<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 = [None]*26
    self.isLeaf = False
  
  def addWord(self, w):
    r = self
    for ch in w:
      idx = ord(ch)-97
      if not r.children[idx]:
        r.children[idx] = TrieNode()
      r = r.children[idx]
    r.isLeaf = True
  
  def prefixLengthMatch(self, w):
    r = self
    c = 0
    res = []
    for ch in w:
      idx = ord(ch)-97
      if not r.children[idx]:
        return res
      r = r.children[idx]

      c += 1
      if r.isLeaf:
        res.append(c)
    return res

class Solution:
    def minExtraChar(self, s: str, dictionary: List[str]) -> int:
      root = TrieNode()
      for w in dictionary:
        root.addWord(w)
      memo = {}
      def rec(s):
        if s in memo:
          return memo[s]
        if not s:
          return 0
        n = len(s)
        r = n
        for i in range(n):
          t = root.prefixLengthMatch(s[i:])
          #print("\t",s[i:], t)
          f = False
          if t:
            if t[-1] == n-i:
              f = True
              r = min(r, i)
            else:
              for j in t:
                r = min(r, i+rec(s[i+j:]))
          if not f:
            r = min(r, i+1+rec(s[i+1:]))
        #print("res", s,r)
        memo[s] = r
        return r
      
      return rec(s)

        
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