I am looking into problems to learn Dynamic Programming. I think this was not a good example of a problem that needs such a thing. In my view, the proposed solution is over engineered as you literally need only a counter. If you want to improve this solution you can pass as a parameter to the trie the string and the two pointers (left, right) so you don't create a new string each time.

import java.util.Map;
import java.util.HashMap;
class TrieNode {
    Map<Character, TrieNode> children = new HashMap<>();  // Representing each character of the alphabet.
    boolean isEnd = false;  // To determine if the current TrieNode marks the end of a word.
}

class Trie { 
    TrieNode root; 

    Trie() {
        root = new TrieNode();
    }

    void insert(String word) {
        var node = root;
        for (char ch: word.toCharArray()) {
            node = node.children.computeIfAbsent(ch, c -> new TrieNode());
        }
        node.isEnd = true;
    }

        boolean searchPrefix(String needle) {
            var node = root;
            for (char ch : needle.toCharArray()) {
                if (!node.children.containsKey(ch)) {
                    return false;
                }
                node = node.children.get(ch);
            }
            return true;
        }

        boolean search(String needle) {
            var node = root;
            for (char ch : needle.toCharArray()) {
                if (!node.children.containsKey(ch)) {
                    return false;
                }
                node = node.children.get(ch);
            }
            return node.isEnd;
        }
}

class Solution {

    // Method to calculate minimum extra characters.
    public int minExtraChar(String str, String[] dictionary) {
        // ToDo: Write Your Code Here.
        var trie = new Trie();
        for (var s : dictionary) {
            trie.insert(s);
        }
        int counter = 0;
        StringBuilder tmp = new StringBuilder();
        for (int right = 0, left = 0; right < str.length(); ) {
            tmp.append(str.charAt(right));
            var s = tmp.toString();
            if (trie.search(s)) {
                tmp.setLength(0);
                right++;
            } else if (!trie.searchPrefix(s)) {
                left++;
                right++;
                counter++;
                tmp.setLength(0);
            } else {
                right++;
            }
        }
        return counter;
    }
}