<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.end_word = False
    self.childeren = {}
    self.length = 0

class Solution:
  def minExtraChar(self, s: str, dictionary: List[str]) -> int:
    # build trie n*c, c=n so n^2 = 2500
    # in my find for trie, i can add a list of intervals for matches with start at that index
    # sort intervals 50 log 50 
    # then doing interval traversing would be n^2 = 2500
    root = TrieNode()
    for word in dictionary:
      self.build_trie(word, root)
    intervals = []
    for i in range(len(s)):
      self.find_intervals(root, s, i, intervals)
    if not intervals:
      return len(s)
    memo = {}
    memo[len(intervals)-1] = intervals[-1][1] - intervals[-1][0] + 1 
    coverage = self.find_max_coverage(0, intervals, memo)
    test = []
    res = 0
    for i in memo:
      res = max(res, memo[i])
    return len(s)-res

  def find_max_coverage(self, index, intervals, memo):
    if index in memo:
      return memo[index]
    memo[index] = intervals[index][1] - intervals[index][0] + 1 
    for i in range(index+1, len(intervals)):
      if intervals[index][0] <= intervals[i][0] <= intervals[index][1]: 
        self.find_max_coverage(i, intervals, memo)
      else:
        best = self.find_max_coverage(i, intervals, memo) + intervals[index][1] - intervals[index][0] + 1 
        memo[index] = max(best, memo[index])
    return memo[index]

  def find_intervals(self, node, word, start_index, intervals):
    for i in range(start_index, len(word)):
      if word[i] in node.childeren:
        node = node.childeren[word[i]]
        if node.end_word:
          intervals.append([start_index, start_index+node.length])
      else:
        return

  def build_trie(self, word, node):
    for i, char in enumerate(word):
      if not char in node.childeren:
        new_node = TrieNode()
        node.childeren[char] = new_node
      node = node.childeren[char]
      if i == len(word)-1:
        node.end_word = True
        node.length = i
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