<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 recur(string s, int r, set<string>& dictionary, vector<int>& dp) {
        if (s.size() <= r) {
            cout << "done" << endl;
            return 0;
        }
        if (dp[r] != -1) return dp[r];
        // cout << r << endl;
        int mini = s.size() - r;
        for (int j = r; j < s.size(); j++)
        {for (int i = j; i < s.size(); i++) {
            // cout << j << " " << i << " : " << s.substr(j, i-j + 1) << endl;
            if (dictionary.find(s.substr(j, i-j + 1)) == dictionary.end()) continue;
            cout << s.substr(j, i-j + 1) << endl;
            int x = recur(s, i+1, dictionary, dp);
            if (x != -1) {
                mini = min(mini, x+(j-r));
            }
        }}
        cout << mini << endl;
        // if (mini == INT_MAX) return -1;
        dp[r] = mini;
        return mini;
    }
    int minExtraChar(string s, vector<string>& dictionary) {
        set<string> dict_set = {dictionary.begin(), dictionary.end()};
        vector<int> dp(s.size(), -1);
        return recur(s, 0, dict_set, dp);
    }
};
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