<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 minExtraChar(string s, vector<string>& dic ) {
        vector<int> dp( s.size()+1 , INT_MAX ) ;
        dp[0] = 0  ; 
        unordered_map<string,int> mp ; 
        for( auto x : dic ){
            mp[x] = 1 ; 
        }
        for( int i=0 ; i<s.size() ; i++ ){
            string temp = "" ;
            dp[i+1] = dp[i]+1 ;
            for( int j=i ; j>=0 ; j-- ){
                temp = s[j] + temp ;
                if( mp.count(temp)==1 ){
                    dp[i+1] = min( dp[i+1] , dp[j] ) ;
                }
            }
        }

        return dp[s.size()] ;
    }
};

//class Solution {
// public:
//     map<string,int> m ;

//     int solve( string & s , int i , vector<int> & dp ){
//         if(i>=s.size()){
//             return 0 ;
//         }
//         if( dp[i]!=-1 ){
//             return dp[i] ;
//         }
//         int ans = solve( s , i+1 , dp ) ;
//         string match = "" ;
//         for( int j=i ; j<s.size() ; j++ ){
//             match = match + s[j] ;
//             if(m[match]==1){
//                 int temp = match.size() + solve(s , j+1, dp) ;
//                 ans = max( ans ,temp ) ;
//             }
//         }
//         dp[i] = ans ;
//         return ans ;
//     }
//     int minExtraChar(string s, vector<string>& dictionary) {
//         int n = s.size() ;
//         vector<int> dp( n , -1 ) ;
//         for( auto x : dictionary ){
//             m[x] = 1 ;
//         }
//         return n-solve( s , 0 , dp ) ;
//     }
// };
89000