<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 {
public:
    int minExtraChar(string s, vector<string>& dictionary) {
        string extract = createExtract(s, dictionary);

        return extract.size();
    }
private:
    unordered_map<string, string> memo;
    string createExtract(string s, vector<string>& dictionary) {
        if (memo.count(s)) return memo[s];
        string extract =  s ;
        for (auto& dic : dictionary) {
            int idx = s.find(dic);
            if (idx != string::npos) {
                string before = s.substr(0, idx);
                string after = s.substr(idx + dic.size(), s.size() - dic.size() - idx + 1);
                //cout << "Found :" << dic << endl;
                //cout << "Before:" << s.substr(0, idx) << endl;
                //cout << "After :" << s.substr(idx + dic.size(), s.size() - dic.size() - idx + 1) << endl << endl;

                before = createExtract(before, dictionary);
                after = createExtract(after, dictionary);

                if (before.size() + after.size() < extract.size() ) {
                    extract = before + after;
                }
            }
        }
        memo[s] = extract;
        return extract;
    }
};
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