472. Concatenated Words

Given a list of words (without duplicates), please write a program that returns all 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 in the given array.

Example:

Input: ["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".

Note:

1. The number of elements of the given array will not exceed 10,000

2. The length sum of elements in the given array will not exceed 600,000.

3. All the input string will only include lower case letters.

4. The returned elements order does not matter.

// Use Dynamic Programming from 139. Word Break
bool helper(string word, unordered_set<string>& dict) {
    if (dict.empty()) return false;
    int n = word.size();
    vector<bool> dp(n + 1, false);
    dp[0] = true;
    for (int i = 1; i <= n; ++i) {
        for (int j = 0; j < i; ++j) {
            if (dp[j] && dict.count(word.substr(j, i - j))) {
                dp[i] = true;
                break;
            }
        }
    }
    return dp.back();
}
vector<string> findAllConcatenatedWordsInADict(vector<string>& words) {
    unordered_set<string> dict;
    vector<string> res;
    sort(words.begin(), words.end(), [](string& w1, string& w2) {
        return w1.length() < w2.length();
    });
    for (string word : words) {
        if (helper(word, dict)) res.push_back(word);
        dict.insert(word);
    }
    return res;
}

// DFS
bool helper(string& word, unordered_set<string>& dict, int pos, int cnt) {
    if (pos >= word.length() && cnt >= 2) return true;
    for (int len = 1; len <= word.length() - pos; ++len) {
        string t = word.substr(pos, len);
        if (dict.count(t) && helper(word, dict, pos + len, cnt + 1))
            return true;
    }
    return false;
}
vector<string> findAllConcatenatedWordsInADict(vector<string>& words) { // time: O(n^2); space: O(n)
    vector<string> res;
    unordered_set<string> dict(words.begin(), words.end());
    for (string word : words) {
        if (word.empty()) continue;
        if (helper(word, dict, 0, 0)) res.push_back(word);
    }
    return res;
}