<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 TrieNode {
public:
    unordered_map<char, TrieNode*> children;
    bool is_word;
    TrieNode() : is_word(false) {}
};
class Solution {
public:
    int minExtraChar(string s, vector<string>& dictionary) {
        unordered_map<int, int> dp;
        TrieNode* root = new TrieNode();
        for (string &s : dictionary) {
            TrieNode* curr = root;
            for (char &c: s) {
                if (!curr->children.count(c)) {
                    curr->children[c] = new TrieNode();
                }
                curr = curr->children[c];
            }
            curr->is_word = true;
        }
        int n = s.size();
        function<int(int)> dfs = [&] (int start) -> int{
            if (start == n) return 0;
            if (dp.count(start)) return dp[start];
            int ans = dfs(start + 1) + 1;
            TrieNode* curr = root;
            for (int i = start; i < n; i++) {
                if (!curr->children.count(s[i])) break;
                curr = curr->children[s[i]];
                if (curr->is_word) {
                    ans = min(ans, dfs(i + 1));
                }
            }
            return dp[start] = ans;
        };
        return dfs(0);
    }
};
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