<p>You are given a <strong>0-indexed</strong> string <code>s</code> and a dictionary of words <code>dictionary</code>. You have to break <code>s</code> into one or more <strong>non-overlapping</strong> substrings such that each substring is present in <code>dictionary</code>. There may be some <strong>extra characters</strong> in <code>s</code> which are not present in any of the substrings.</p>

<p>Return <em>the <strong>minimum</strong> number of extra characters left over if you break up </em><code>s</code><em> optimally.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;leetscode&quot;, dictionary = [&quot;leet&quot;,&quot;code&quot;,&quot;leetcode&quot;]
<strong>Output:</strong> 1
<strong>Explanation:</strong> We can break s in two substrings: &quot;leet&quot; from index 0 to 3 and &quot;code&quot; from index 5 to 8. There is only 1 unused character (at index 4), so we return 1.

</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;sayhelloworld&quot;, dictionary = [&quot;hello&quot;,&quot;world&quot;]
<strong>Output:</strong> 3
<strong>Explanation:</strong> We can break s in two substrings: &quot;hello&quot; from index 3 to 7 and &quot;world&quot; from index 8 to 12. The characters at indices 0, 1, 2 are not used in any substring and thus are considered as extra characters. Hence, we return 3.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 50</code></li>
	<li><code>1 &lt;= dictionary.length &lt;= 50</code></li>
	<li><code>1 &lt;= dictionary[i].length &lt;= 50</code></li>
	<li><code>dictionary[i]</code>&nbsp;and <code>s</code> consists of only lowercase English letters</li>
	<li><code>dictionary</code> contains distinct words</li>
</ul>

static int io_opt = []() {
  std::ios::sync_with_stdio(false);
  return 0;
}();

struct TrieNode {
  unordered_map<char, TrieNode*> children;
  bool is_word = false;
};

class Solution {
 public:
  int minExtraChar(string s, vector<string>& dictionary) {
    TrieNode root;
    TrieNode* p = &root;
    for (auto& str : dictionary) {
      p = &root;
      for (char c : str) {
        if (nullptr == p->children[c]) {
          p->children[c] = new TrieNode();
        }
        p = p->children[c];
      }
      p->is_word = true;
    }
    int size = s.size();
    int j;
    vector<int> dp(size + 1, INT_MAX);
    dp[0] = 0;
    for (int i = 1; i <= size; ++i) {
      p = &root;
      for (j = i - 1; j < size; ++j) {
        if (p->children.end() == p->children.find(s[j])) break;
        p = p->children[s[j]];
        if (p->is_word) dp[j + 1] = min(dp[j + 1], dp[i - 1]);
      }
      dp[i] = min(dp[i], dp[i - 1] + 1);
    }

    return dp.back();
  }
};
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