<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) {
        auto m = build(s, dictionary);
        int n = s.size();
        vector<int> dp(n + 1, 0);
        int rs = n;
        for(int i = n-1;i>=0;i--){
            dp[i] = n - i;
            for(auto it: m[i]){
                cout <<i <<" "<<it<<endl;
                if(i +it <=n) dp[i] = min(dp[i], dp[i + it]);
            }
            for(int j = i+1;j <n;j++) dp[i] = min(dp[i], j -i + dp[j]);
            rs = min(rs, i + dp[i]);
        }
        return rs;
    }

    map<int, set<int>> build(string s, vector<string> &a){
        map<int, set<int>> m;
        for(int i = 0 ; i<s.size();i++){
            for(string str: a){
                if(i + str.size() > s.size()) continue;
                if(s.substr(i, str.size()) == str) m[i].insert(str.size());
            }
        }
        return m;
    }
};
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