<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>

struct Node{
    Node* children[26]{};
    bool isEnd = false;
};
class Solution {
public:
    int minExtraChar(string s, vector<string>& dictionary) {
        //动态规划，在第i个位置，最少剩下几个字符
        int n = s.size(), m = dictionary.size();
        Node* root = new Node();
        function <void(string &)> insert = [&](string &s) -> void{
            Node *node = root;
            for(int i = s.size() - 1; i >= 0; i--){
                int idx = s[i] - 'a';
                if(node -> children[idx] == nullptr) node -> children[idx] = new Node();
                node = node -> children[idx]; 
            }
            node -> isEnd = true;
        };
        for(int i = 0; i < m; i++) insert(dictionary[i]);
        int dp[n + 1];
        dp[0] = 0;
        for(int i = 0; i < n; i++){
            //不对当前字符处理，即不分割，就会多剩下一个字符
            dp[i + 1] = dp[i] + 1;
            Node* p = root;
            for(int j = i; j >= 0; j--){
                p = p -> children[s[j] - 'a'];
                if(p == nullptr)break;
                if(p -> isEnd) dp[i + 1] = min(dp[i + 1], dp[j]);
            }
        }
        return dp[n];
    }
};
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