<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 Trie
{
public:
    Trie(char c): c(c), isWord(false)
    {
        children.reserve(26);
        for(int i = 0; i < 26; i++) children.push_back(nullptr);
    }
    
    char c;
    bool isWord;
    vector<Trie*> children;
};

class Solution {
public:

void insertWord(const string& word, Trie* root)
{
    if(!root) return;
    Trie* cur = root;
    for(char c : word)
    {
        if(cur->children[c - 'a'])
        {
            cur = cur->children[c - 'a'];
        }
        else
        {
            Trie* newNode = new Trie(c);
            cur->children[c - 'a'] = newNode;
            cur = cur->children[c - 'a'];
        }
    }
    cur->isWord = true;  
}

Trie* buildTrie(const vector<string>& dict)
{
    Trie* root = new Trie('#');
    
    for(int i = 0; i < dict.size(); i++)
    {
        insertWord(dict[i], root);
    }
    
    return root;
}

int minExtraChar(string str, vector<string> dict) 
{
    if(dict.size() == 0) return str.size();
    
    int n = str.length();
    Trie* root = buildTrie(dict);
    unordered_map<int, int> memo;

    function<int(int)> dp = [&](int start) 
    {
        if (start == n) return 0;
        if (memo.count(start)) return memo[start];

        int ans = dp(start + 1) + 1;
        Trie* node = root;
        for (int end = start; end < n; end++) 
        {
            char c = str[end];
            if (!node->children[c - 'a']) break;
            
            node = node->children[c - 'a'];
            if (node->isWord) ans = min(ans, dp(end + 1));
        }
        return memo[start] = ans;
    };

    return dp(0);
}
};
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