<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.isEnd = False

    def add_word(self, word):
        curr = self
        for c in word: 
            if c not in curr.children: 
                curr.children[c] = TrieNode()
            curr = curr.children[c]
        curr.isEnd = True

    def check_word(self, word):
        curr = self
        for c in word: 
            if c not in curr.children: 
                return False
            curr = curr.children[c]
        
        return curr.isEnd 

class Solution:
    def minExtraChar(self, s: str, dictionary: List[str]) -> int:
        """
        s should be divided into pieces 
        and you dont need all char covered 
        最小化number of extra characters left

        如果从后往前考虑切分 是不是会更
        leet code leetcode sc

        i think there is some dfs/dp aspect in here  
        """
        root = TrieNode()
        n = len(s)
        for word in dictionary: 
            root.add_word(word)
        
        cache = [0] * len(s)

        def dfs(i):
            # use this as the maximum number 
            # this substring from i to end 
            # could cover using the trie 
            if i == len(s):
                return 0 
            if cache[i] != 0: 
                return cache[i] 
            
            max_cover = 0 
            for k in range(i, n):
                for z in range(k+1, n+1):
                    substr = s[k:z]
                    # print(substr)
                    if root.check_word(substr):
                        max_cover = max(max_cover, len(substr) + dfs(z))
            
            cache[i] = max_cover
            return max_cover
        
        return len(s) - dfs(0)
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