<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 tries{
    public:
    char ch;
    tries*children[26];
    bool isterminal;
    tries(char ch){
        this->ch=ch;
        for(int i=0; i<26; i++){
            children[i]=NULL;
        }
        isterminal=false;
    }
};
void insert(tries*root, string str){
    if(str.size()==0){
        root->isterminal=true;
        return;
    }
    char ch=str[0];
    int index=ch-'a';
    tries*child;
    if(root->children[index]!=NULL){
        child=root->children[index];
    }
    if(root->children[index]==NULL){
        child=new tries(ch);
        root->children[index]=child;
    }
    insert(child, str.substr(1));
}
void findprefix(tries*root, string s, int& len){
    if(root->isterminal){
        return;
    }
    if(s.empty()){
        return;
    }
    int index=s[0]-'a';
    tries*child=root->children[index];
    if(child!=NULL){
        len++;
        findprefix(child, s.substr(1), len);
    }
    else{
        len=0;
        return;
    }
}
class Solution {
public:
    string replaceWords(vector<string>& dictionary, string sentence) {
        tries*root=new tries('-');
        string ans="";
        for(int i=0; i<dictionary.size(); i++){
            insert(root, dictionary[i]);
        }
        for(int i=0; i<sentence.size(); i++){
            string s="";
            while(i<sentence.size()&&sentence[i]!=' '){
                s.push_back(sentence[i]);
                i++;
            }
            string prefix="";
            int len=0;
            findprefix(root, s, len);
            if(len==0){
                ans=ans+s;
            }
            else{
                ans=ans+s.substr(0, len);
            }
            ans=ans+" ";
        }
        ans.pop_back();
        return ans;
    }
};
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