<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 trie {
private:
  static const int MAX_CHARS = 26;
  bool isLeaf;
  trie *child[MAX_CHARS];

public:
  trie() {
    memset(child, 0, sizeof(child));
    isLeaf = false;
  }
  void insert(const string &str, int i = 0) {
    for (trie *cur = this; cur; i++) {
      if (str[i] == '\0') {
        cur->isLeaf = true;
        break;
      }
      int pos = str[i] - 'a';
      if (!cur->child[pos])
        cur->child[pos] = new trie();
      cur = cur->child[pos];
    }
  }

  bool word_exist(const string &str, int i = 0) {
    for (trie *cur = this; cur; i++) {
      if (str[i] == '\0')
        return cur->isLeaf;

      int pos = str[i] - 'a';
      if (!cur->child[pos])
        return false;
      cur = cur->child[pos];
    }
    return true;
  }

  bool prefix_exist(const string &str, int i = 0) {
    for (trie *cur = this; cur; i++) {
      if (str[i] == '\0')
        return true;

      int pos = str[i] - 'a';
      if (!cur->child[pos])
        return false;
      cur = cur->child[pos];
    }
    return true;
  }

  string first_word_prefix(const string &str, int i = 0) {
    for (int i = 1; i <= str.size(); ++i)
      if (word_exist(str.substr(0, i)))
        return str.substr(0, i);

    return str;
  }
};

class Solution {
public:
  string replaceWords(vector<string>& dictionary, string sentence) {
    string ans;
    trie root;
    for (int i = 0; i < dictionary.size(); ++i)
      root.insert(dictionary[i]);

    int pos = find(sentence.begin(), sentence.end(), ' ') - sentence.begin();
    ans += root.first_word_prefix(sentence.substr(0, pos));

    for (int i = pos + 1; i < sentence.size();) {
      pos = find(sentence.begin() + i, sentence.end(), ' ') - sentence.begin();
      ans += ' ';
      ans += root.first_word_prefix(sentence.substr(i, pos - i));
      i = pos + 1;
    }
    return ans;
  }
};
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