<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 Solution {
public:
    // // Apprach1: Using Brute force
    // int Solve(string s, vector<string>& dict,int i,unordered_set<string>&hash,int n){
    //      if(i==n){
    //         return 0;
    //      }
    //      int skip=1+Solve(s,dict,i+1,hash,n,dp);
    //      for(int j=i;j<n;j++){
    //         if(hash.count(s.substr(i,j+1-i))){
    //             skip=min(skip,Solve(s,dict,j+1,hash,n,dp));
    //         }
    //      }
    //      return skip;
    // }
    int Solve(string s, vector<string>& dict,int i,unordered_set<string>&hash,int n,vector<int>&dp){
         if(i==n){
            return 0;
         }
         if(dp[i]!=-1) return dp[i];
         int skip=1+Solve(s,dict,i+1,hash,n,dp);
         for(int j=i;j<n;j++){
            if(hash.count(s.substr(i,j+1-i))){
                skip=min(skip,Solve(s,dict,j+1,hash,n,dp));
            }
         }
         return dp[i]=skip;
    }
    int minExtraChar(string s, vector<string>& dict) {
        unordered_set<string> hash(dict.begin(),dict.end());
        int n=s.length();
        vector<int>dp(n+1,-1);
        return Solve(s,dict,0,hash,n,dp);
    }
};
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