is the length of s and m is the total length of words. time and uses O(m) space. This complexity arises because we process each character in `s` and each word only once, making it efficient for larger inputs.

What is the brute force solution for Number of Matching Subsequences?

The brute force approach for Number of Matching Subsequences is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks each word in the list to see if it can be formed by deleting some characters from the string `s` without changing the order. This is straightforward but inefficient for larger inputs. The optimal Optimal Solution approach improves this to O(n + m), where n is the length of s and m is the total length of words..

What are the interview tips for Number of Matching Subsequences?

Always clarify the definition of a subsequence with the interviewer. Discuss your thought process and potential edge cases before diving into coding. Consider the efficiency of your approach and be ready to explain how you can optimize it.

What are common mistakes when solving Number of Matching Subsequences?

Not considering edge cases where words are longer than `s`. Confusing subsequences with substrings; remember that order matters but continuity does not.

#792

Number of Matching Subsequences

Medium

ArrayHash TableStringBinary SearchDynamic ProgrammingTrieSortingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n + m), where n is the length of s and m is the total length of words.
Space	O(1)	O(m)

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Intuition

Time O(n + m), where n is the length of s and m is the total length of words.Space O(m)

The optimal approach uses a hashmap to group words by their first character and a queue to efficiently check subsequences. This reduces the number of checks needed for each character in `s`.

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Algorithm

3 steps

1Step 1: Create a hashmap where each key is a character from `s` and the value is a list of words that start with that character.
2Step 2: For each character in `s`, process the words that start with that character by checking if they can be formed as subsequences.
3Step 3: Use a queue to track the words and their next character index to check against `s`.

solution.py16 lines

1from collections import defaultdict, deque
2
3def numMatchingSubseq(s, words):
4    waiting = defaultdict(deque)
5    for word in words:
6        waiting[word[0]].append(word)
7    count = 0
8    for char in s:
9        if char in waiting:
10            for _ in range(len(waiting[char])):
11                word = waiting[char].popleft()
12                if len(word) == 1:
13                    count += 1
14                else:
15                    waiting[word[1]].append(word[1:])
16    return count

ℹ

Complexity note: This complexity arises because we process each character in `s` and each word only once, making it efficient for larger inputs.

1Understanding subsequences is crucial; they maintain order but can skip characters.
2Using data structures like hashmaps and queues can significantly optimize the search process.

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