<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 dp[51];
set<string> d;
int n;
    int minExtraChar(string s, vector<string>& dictionary) {
        memset(dp, -1, sizeof(dp));
        n=s.length();
        vector<vector<string>> subS(n+1, vector<string>(n+1, ""));
        for(auto word:dictionary)d.insert(word);
        string temp="";
        for(int i=0;i<n;i++){
            temp="";
            for(int j=i;j<n;j++){
                temp += s[j];
                subS[i][j]=temp;
            }
        }
        return solve(subS, s, 0);
    }

    int solve(vector<vector<string>>& subS, string& s, int index){
        if(index>=n)return 0;
        if(dp[index]!=-1)return dp[index];
        int ans=51;
        for(int i=index;i<n;i++){
            if(d.find(subS[index][i]) != d.end()){
                ans = min(ans, solve(subS, s, i+1));
            }
            else{
                ans = min(ans, i-index+1+solve(subS, s, i+1));
            }
        }
        return dp[index]=ans;
    }
};
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