<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>

struct TrieNode {
    bool isEnd;
    vector<TrieNode*> next;

    TrieNode() {
        isEnd = false;
        next = vector<TrieNode*>(26);
    }
};

class Solution {
public:
    int minExtraChar(string s, vector<string>& dictionary) {
        int n = s.size();
        vector<int> dp(n+1, 0);
        for (int i = 0; i <= n; i++) {
            dp[i] = i;
        }

        TrieNode* trie = new TrieNode();

        // add dict
        for (string &str: dictionary) {
            TrieNode* ptr = trie;

            for (char &c: str) {
                if (ptr->next[c-'a'] == NULL) {
                    ptr->next[c-'a'] = new TrieNode();
                }
                ptr = ptr->next[c-'a'];
            }
            ptr->isEnd = true;
        }

        for (int i = 0; i < n; i++) {
            TrieNode* ptr = trie;
            for (int j = i; j < n; j++) {
                char c = s[j];
                if (ptr->next[c-'a'] == NULL) {
                    break;
                }
                ptr = ptr->next[c-'a'];
                
                if (ptr->isEnd == true) {
                    dp[j+1] = min(dp[j+1], dp[i]);
                }
            }
            dp[i+1] = min(dp[i+1], dp[i]+1);
        }

        return dp[n];
    }
};
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