<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>

// Node structure for Trie
struct Node {
    // Array to store links to child nodes,
    // each index represents a letter
    Node* links[26];
    // Flag indicating if the node
    // marks the end of a word
    bool flag = false;

    // Check if the node contains
    // a specific key (letter)
    bool containsKey(char ch) {
        return links[ch - 'a'] != NULL;
    }

    // Insert a new node with a specific
    // key (letter) into the Trie
    void put(char ch, Node* node) {
        links[ch - 'a'] = node;
    }

    // Get the node with a specific
    // key (letter) from the Trie
    Node* get(char ch) {
        return links[ch - 'a'];
    }

    // Set the current node
    // as the end of a word
    void setEnd() {
        flag = true;
    }

    // Check if the current node
    // marks the end of a word
    bool isEnd() {
        return flag;
    }
};

// Trie class
class Trie {
    
private:
    Node* root;

public:
    // Constructor to initialize the
    // Trie with an empty root node
    Trie() {
        root = new Node();
    }

    // Inserts a word into the Trie
    // Time Complexity O(len), where len
    // is the length of the word
    void insert(string word) {
        Node* node = root;
        for (int i = 0; i < word.length(); i++) {
            if (!node->containsKey(word[i])) {
                // Create a new node for
                // the letter if not present
                node->put(word[i], new Node());
            }
            // Move to the next node
            node = node->get(word[i]);
        }
        // Mark the end of the word
        node->setEnd();
    }

    // Returns if the word
    // is in the trie
    bool search(string word) {
        Node* node = root;
        for (int i = 0; i < word.length(); i++) {
            if (!node->containsKey(word[i])) {
                // If a letter is not found,
                // the word is not in the Trie
                return false;
            }
            // Move to the next node
            node = node->get(word[i]);
        }
        // Check if the last node
        // marks the end of a word
        return node->isEnd();
    }

    // Returns if there is any word in the
    // trie that starts with the given prefix
    bool startsWith(string prefix) {
        Node* node = root;
        for (int i = 0; i < prefix.length(); i++) {
            if (!node->containsKey(prefix[i])) {
                // If a letter is not found, there is
                // no word with the given prefix
                return false;
            }
            // Move to the next node
            node = node->get(prefix[i]);
        }
        // The prefix is found in the Trie
        return true;
    }
};

class Solution {
public:
    int minExtraChar(string s, vector<string>& dictionary) {

        Trie trie;

        for(auto word : dictionary) trie.insert(word);


        int size = s.size();
        vector<int> dp(size+1);

        for(int i = size-1; i >=0; i--){
            dp[i] = dp[i+1] + 1;
            for(int j = i; j < size; j++){
                string word = "";
                for(int k = i; k <= j; k++){
                    word.push_back(s[k]);
                }
                if(trie.search(word)){
                    dp[i] = min(dp[i], dp[j+1]);
                }
            }
        }

        return dp[0];
    }
};
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