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472. Concatenated Words

DifficultyHard

Description

Given an array of strings words (without duplicates), return all the concatenated words in the given list of words.

A concatenated word is defined as a string that is comprised entirely of at least two shorter words (not necessarily distinct) in the given array.

 

Example 1:

Input: words = ["cat","cats","catsdogcats","dog","dogcatsdog","hippopotamuses","rat","ratcatdogcat"]
Output: ["catsdogcats","dogcatsdog","ratcatdogcat"]
Explanation: "catsdogcats" can be concatenated by "cats", "dog" and "cats"; 
"dogcatsdog" can be concatenated by "dog", "cats" and "dog"; 
"ratcatdogcat" can be concatenated by "rat", "cat", "dog" and "cat".

Example 2:

Input: words = ["cat","dog","catdog"]
Output: ["catdog"]

 

Constraints:

  • 1 <= words.length <= 104
  • 1 <= words[i].length <= 30
  • words[i] consists of only lowercase English letters.
  • All the strings of words are unique.
  • 1 <= sum(words[i].length) <= 105

Solutions

Solution 1

Thinking

A concatenated word is at least two shorter dictionary words. Trying every split and a set lookup grows with both \(n\) and length.

Sort by length and insert shorter words into a trie first. DFS the current word on the trie: at an end-of-word node recurse on the suffix; if the whole word splits, it is concatenated, otherwise insert it.

Concatenated words stay out of the trie, since a longer word can always fall back to atomic pieces. The empty suffix is the success base case.

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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

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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;
    }
}

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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;
    }
};

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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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