# 648 Replace Words

## 648. [Replace Words](https://leetcode.com/problems/replace-words/description/)

## 1. Question

In English, we have a concept called`root`, which can be followed by some other words to form another longer word - let's call this word`successor`. For example, the root`an`, followed by`other`, which can form another word`another`.

Now, given a dictionary consisting of many roots and a sentence. You need to replace all the`successor`in the sentence with the`root`forming it. If a`successor`has many`roots`can form it, replace it with the root with the shortest length.

You need to output the sentence after the replacement.

**Example 1:**

```
Input: dict = ["cat", "bat", "rat"]
sentence = "the cattle was rattled by the battery"

Output: "the cat was rat by the bat"
```

**Note:**

1. The input will only have lower-case letters.
2. 1 <= dict words number <= 1000
3. 1 <= sentence words number <= 1000
4. 1 <= root length <= 100
5. 1 <= sentence words length <= 1000

## 2. Implementation

**(1) Trie**

思路: 很典型的查询prefix的应用，所以用trie很适合，只要在trieNode的基础上加入字典里的string所对应的index即可

```java
class Solution {
    class TrieNode {
        char val;
        int index;
        boolean isWord;
        TrieNode[] childNode;

        public TrieNode(char val) {
            this.val = val;
            index = -1;
            childNode = new TrieNode[26];
        }
    }

    class Trie {
        TrieNode root;

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

        public void insert(String word, int index) {
            if (word == null || word.length() == 0) {
                return;
            }

            TrieNode curNode = root;

            for (char c : word.toCharArray()) {
                int pos = c - 'a';
                if (curNode.childNode[pos] == null) {
                    curNode.childNode[pos] = new TrieNode(c);
                }
                curNode = curNode.childNode[pos];
            }
            curNode.isWord = true;
            curNode.index = index;
        }

        public int searchPrefix(String word) {
            if (word == null || word.length() == 0) {
                return -1;
            }

            TrieNode curNode = root;

            for (char c : word.toCharArray()) {
                int pos = c - 'a';
                if (curNode.childNode[pos] == null) {
                    return -1;
                }
                else if (curNode.childNode[pos].isWord){
                    return curNode.childNode[pos].index;
                }
                curNode = curNode.childNode[pos];
            }
            return curNode.index;
        }
    }
    public String replaceWords(List<String> dict, String sentence) {
        if (dict.size() == 0 || sentence == null || sentence.length() == 0) {
            return "";
        }

        Trie trie = new Trie();

        for (int i = 0; i < dict.size(); i++) {
            trie.insert(dict.get(i), i);
        }

        String[] words = sentence.split("\\s+");

        for (int i = 0; i < words.length; i++) {
            int index = trie.searchPrefix(words[i]);

            if (index != -1) {
                words[i] = dict.get(index);
            }
        }

        StringBuilder res = new StringBuilder();

        for (int i = 0; i < words.length - 1; i++) {
            res.append(words[i]);
            res.append(" ");
        }

        res.append(words[words.length - 1]);
        return res.toString();
    }
}
```

## 3. Time & Space Complexity

**Trie:** 时间复杂度O(n \* m), n是dict里面string的个数，m是string的平均长度。 空间复杂度O(nm)


---

# Agent Instructions: Querying This Documentation

If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question.

Perform an HTTP GET request on the current page URL with the `ask` query parameter:

```
GET https://protegejj.gitbook.io/oj-practices/chapter1/trie/648-replace-words.md?ask=<question>
```

The question should be specific, self-contained, and written in natural language.
The response will contain a direct answer to the question and relevant excerpts and sources from the documentation.

Use this mechanism when the answer is not explicitly present in the current page, you need clarification or additional context, or you want to retrieve related documentation sections.