<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) : 
        self.children = {}
        self.isWord = False 
class Trie : 
    def __init__(self) : 
        self.root = TrieNode() 
    def addWord(self,word) : 
        node = self.root
        for letter in word : 
            if letter not in node.children : 
                node.children[letter] = TrieNode() 
            node = node.children[letter]
        node.isWord = True 
    def findWord(self,word) : 
        node = self.root 
        for letter in word : 
            if letter not in node.children :
                return False
            node = node.children[letter]
        return node.isWord

class Solution:
    def minExtraChar(self, s: str, dictionary: List[str]) -> int:
        
        t = Trie() 

        for word in dictionary : 
            t.addWord(word)
        dp = {len(s) : 0} 
        def dfs(i) : 
            nonlocal t 
            if i in dp :
                return dp[i] 
            res = 1 + dfs(i+1)
            curr = t.root
            for j in range(i,len(s)) : 
                if s[j] not in curr.children :
                    break
                curr = curr.children[s[j]]

                if curr.isWord : 
                    res = min(res, dfs(j+1))
            dp[i] = res
            return res
        return dfs(0)
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