<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 Node{
    public:
    vector<Node*>arr;
    bool isEnd;
    Node(){
        arr.resize(26,NULL);
        isEnd=false;
    }
};

class Trie{
    public:
    Node* root=new Node();
    void insert(string word){
        Node* cur=root;
        for(auto ch: word){
            if(!cur->arr[ch-'a'])cur->arr[ch-'a']=new Node();
            cur=cur->arr[ch-'a'];
        }
        cur->isEnd=true;
    }
    bool check(string word){
        Node* cur=root;
        for(auto ch: word){
            if(!cur->arr[ch-'a'])return false;
            cur=cur->arr[ch-'a'];
        }
        return cur->isEnd;
    }
};

class Solution {
public:
    int solve(string s, int ind, Trie* t,vector<int>&dp){
        int n=s.size(); 
        if(ind==n)return 0;
        if(dp[ind]!=-1)return dp[ind];
        int x=INT_MAX;
        for(int i=ind;i<n;i++){
            string str = s.substr(ind,i-ind+1);
            if(t->check(str))x=min(x,solve(s,i+1,t,dp));
        }
        return dp[ind]=min(x,1+solve(s,ind+1,t,dp));
    }
    int minExtraChar(string s, vector<string>& dictionary) {
        int n=s.size();
        Trie* t = new Trie();
        for(auto it: dictionary)t->insert(it);
        vector<int>dp(n+1,-1);
        return solve(s,0,t,dp);
    }
};
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