<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 ch;
        bool isTerminal;
        TrieNode* children[26];

        TrieNode(char ch) : ch(ch),isTerminal(false){
            for(int i=0 ; i < 26 ; ++i){
                children[i] = NULL;
            }
        } 

};


class Trie {
public:

    TrieNode* root;
  
    Trie() {
        root = new TrieNode('-');
    }

    void insertUtil(TrieNode* root,string word,int i){
        if(i >= word.length()){
            root->isTerminal = true;
            return;
        }

        int index = word[i] - 'a';

        if(!root->children[index]){
            root->children[index] = new TrieNode(word[i]);
        }

        insertUtil(root->children[index],word,i+1);
    }

    bool searchUtil(TrieNode* root,string word,int &i){

        if(root->isTerminal) return true;

        if(i>= word.length()){
            return root->isTerminal;
        }

        int index = word[i++] - 'a';

        if(root->children[index]){
            return searchUtil(root->children[index],word,i);
        }

        return false;

    }

    bool startsWithUtil(TrieNode* root,string word,int i){
        if(i>=word.length()){
            return true;
        }

        int index = word[i] - 'a';
        if(root->children[index]){
            return startsWithUtil(root->children[index],word,i+1);
        }

        return false;
    }
    
    void insert(string word) {
        insertUtil(root,word,0);
    }

    
    int search(string word) {
        int i = 0;
        bool flag = searchUtil(root,word,i);
        if(flag == false) return -1;
        return i;
    }
    
    bool startsWith(string prefix) {
        return startsWithUtil(root,prefix,0);
    }
};

/**
 * Your Trie object will be instantiated and called as such:
 * Trie* obj = new Trie();
 * obj->insert(word);
 * bool param_2 = obj->search(word);
 * bool param_3 = obj->startsWith(prefix);
 */




class Solution {
public:
    string replaceWords(vector<string>& dictionary, string sentence) {
       string ans = "";
       int i = 0 , j = 0 ;

       int n = sentence.length();

       Trie tr;
       for(auto i : dictionary) tr.insert(i);

       while(j < n ){
         if(sentence[j] == ' ' || j == n-1){
            int len = j == n-1 ? j-i+1 : j-i;
            string word = sentence.substr(i,len);
            int trieIndex = tr.search(word);
            ans += trieIndex == -1 ? word : word.substr(0,trieIndex);
            if(sentence[j] == ' ') ans+=' ';
            i = j + 1;

         }
         ++j;
       }

       return ans;
        
    }
};
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