LeetCode 916: Word Subsets Solution

Master LeetCode problem 916 (Word Subsets), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

916. Word Subsets

Medium

Array Hash Table String

Problem Explanation

Explanation

To solve this problem, we need to find the universal strings from words1 which contain all the letters present in any word in words2. We can achieve this by first finding the counts of each letter in all words in words2, then finding the maximum count of each letter across all words2 words. Finally, we iterate over words1 and check if each word contains all the letters with counts greater than or equal to the max counts found in words2.

Algorithm:

Create a frequency array for each letter in all words in words2.
Find the maximum count of each letter across all words2 words.
Iterate over words1, for each word:
- Create a frequency array for the current word.
- Check if the current word contains all the letters with counts greater than or equal to the max counts found in words2.
- If yes, add the word to the result list.

Time Complexity

The time complexity of this solution is O(n + m), where n is the total number of characters in words1 and words2, and m is the total number of unique characters in words2.

Space Complexity

The space complexity of this solution is O(1) since the frequency arrays have a constant size of 26 for lowercase English letters.

Solution Code

class Solution {
    public List<String> wordSubsets(String[] words1, String[] words2) {
        List<String> result = new ArrayList<>();
        
        int[] maxFreq = new int[26];
        for (String word : words2) {
            int[] freq = new int[26];
            for (char c : word.toCharArray()) {
                freq[c - 'a']++;
            }
            for (int i = 0; i < 26; i++) {
                maxFreq[i] = Math.max(maxFreq[i], freq[i]);
            }
        }
        
        for (String word : words1) {
            int[] freq = new int[26];
            for (char c : word.toCharArray()) {
                freq[c - 'a']++;
            }
            boolean isUniversal = true;
            for (int i = 0; i < 26; i++) {
                if (freq[i] < maxFreq[i]) {
                    isUniversal = false;
                    break;
                }
            }
            if (isUniversal) {
                result.add(word);
            }
        }
        
        return result;
    }
}

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