<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>

class Solution {
public:
    struct Node{
        Node* son[26];
        bool is_end;

        Node() {
            for (int i = 0; i < 26; i ++)
                son[i] = NULL;
            is_end = false;
        }

        void insert(string word) {
            auto p = this;
            for (auto c: word) {
                int u = c - 'a';
                if (!p->son[u]) p->son[u] = new Node();
                p = p->son[u];
            }
            p->is_end = true;
        }

        bool search(string word) {
            auto p = this;
            for (auto c: word) {
                int u = c - 'a';
                if (!p->son[u]) return false;
                p = p->son[u];
            }
            return p->is_end;
        }
    }*root;

    int minExtraChar(string s, vector<string>& dictionary) {
        // 将字典中的每一个单词存入字典树
        root = new Node();
        for (auto word: dictionary) root->insert(word);

        // 动态规划
        int n = s.size();
        vector<int> dp(n + 1, INT_MAX);
        dp[0] = 0;
        s = " " + s; // 将s改成1-based
        for (int i = 1; i <= n; i ++) {
            dp[i] = min(dp[i], dp[i - 1] + 1);
            for (int j = 1; j <= i; j ++) 
                if (root->search(s.substr(j, i - j + 1)))
                    dp[i] = min(dp[i], dp[j - 1]);
        }
        return dp[n];
    }
};

/*
字典树Trie + 动态规划DP

f[i]: 前i个字符的最少额外字符数（1 <= i <= n)
求f[n].

状态转移：
1. 第i个字符是额外字符：f[i] = f[i - 1] + 1
2. 在[1, i]范围中找到一个j，使得子串s[j ... i]属于字典树，则f[i] = min(f[i], f[j - 1])

初始化：f[0] = 0, f[others] = INT_MAX
*/
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