<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>

// Time:  O((n + m) * l), l is max(len(w) for w in dictionary)
// Space: O(n + t)

// trie, dp
class Solution {
public:
    int minExtraChar(string s, vector<string>& dictionary) {
        Trie trie;
        for (const auto& w : dictionary) {
            trie.insert(w);
        }
        vector<int> dp(size(s) + 1, numeric_limits<int>::max());
        dp[0] = 0;
        for (int i = 0; i < size(s); ++i) {
            dp[i + 1] = min(dp[i + 1], dp[i] + 1);
            for (int j = i, curr = 0; j < size(s); ++j) {
                curr = trie.child(curr, s[j]);
                if (!curr) {
                    break;
                }
                if (trie.is_string(curr)) {
                    dp[j + 1] = min(dp[j + 1], dp[i]);
                }
            }
        }
        return dp.back();
    }

private:
class Trie {
    public:
        Trie() {
            create_node(); 
        }

        void insert(const string& s) {
            int curr = 0;
            for (const auto& c : s) {
                if (!nodes_[curr][c - 'a']) {
                    nodes_[curr][c - 'a'] = create_node();
                }
                curr = nodes_[curr][c - 'a'];
            }
            ++nodes_[curr].back();
        }

        int child(int curr, char c) const {
            return nodes_[curr][c - 'a'];
        }

        bool is_string(int curr) const {
            return nodes_[curr].back();
        }

    private:
        int create_node() {
            nodes_.emplace_back(26 + 1);
            return size(nodes_) - 1;
        }

        using TrieNode = vector<int>;
        vector<TrieNode> nodes_;
    };
};
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