<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 TrieNode{
   public:
    bool iscompleteword;
    int count;
    TrieNode* children[26];
    TrieNode(){
        iscompleteword=false;
        count=0;
        memset(children,0,sizeof(children));
    }
};
class Trie{
    public:
     TrieNode* root;
     Trie(){
        root=new TrieNode();
     }
     void insert(string word){
        TrieNode* node=root;
        for(char ch:word){
            int idx=ch-'a';
            if(node->children[idx]==NULL){
                node->children[idx]=new TrieNode();
            }
            node=node->children[idx];
        }
        node->iscompleteword=true;
     }
};

class Solution {
public:

    int minExtraChar(string s, vector<string>& d) {
        Trie* t=new Trie();
        vector<int>dp(s.size(),-1);
        for(int j=0;j<d.size();j++){
            t->insert(d[j]);
        }
        function<int(int)>helper=[&](int ind)->int{
            if(ind==s.size())
            return 0;
            if(dp[ind]!=-1){
                return dp[ind];
            }
            TrieNode* node=t->root;
            char ch=s[ind];
            int ans=INT_MAX/2;
            
            for(int j=ind;j<s.size();j++){
                ch=s[j];
                if(node!=NULL)
                node=node->children[ch-'a'];
                if(node!=NULL&&node->iscompleteword){
                    ans=min(ans,helper(j+1));
                }
                else{
                    ans=min(ans,j-ind+1+helper(j+1));
                }
            }
            return dp[ind]=ans;
        };
        return helper(0);
    }
};
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