<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:
    def __init__(self) -> None:
        self.children: list[TrieNode] = [None] * 26
        self.is_word = False

class Trie:
    def __init__(self) -> None:
        self.root = TrieNode()
    
    def add(self, word: str) -> None:
        temp = self.root
        for ch in word:
            idx = ord(ch) - ord("a")
            if temp.children[idx] is None:
                temp.children[idx] = TrieNode()
            temp = temp.children[idx]
        temp.is_word = True

    def search(self, word: str) -> bool:
        temp = self.root
        for ch in word:
            idx = ord(ch) - ord("a")
            if temp.children[idx] is None:
                return False
            temp = temp.children[idx]
        return temp.is_word
    
    def search_till_word(self, s: str, idx: int) -> list[int]:
        res = []
        temp = self.root
        for i in range(idx, len(s)):
            idx = ord(s[i]) - ord("a")
            child = temp.children[idx]
            if child is None:
                return res
            if child.is_word:
                res.append(i + 1)
            temp = child
        return res
    
    def __str__(self) -> bool:
        res = []
        def recurse(node: TrieNode, curr: str) -> None:
            if node.is_word:
                res.append(curr)
            for i in range(len(node.children)):
                if node.children[i] is None:
                    continue
                recurse(node.children[i], curr + chr(i + ord("a")))
        recurse(self.root, "")
        return "\n".join(res) if res else ""
    
class Solution:
    def minExtraChar(self, s: str, dictionary: List[str]) -> int:
        @cache
        def recurse(index: int) -> int:
            if index == n:
                return 0
            next_indices = trie.search_till_word(s, index)
            min_extras = 1 + recurse(index + 1)
            for next_index in next_indices:
                min_extras = min(min_extras, recurse(next_index))
            return min_extras
        
        n = len(s)
        trie = Trie()
        for word in dictionary:
            trie.add(word)

        return recurse(0)
229