<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:
    class Node {
        public:
        bool end;
        Node* next[26];

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

    class Trie {
        public:
        Node* trie;

        Trie(){
            trie = new Node();
        }

        void insert(string word){
            int n = word.size();
            Node* it = trie;
            for(int i = 0; i < n; i++){
                if(it->next[word[i] - 'a'] == NULL)
                    it->next[word[i] - 'a'] = new Node();
                it = it->next[word[i] - 'a'];
            }
            it->end = true;
        }
    };

    int util(string &s, int idx, Node* t, vector<int>& dp){
        int n = s.size();
        if(idx == n) return 0;
        if(dp[idx] != -1) return dp[idx];
        int ans = 1 + util(s, idx+1, t, dp);
        int i = idx;
        Node* it = t;
        while(i < n && it->next[s[i] - 'a'] != NULL){
            it = it->next[s[i] - 'a'];
            if(it->end) ans = min(ans, util(s, i+1, t, dp));
            i++;
        }
        return dp[idx] = ans;
    }

    int minExtraChar(string s, vector<string>& dictionary) {
        Trie* t = new Trie();

        for(string word: dictionary){
            t->insert(word);
        }
        vector<int> dp(s.size(), -1);
        return util(s, 0, t->trie, dp);
    }
};
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