The current C++ suggested solution uses Greedy algorithm with Trie structure but that doesn't cover all cases.

For example, when the input string s is "ecolloycollotkvzqpdaumuqgs" and the dictionary is ["flbri","uaaz","numy","laper","ioqyt","tkvz","ndjb","gmg","gdpbo","x","collo","vuh","qhozp","iwk","paqgn","m","mhx","jgren","qqshd","qr","qpdau","oeeuq","c","qkot","uxqvx","lhgid","vchsk","drqx","keaua","yaru","mla","shz","lby","vdxlv","xyai","lxtgl","inz","brhi","iukt","f","lbjou","vb","sz","ilkra","izwk","muqgs","gom","je"], the current Greedy solution returns 14 instead of 2.

This is because, in Greedy approach, the mapping characters are "c" "c" "tkvz", "qpdau" and "m".

However, this is not the maximum mapping character case: "collo", "collo", "tkvz", "qpdau" "muqgs" which left only 2 unmapped characters ("e" and "y")

class TrieNode {
public:
    TrieNode() : children(26, nullptr), isWord(false) {}

    vector<TrieNode*> children;
    bool isWord;
};

class Solution {
public:
    int minExtraChar(string s, vector<string>& dictionary) {
        int n = s.length();
        vector<int> dp(n + 1, 0);
        TrieNode* root = buildTrie(dictionary);
        
        for(int start = n - 1; start >= 0; start--)
        {
            TrieNode* node = root;
            
            // base case: skip the current letter and check from the next letter
            dp[start] = dp[start + 1] + 1;

            // check all substrings that starts with the current letter
            for(int end = start; end < n; end++)
            {
                if (!node->children[s[end] - 'a'])
                {
                    break;
                }

                node = node->children[s[end] - 'a'];

                if (node->isWord)
                {
                    dp[start] = min(dp[start], dp[end + 1]);
                }
            }
        }

        return dp[0];
    }

private:
    // helper function to convert the given dictionary to trie form
    TrieNode* buildTrie(vector<string>& dictionary)
    {
        TrieNode* root = new TrieNode();

        for(string& word : dictionary)
        {
            TrieNode* node = root;
            for(char& c : word)
            {
                if (!node->children[c - 'a'])
                {
                    node->children[c - 'a'] = new TrieNode();
                }

                node = node->children[c - 'a'];
            }

            node->isWord = true;
        }

        return root;
    }
};