How do you solve Word Subsets on LeetCode?

Word Subsets (LeetCode #916) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n + m) time complexity and O(n) space complexity. Key insight: Understanding character frequency is key to solving subset problems.

What is the brute force solution for Word Subsets?

The brute force approach for Word Subsets is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks each word in words1 against all words in words2 to see if it contains all the letters required by each word in words2. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n + m).

What algorithm or data structure is used to solve Word Subsets?

Word Subsets uses the following concepts: Array, Hash Table, String. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Word Subsets?

Always clarify the problem requirements before jumping into coding. Discuss your thought process and potential optimizations with the interviewer. Practice writing clean and efficient code to improve your speed during interviews.

What are common mistakes when solving Word Subsets?

Not considering character multiplicity when checking subsets. Failing to optimize checks by pre-computing character requirements.

#916

Word Subsets

Medium

ArrayHash TableStringHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n + m)Space O(n)

The optimal solution builds a frequency count of all characters needed from words2 and then checks each word in words1 against this count. This reduces redundant checks and speeds up the process.

⚙️

Algorithm

3 steps

1Step 1: Create a frequency count of the maximum occurrences of each character required by all words in words2.
2Step 2: For each word in words1, create a frequency count of its characters.
3Step 3: Compare the frequency count of the current word in words1 with the frequency count from words2. If it meets or exceeds all counts, add it to the result.

solution.py15 lines

1from collections import Counter
2
3def wordSubsets(words1, words2):
4    max_count = Counter()
5    for word in words2:
6        count = Counter(word)
7        for char in count:
8            max_count[char] = max(max_count[char], count[char])
9
10    result = []
11    for word1 in words1:
12        count1 = Counter(word1)
13        if all(count1[char] >= max_count[char] for char in max_count):
14            result.append(word1)
15    return result

ℹ

Complexity note: The time complexity is O(n + m) where n is the total number of characters in words1 and m is the total number of characters in words2. We only traverse each word a couple of times, making this approach efficient.

1Understanding character frequency is key to solving subset problems.
2Using hash maps can significantly reduce the complexity of character counting.

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