<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{
        char val;
        unordered_map<char, Node*> m;
        bool end = false;
    };
    void insert(Node* root, string s, int idx = 0){
        if(idx == s.size())
        {
            root -> end = true;
            return;
        }
        if((root -> m).find(s[idx]) == (root -> m).end())
        {
            Node* next_node = new Node();
            next_node -> val = s[idx];
            (root -> m)[s[idx]] = next_node;
        }
        insert((root -> m)[s[idx]], s, idx + 1);
        return;
    }
    bool check(Node* root, string s, int idx = 0){
        if(idx == s.size())
            return (root -> end)? true : false;
        if((root -> m).find(s[idx]) == (root -> m).end())
            return false;
        return check((root -> m)[s[idx]], s, idx + 1);
    }

    int minExtraChar(string s, vector<string>& dictionary) {
        Node* trie = new Node();
        for(auto w: dictionary)
            insert(trie, w);
        vector<int> dp(s.size() + 1, INT_MAX);
        dp.back() = 0;
        for(int i = s.size() - 1; i > -1; i--)
        {
            dp[i] = dp[i + 1] + 1;
            for(int j = i + 1; j <= s.size(); j++)
                if(check(trie, s.substr(i, j - i)))
                    dp[i] = min(dp[i], dp[j]);
        }
        return dp[0];
    }
};
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