<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:
        
        """
            similar to what we do in word break
            starting from i=0, we check if we can build any of word in dict
            if not, count ++, move on to the next char

            if yes, we recurse on the rest of the string
        """

        dp = [0]*(len(s)+1)
        
        for i in range(len(s)-1, -1, -1):
            min_extra = 1 + dp[i+1] # case: skipping over a char
            for word in dictionary:
                offset = len(word)
                if (i+offset <= len(s) and word == s[i:i+offset]):
                    min_extra = min(min_extra, dp[i+offset]) # case: found a matching dictionary word
            dp[i] = min_extra
        
        return dp[0]


        # memo = [None]*(len(s)+1)
        # # memo[len(s)] = 0

        # def recursion(startIndex):
        #     if startIndex == len(s):
        #         return 0 

        #     if memo[startIndex]:
        #         return memo[startIndex]

        #     min_extra = 1 + recursion(startIndex+1)

        #     for word in dictionary:
        #         offset = len(word)
        #         if word == s[startIndex:startIndex+offset]:
        #             min_extra = min(min_extra, recursion(startIndex+offset))

        #     memo[startIndex] = min_extra

        #     return min_extra
            
        #     # if not match:
        #     #     # print("no match count", count)

        # return recursion(0)
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