<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>

from collections import defaultdict
from functools import cache

class Solution:
    def minExtraChar(self, s: str, dictionary: List[str]) -> int:
        class TrieNode:
            def __init__(self):
                self.children = defaultdict(TrieNode)
                self.isWord = False

        class Trie:
            def __init__(self):
                self.root = TrieNode()
            
            def add(self, word):
                cur = self.root
                for ch in word:
                    cur = cur.children[ch]
                cur.isWord = True
        
        n = len(s)

        trie = Trie()
        for word in dictionary:
            trie.add(word)

        # chars left over
        @cache
        def backtrack(start):
            if start == n:
                return 0
            
            res = 1+backtrack(start+1)

            cur = trie.root
            for i in range(start, n):
                ch = s[i]
                if ch not in cur.children:
                    break

                cur = cur.children[ch]
                if cur.isWord:
                    res = min(res, backtrack(i+1))
            
            return res
        
        return backtrack(0)

                




                
            

        
19200