<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>

struct Node{
    Node* links[26];
    bool flag = false;

    bool containsKey(char ch){
        return (links[ch - 'a'] != nullptr);
    }

    void put(char ch, Node* node){
        links[ch - 'a'] = node;
    }

    Node* get(char ch){
        return links[ch - 'a'];
    }

    bool isEnd(){
        return flag;
    }

    void setflag(){
        flag = true;
    }
};

class Trie {
public:
    Node* root;
    Trie() {
        root = new Node();
    }
    
    void insert(string word) {
        Node* node = root;
        for(int i=0;i<word.size();i++){
            if(!node->containsKey(word[i])){
                node->put(word[i], new Node());
            }
            node = node->get(word[i]);
        }
        node->setflag();
    }
   
    int search(string word) {
        Node* node = root;
        int i;
        for(i = 0;i<word.size();i++){
            if(!node->containsKey(word[i])){
                return 0;
            }
            node = node->get(word[i]);
        }
        return node->isEnd();
    }
};

class Solution {
public:
    Trie trie;
    vector<int> dp;

    int backtrack(int i, string& s, int n) {
        if (i == n) {
            return 0;  
        }

        if (dp[i] != -1) {
            return dp[i];
        }
        
        int take = INT_MAX;
        for (int j = i; j < n; j++) {
            if (trie.search(s.substr(i, j - i + 1))) { 
                take = min(take, backtrack(j + 1, s, n));
            }
        }

        int nottake = 1 + backtrack(i + 1, s, n);
    
        return dp[i] = min(take, nottake);
    }

    int minExtraChar(string s, vector<string>& dictionary) {
        for (auto it : dictionary) {
            trie.insert(it);
        }
        
        int n = s.size();
        dp.resize(n, -1);  
        
        return backtrack(0, s, n);
    }
};

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