<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 Solution:
#     def minExtraChar(self, s: str, dictionary: List[str]) -> int:
        
#         dp = {}
#         def dfs(i):
#             if i in dp:
#                 return dp[i]
#             for word in dictionary:
#                 length = len(word)
#                 if i + length >len(s):
#                     dp[i] = dp.get(i,0)
#                     continue
#                 if s[i:i+length] == word:
#                     dp[i] = max(dp.get(i,0),length + dfs(i+length),dfs(i+1))
#                 else:
#                     dp[i] = max(dp.get(i,0),dfs(i+1))
                
                    
#             return dp.get(i,0)
#         return len(s) - dfs(0)


class Solution:
    def minExtraChar(self, s: str, dictionary: List[str]) -> int:
        
        dp = {}
        dictionary = set(dictionary)
        def dfs(i):
            if i in dp:
                return dp[i]
            maxLength = 0
            for word in dictionary:
                length = len(word)
                if i + length >len(s):
                    continue
                if s[i:i+length] == word:
                    maxLength = max(maxLength,length + dfs(i+length))
                dp[i] = max(maxLength,dfs(i+1))
                
                    
            return dp.get(i,0)
        return len(s) - dfs(0)
444