<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>

class TrieNode{
    public:
        char letter;
        bool terminal;
        unordered_map<char,TrieNode*>children;
        TrieNode(char letter){
            this->letter=letter;
            this->terminal=false;
        }
        void makeTerminal(){
            this->terminal=true;
        }
        bool isTerminal(){
            return this->terminal;
        }
};
class Solution {
public:
    void makeTrie(vector<string>&dict,TrieNode* root){
        for(int i=0;i<dict.size();i++){
            string s=dict[i];
            TrieNode* curr=root;
            for(int j=0;j<s.size();j++){
                if(curr->children.find(s[j])!=curr->children.end()){
                    //already exist
                    curr=curr->children[s[j]];
                }else{
                    //create new node
                    TrieNode* newNode=new TrieNode(s[j]);
                    curr->children[s[j]]=newNode;
                    curr=newNode;
                }
            }
            //make last node as terminal node
            curr->makeTerminal();
        }
    }
    string getWord(TrieNode* root,string w){
        TrieNode* curr=root;
        string t="";
        for(int i=0;i<w.size();i++){
            if(curr->children.find(w[i])!=curr->children.end()){
                //it exist
                curr=curr->children[w[i]];
                t+=curr->letter;
                if(curr->isTerminal())return t;
            }else{
                //char does not exist
                return w;
            }
        }
        //check if it s a terminal 
        if(curr->isTerminal())return t;
        else return w;
    }
    string replaceWords(vector<string>& dictionary, string sentence) {
        TrieNode* root=new TrieNode('\0');
        //make prefix tree for all dictionary words
        makeTrie(dictionary,root);
        string ans="";
        stringstream words(sentence);
        string w;
        while(words>>w){
            //w contains word to be checked
            string replace=getWord(root,w);
            ans=ans+replace;
            ans+=" ";
        }
        ans.pop_back();
        return ans;
    }
};
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