<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:

    int solve(int index ,  string  &  str ,  map<string , int>&mp , vector<int>&dp){
        if(index >= str.length()){
            return  0 ;
        }
        if(dp[index] != -1){
            return  dp[index] ;
        }
        string  temp  = "" ;
        int  ans = INT_MAX ;
        for (int i =  index ; i<str.length()  ;  i++){
            temp += str[i] ;
            if(mp[temp] == 1){
                int a =  solve(i+1 ,  str ,  mp ,  dp  ) ;
                ans =  min(ans ,  a ) ;
            }
            int d =  temp.length() +  solve(i +1 ,  str ,  mp ,dp ) ;
            ans = min(ans  ,  d) ;
        }
        return dp[index] =   ans ;
    }
    int minExtraChar(string s, vector<string>& dictionary) {
        map<string ,  int>mp ;
        for (auto  it :  dictionary){
            mp[it] = 1 ;
        }
        vector<int>dp(s.length() , -1) ;
        int ans  =  solve(0 ,  s ,  mp , dp) ;
        return  ans ;
    }
};
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