<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) {
        map<string,bool>vis;
        for(int i=0;i<dictionary.size();i++){
            vis[dictionary[i]]=1;
        }
    vector<pair<int,int>>vect;
    map<pair<int,int>,int>dp;
    for(int i=0;i<s.length();i++){
        for(int j=i;j<s.length();j++){
            if(vis[s.substr(i,j-i+1)]==1){
                vect.push_back(make_pair(i,j));
                dp[make_pair(i,j)]=j-i+1;
            }
        }
    }
    
    for(int i=vect.size()-2;i>=0;i--){
        int ans=1e9;
        for(int j=i+1;j<vect.size();j++){
            if(ans<vect[j].first)break;
            if(vect[i].second<vect[j].first){
               
                dp[make_pair(vect[i].first,vect[i].second)]=max(dp[make_pair(vect[i].first,vect[i].second)],dp[make_pair(vect[j].first,vect[j].second)]+vect[i].second-vect[i].first+1);
ans=min(ans,vect[j].second);
            }
        }
    }
int ans1=0;
for(auto v:dp){
    ans1=max(ans1,v.second);
}
return s.length()-ans1;



    
    
    
    
    
    
    }
};
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