<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 Trie{
    struct TrieNode{
        TrieNode* child[26];
        bool isEnd;
    };
    TrieNode* root;
public:
    Trie(){
        root=new TrieNode();
        root->isEnd=false;
    }
    void insert(string s)
    {
        TrieNode* curr=root;
        for(char c:s)
        {
            if(curr->child[c-'a']==NULL){
                curr->child[c-'a']=new TrieNode();
            }
            curr=curr->child[c-'a'];
        }
        curr->isEnd=true;
    }
    bool find(string s){
        TrieNode* curr=root;
        for(char c:s)
        {
            if(curr->child[c-'a']==NULL)
                return false;
            curr=curr->child[c-'a'];
        }
        return curr->isEnd;
        
    }
    bool startsWith(string s)
    {
        TrieNode* curr=root;
        for(char c:s)
        {
            if(curr->child[c-'a']==NULL)
                return false;
            curr=curr->child[c-'a'];
        }
        return true;
    }
};
class Solution {
public:
    int findans(string s,int idx,vector<int> &dp,Trie &trie)
    {
        if(idx>=s.length())
            return 0;
        if(dp[idx]!=-1)
            return dp[idx];
        cout<<idx<<endl;
        int nottake=1+findans(s,idx+1,dp,trie);
        int take=INT_MAX;
        for(int i=idx+1;i<=s.length();i++)
        {
            string temp=s.substr(idx,i-idx);
            if(trie.startsWith(temp)==false)
                break;
            if(trie.find(temp))
            {
                //cout<<temp<<endl;
                take=min(take,findans(s,i,dp,trie));
            }
        }
        //cout<<idx<<" "<<take<<" "<<nottake<<endl;
        return dp[idx]=min(take,nottake);
    }
    int minExtraChar(string s, vector<string>& dictionary) {
        Trie trie;
        for(string str:dictionary)
        {
            trie.insert(str);
        }
        vector<int> dp(s.length(),-1);
        return findans(s,0,dp,trie);
    }
};
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