<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>& dictionary) {
        int n = s.size();
        unordered_set<string> dict(begin(dictionary), end(dictionary));
        
        vector<vector<int>> dp(n + 1, vector<int>(n + 1, INT_MAX));

        for(int partitionLength = 0; partitionLength <= n; ++partitionLength)
            dp[n][partitionLength] = n - partitionLength;

        for(int startIndex = n-1; startIndex >= 0; --startIndex) {
            for(int partitionLength = n; partitionLength >= 0; --partitionLength) {
                int minExtra = dp[startIndex + 1][partitionLength];
                string word;
                for(int index = startIndex; index < n; ++index) {
                    word += s[index];
                    if(dict.count(word)) {
                        int newIndex = partitionLength + (index - startIndex + 1);
                        if(newIndex >= 0 && newIndex <= n) {
                            minExtra = min(minExtra, dp[index + 1][newIndex]);
                        }
                    }
                }
                dp[startIndex][partitionLength] = minExtra;
            }
        }

        return dp[0][0];
    }
};
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