<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:
        unordered_map<char,TrieNode*> nodeMap;
        bool isWord ;
        
        TrieNode(){
            isWord = false;
        }
};


class Solution {
public:
            
    TrieNode* root;
    int dp[51];
    
    void addWord(string word){
        TrieNode* currentNode = root;
        for(int i=0;i<word.size();i++){
            if(currentNode->nodeMap.count(word[i])==0) currentNode->nodeMap[word[i]] = new TrieNode();
            
            currentNode = currentNode->nodeMap[word[i]];
        }    
        currentNode->isWord = true;
    }
    
    
    int getMinChars(int index,string &s){
        if(index == s.size()) return 0;
        if(dp[index]!=-1) return dp[index];
        int ans = 1+getMinChars(index+1,s);
        
        TrieNode* currentNode = root;
        for(int k=index;k<s.size();k++){
            if(currentNode->nodeMap.count(s[k])==0) break;
            currentNode = currentNode->nodeMap[s[k]];
            if(currentNode->isWord) ans = min(ans,getMinChars(k+1,s));
        }
        
        dp[index] = ans;
        return ans;
    }
    
    int minExtraChar(string s, vector<string>& dictionary) {
        // s = "leetscode"
        // -> ""
        // word break , word break - 2 is also very important problem
        // -> 
        // if we found that current word == isWord
        // -> isWord -> (scode) -> getMinChar(root,scode) -> word.size(); is the maximum extra chars that are left
        // while for root-> we can check 
        // basically a trie-recursive problem this is
        // getMinChars(TrieNode* newNode,currentNode){
        // -> "leetscode"
            // int getMinChars(string word,TrieNode* startNode){
                        // if nothing matches then return ans;
                        // if the given word is not matching we can skip the char and move 
        // so if we are doing "leetscode"-> "
        // (l e e t s  -> (scode) -> 
        // }
        // 
        root = new TrieNode();
        for(string word:dictionary){
            addWord(word);
        }
        memset(dp,-1,sizeof(dp));    
        return getMinChars(0,s);
        
    }
};
102904