# [472. Concatenated Words](https://leetcode.com/problems/concatenated-words)

[中文文档](/solution/0400-0499/0472.Concatenated%20Words/README.md)

## Description

<p>Given an array of strings <code>words</code> (<strong>without duplicates</strong>), return <em>all the <strong>concatenated words</strong> in the given list of</em> <code>words</code>.</p>

<p>A <strong>concatenated word</strong> is defined as a string that is comprised entirely of at least two shorter words (not necesssarily distinct)&nbsp;in the given array.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> words = [&quot;cat&quot;,&quot;cats&quot;,&quot;catsdogcats&quot;,&quot;dog&quot;,&quot;dogcatsdog&quot;,&quot;hippopotamuses&quot;,&quot;rat&quot;,&quot;ratcatdogcat&quot;]
<strong>Output:</strong> [&quot;catsdogcats&quot;,&quot;dogcatsdog&quot;,&quot;ratcatdogcat&quot;]
<strong>Explanation:</strong> &quot;catsdogcats&quot; can be concatenated by &quot;cats&quot;, &quot;dog&quot; and &quot;cats&quot;; 
&quot;dogcatsdog&quot; can be concatenated by &quot;dog&quot;, &quot;cats&quot; and &quot;dog&quot;; 
&quot;ratcatdogcat&quot; can be concatenated by &quot;rat&quot;, &quot;cat&quot;, &quot;dog&quot; and &quot;cat&quot;.</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> words = [&quot;cat&quot;,&quot;dog&quot;,&quot;catdog&quot;]
<strong>Output:</strong> [&quot;catdog&quot;]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= words.length &lt;= 10<sup>4</sup></code></li>
	<li><code>1 &lt;= words[i].length &lt;= 30</code></li>
	<li><code>words[i]</code> consists of only lowercase English letters.</li>
	<li>All the strings of <code>words</code> are <strong>unique</strong>.</li>
	<li><code>1 &lt;= sum(words[i].length) &lt;= 10<sup>5</sup></code></li>
</ul>

## Solutions

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### **Python3**

```python
class Trie:
    def __init__(self):
        self.children = [None] * 26
        self.is_end = False

    def insert(self, w):
        node = self
        for c in w:
            idx = ord(c) - ord('a')
            if node.children[idx] is None:
                node.children[idx] = Trie()
            node = node.children[idx]
        node.is_end = True


class Solution:
    def findAllConcatenatedWordsInADict(self, words: List[str]) -> List[str]:
        def dfs(w):
            if not w:
                return True
            node = trie
            for i, c in enumerate(w):
                idx = ord(c) - ord('a')
                if node.children[idx] is None:
                    return False
                node = node.children[idx]
                if node.is_end and dfs(w[i + 1 :]):
                    return True
            return False

        trie = Trie()
        ans = []
        words.sort(key=lambda x: len(x))
        for w in words:
            if dfs(w):
                ans.append(w)
            else:
                trie.insert(w)
        return ans
```

### **Java**

```java
class Trie {
    Trie[] children = new Trie[26];
    boolean isEnd;

    void insert(String w) {
        Trie node = this;
        for (char c : w.toCharArray()) {
            c -= 'a';
            if (node.children[c] == null) {
                node.children[c] = new Trie();
            }
            node = node.children[c];
        }
        node.isEnd = true;
    }
}

class Solution {
    private Trie trie = new Trie();

    public List<String> findAllConcatenatedWordsInADict(String[] words) {
        Arrays.sort(words, (a, b) -> a.length() - b.length());
        List<String> ans = new ArrayList<>();
        for (String w : words) {
            if (dfs(w)) {
                ans.add(w);
            } else {
                trie.insert(w);
            }
        }
        return ans;
    }

    private boolean dfs(String w) {
        if ("".equals(w)) {
            return true;
        }
        Trie node = trie;
        for (int i = 0; i < w.length(); ++i) {
            int idx = w.charAt(i) - 'a';
            if (node.children[idx] == null) {
                return false;
            }
            node = node.children[idx];
            if (node.isEnd && dfs(w.substring(i + 1))) {
                return true;
            }
        }
        return false;
    }
}
```

### **C++**

```cpp
class Trie {
public:
    vector<Trie*> children;
    bool isEnd;
    Trie()
        : children(26)
        , isEnd(false) {}

    void insert(string w) {
        Trie* node = this;
        for (char c : w) {
            c -= 'a';
            if (!node->children[c]) node->children[c] = new Trie();
            node = node->children[c];
        }
        node->isEnd = true;
    }
};

class Solution {
public:
    Trie* trie = new Trie();

    vector<string> findAllConcatenatedWordsInADict(vector<string>& words) {
        sort(words.begin(), words.end(), [&](const string& a, const string& b) {
            return a.size() < b.size();
        });
        vector<string> ans;
        for (auto& w : words) {
            if (dfs(w))
                ans.push_back(w);
            else
                trie->insert(w);
        }
        return ans;
    }

    bool dfs(string w) {
        if (w == "") return true;
        Trie* node = trie;
        for (int i = 0; i < w.size(); ++i) {
            int idx = w[i] - 'a';
            if (!node->children[idx]) return false;
            node = node->children[idx];
            if (node->isEnd && dfs(w.substr(i + 1))) return true;
        }
        return false;
    }
};
```

### **Go**

```go
type Trie struct {
	children [26]*Trie
	isEnd    bool
}

func newTrie() *Trie {
	return &Trie{}
}
func (this *Trie) insert(word string) {
	node := this
	for _, c := range word {
		c -= 'a'
		if node.children[c] == nil {
			node.children[c] = newTrie()
		}
		node = node.children[c]
	}
	node.isEnd = true
}

func findAllConcatenatedWordsInADict(words []string) (ans []string) {
	sort.Slice(words, func(i, j int) bool { return len(words[i]) < len(words[j]) })
	trie := newTrie()
	var dfs func(string) bool
	dfs = func(w string) bool {
		if w == "" {
			return true
		}
		node := trie
		for i, c := range w {
			c -= 'a'
			if node.children[c] == nil {
				return false
			}
			node = node.children[c]
			if node.isEnd && dfs(w[i+1:]) {
				return true
			}
		}
		return false
	}
	for _, w := range words {
		if dfs(w) {
			ans = append(ans, w)
		} else {
			trie.insert(w)
		}
	}
	return
}
```

### **...**

```

```

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