LeetCode 648: Replace Words Solution

Master LeetCode problem 648 (Replace Words), 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.

648. Replace Words

Medium

Array Hash Table String Trie

Problem Explanation

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.

Solution Code

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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