<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(string& s, unordered_map<string, int>&mp, vector<int>&dp, int index)
    {
        if (index >= s.size()) return 0;
        if (dp[index] != -1) return dp[index]; //use the stored results
        
        string currStr = "";
        int minExtra = s.size();
        for (int cutIdx = index; cutIdx < s.size(); cutIdx++)
        {
            currStr.push_back(s[cutIdx]);
            //currStr will be a string from (index to cutIdx)
            
            //if string not in dictionary, we have to delete as they are extra chars
            int currExtra = (mp.count(currStr))? 0 : currStr.size();
            int nextExtra = solve(s, mp, dp, cutIdx + 1);
            int totalExtra = currExtra + nextExtra;
            
            minExtra = min(minExtra, totalExtra);
        }
        return dp[index] = minExtra;
    }
    int minExtraChar(string s, vector<string>& dictionary) 
    {
        vector<int>dp(s.size(), -1);
        unordered_map<string, int>mp;
        for (string& word : dictionary) mp[word]++;
        
        int ans = solve(s, mp, dp, 0);
        return ans;
        
    }
};
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