<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:
    TrieNode* children[26];
    bool isEnd = false;
};

class Solution {
public:
    vector<int> memo;
    void insert(TrieNode* root, string word){
        TrieNode* curr = root;
        for(const char& c: word){
            if(curr->children[c-'a'] == nullptr){
                curr->children[c-'a'] = new TrieNode();
            }
            curr = curr->children[c-'a'];
        }
        curr->isEnd = true;
    }
    int eChar(string s, TrieNode* root, int index){
        if(index >= s.length()){
            return 0;
        }
        if(memo[index] != -1){
            return memo[index];
        }
        int count = 1+eChar(s, root, index+1);
        int j = index;
        TrieNode* curr = root;
        while(j<s.length() && curr->children[s[j]-'a'] != nullptr){
            curr = curr->children[s[j]-'a'];
            if(curr->isEnd){
                count = min(count, eChar(s, root, j+1));
            }
            j++;
        }

        return memo[index] = count;
    }

    int minExtraChar(string s, vector<string>& dictionary) {
        TrieNode* root = new TrieNode();
        for(string s : dictionary){
            insert(root, s);
        }
        memo = vector<int>(s.length(), -1);
        return eChar(s, root, 0);
    }
};
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