LeetCode 748: Shortest Completing Word Solution

Master LeetCode problem 748 (Shortest Completing Word), a easy 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.

748. Shortest Completing Word

Easy

Array Hash Table String

Problem Explanation

Explanation

To solve this problem, we need to find the shortest word in the given list of words that contains all the letters in the license plate. We first create a frequency map of letters in the license plate and then iterate through each word in the list to check if it is a completing word. For each word, we create a frequency map and compare it with the license plate frequency map to determine if it contains all required letters. We then update the result with the shortest completing word found so far.

Algorithm:

Create a frequency map for the letters in the license plate.
Initialize a variable result to store the shortest completing word found so far.
Iterate through each word in the list of words:
- Create a frequency map for the letters in the current word.
- Check if the current word is a completing word by comparing its frequency map with the license plate frequency map.
- If it is a completing word and shorter than the current result, update the result.
Return the result as the shortest completing word.

Time complexity: O(n * m), where n is the number of words and m is the length of the longest word. Space complexity: O(1) since the frequency maps have a fixed size.

Solution Code

class Solution {
    public String shortestCompletingWord(String licensePlate, String[] words) {
        int[] lpFreq = new int[26];
        for (char c : licensePlate.toLowerCase().toCharArray()) {
            if (Character.isLetter(c)) {
                lpFreq[c - 'a']++;
            }
        }
        
        String result = "";
        for (String word : words) {
            if (word.length() >= result.length() && (result.isEmpty() || word.length() < result.length())) {
                int[] wordFreq = new int[26];
                for (char c : word.toCharArray()) {
                    wordFreq[c - 'a']++;
                }
                
                boolean isCompleting = true;
                for (int i = 0; i < 26; i++) {
                    if (lpFreq[i] > wordFreq[i]) {
                        isCompleting = false;
                        break;
                    }
                }
                
                if (isCompleting) {
                    result = word;
                }
            }
        }
        
        return result;
    }
}

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