<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>

#define ll long long
class Solution {
public:
    string replaceWords(vector<string>& dictionary, string sentence) {
        vector<string> words;
        string cur = "";
        ll i = 0;
        while(i<sentence.length()){
            if(sentence[i]==' '){
                words.push_back(cur);
                cur = "";
                i++;
            }else{
                cur += sentence[i];
                i++;
            }
        }
        words.push_back(cur);
        vector<vector<ll > > trie;
        vector<ll> f(26);
        trie.push_back(f);
        vector<vector<ll > > valid;
        valid.push_back(f);
        ll cnt = 0;
        for(ll j=0; j<dictionary.size(); j++){
            string s = dictionary[j];
            ll n = s.length();
            ll cur = 0;
            for(ll i=0; i<n; i++){
                ll k = dictionary[j][i]-'a';
                if(trie[cur][k]!=0){
                    if(i==(n-1)){
                        valid[cur][k] = 1;
                    }
                    cur = trie[cur][k];
                    continue;
                }
                cnt++;
                trie.push_back(f);
                valid.push_back(f);
                trie[cur][k] = cnt;
                if(i==(n-1)){
                    valid[cur][k] = 1;
                }
                cur = trie[cur][k];
            }
        }
        vector<string> ans;
        for(ll j=0; j<words.size(); j++){
            string s = words[j];
            ll n = s.length();
            ll cur = 0;
            ll notp = 0;
            string include = "";
            for(ll i=0; i<n; i++){
                ll k = s[i]-'a';
                include += s[i];
                if(trie[cur][k]==0){
                    notp = 1;
                    break;
                }
                if(valid[cur][k]==1){
                    break;
                }
                cur = trie[cur][k];
            }
            if(notp){
                ans.push_back(s);
            }else{
                ans.push_back(include);
            }
        }
        string retans = "";
        for(ll i=0; i<ans.size(); i++){
            retans += ans[i] + " ";
        }
        retans.pop_back();
        return retans;


    }
};
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