648. Replace Words

Explanation

To solve this problem, we can follow these steps:

Create a trie data structure to store all the roots from the dictionary.
For each word in the sentence, check if it has a root in the trie.
Replace the word with the shortest root if it has multiple roots.
Return the modified sentence.

Time complexity: O(n * m) where n is the number of words in the sentence and m is the average length of a word. Space complexity: O(n * m) to store the trie.

View Leetcode Problem

class TrieNode {
    Map<Character, TrieNode> children;
    boolean isEndOfWord;

    TrieNode() {
        children = new HashMap<>();
        isEndOfWord = false;
    }
}

class Trie {
    TrieNode root;

    Trie() {
        root = new TrieNode();
    }

    public void insert(String word) {
        TrieNode node = root;
        for (char c : word.toCharArray()) {
            node.children.putIfAbsent(c, new TrieNode());
            node = node.children.get(c);
        }
        node.isEndOfWord = true;
    }

    public String searchRoot(String word) {
        TrieNode node = root;
        StringBuilder rootWord = new StringBuilder();

        for (char c : word.toCharArray()) {
            if (!node.children.containsKey(c) || node.isEndOfWord) {
                break;
            }
            rootWord.append(c);
            node = node.children.get(c);
        }

        return node.isEndOfWord ? rootWord.toString() : word;
    }
}

class Solution {
    public String replaceWords(List<String> dictionary, String sentence) {
        Trie trie = new Trie();
        for (String root : dictionary) {
            trie.insert(root);
        }

        StringBuilder result = new StringBuilder();
        for (String word : sentence.split(" ")) {
            result.append(trie.searchRoot(word)).append(" ");
        }

        return result.toString().trim();
    }
}

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Explanation

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