<p>In English, we have a concept called <strong>root</strong>, which can be followed by some other word to form another longer word - let&#39;s call this word <strong>derivative</strong>. For example, when the <strong>root</strong> <code>&quot;help&quot;</code> is followed by the word <code>&quot;ful&quot;</code>, we can form a derivative <code>&quot;helpful&quot;</code>.</p>

<p>Given a <code>dictionary</code> consisting of many <strong>roots</strong> and a <code>sentence</code> consisting of words separated by spaces, replace all the derivatives in the sentence with the <strong>root</strong> forming it. If a derivative can be replaced by more than one <strong>root</strong>, replace it with the <strong>root</strong> that has <strong>the shortest length</strong>.</p>

<p>Return <em>the <code>sentence</code></em> after the replacement.</p>

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

<pre>
<strong>Input:</strong> dictionary = [&quot;cat&quot;,&quot;bat&quot;,&quot;rat&quot;], sentence = &quot;the cattle was rattled by the battery&quot;
<strong>Output:</strong> &quot;the cat was rat by the bat&quot;
</pre>

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

<pre>
<strong>Input:</strong> dictionary = [&quot;a&quot;,&quot;b&quot;,&quot;c&quot;], sentence = &quot;aadsfasf absbs bbab cadsfafs&quot;
<strong>Output:</strong> &quot;a a b c&quot;
</pre>

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

<ul>
	<li><code>1 &lt;= dictionary.length &lt;= 1000</code></li>
	<li><code>1 &lt;= dictionary[i].length &lt;= 100</code></li>
	<li><code>dictionary[i]</code> consists of only lower-case letters.</li>
	<li><code>1 &lt;= sentence.length &lt;= 10<sup>6</sup></code></li>
	<li><code>sentence</code> consists of only lower-case letters and spaces.</li>
	<li>The number of words in <code>sentence</code> is in the range <code>[1, 1000]</code></li>
	<li>The length of each word in <code>sentence</code> is in the range <code>[1, 1000]</code></li>
	<li>Every two consecutive words in <code>sentence</code> will be separated by exactly one space.</li>
	<li><code>sentence</code> does not have leading or trailing spaces.</li>
</ul>

struct Node{
	Node* links[26];
	bool flag = false;
	bool containsKey(int ind) {
		return (links[ind] != NULL);
	}
	void put(int ind, Node* node) {
		links[ind] = node;
	}
	Node* get(int ind) {
		return links[ind];
	}
	void setEnd() {
		flag = true;
	}
	bool isEnd() {
		return flag;
	}
};
class Solution {
private:
	string dfs(Node* node, string word, int ind) {
		string currWord = "";
		for(int i = ind; i < word.size(); i++) {
			if(node->containsKey(word[i] - 'a')) {
                // cout << word[i] << " ";
				currWord += word[i];
				node = node->get(word[i] - 'a');
				if(node->isEnd()) {
                    // cout << endl << "end -> " << currWord << " " << word << endl;
					return currWord;
				}
			} else {
                break;
            }
		}
		if(node->isEnd())
			return currWord;
		return "";
	}
public:
    string replaceWords(vector<string>& dictionary, string sentence) {
        Node* root = new Node();
        for(auto &i : dictionary) {
        	Node* node = root;

        	for(auto &j : i) {
        		if(!node->containsKey(j - 'a')) {
        			node->put(j - 'a', new Node());
        		}
        		node = node->get(j - 'a');
        	}
        	node->setEnd();
        }

        vector<string> words;
        string currWord = "";
        for(auto &i : sentence){
        	if(i == ' ') {
        		words.push_back(currWord);
        		currWord = "";
        	} else {
        		currWord += i;
        	}
        }
        if(currWord.size() > 0) 
        	words.push_back(currWord);

       string ans = "";
        for(auto word : words) {
        	string currWord = word;
        	Node* node = root;
        	string temp = dfs(node, word, 0);
        	if(temp.size() != 0 && temp.size() < currWord.size()) {
        		currWord = temp;
        	}
        	ans += (currWord + " ");
        }
        ans.pop_back();
        return ans;
    }
};
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