Bloomberg SSE screen round

1. Introductions (10 mins)

2. Tell me about the most difficult production issue you faced and how you solved it. (10 mins)

3. validate binary search tree -> done with coding in 5 mins

4. given a dictionary of words, find all words having matching prefix

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
class Dictionary {
TrieNode root;
public Dictionary() {
    this.root = new TrieNode('\0');
}

public void add(String word) {
    TrieNode current = root;
    for (int i = 0; i < word.length(); i++) {
        Character key = word.charAt(i);
        if (root.next.containsKey(key)) {
            current = root.next.get(key);
        } else {
            TrieNode newNode = new TrieNode(key);
            root.next.put(key, newNode); // mistake -> should have been current
            current = newNode;
        }
    }
    current.isEnd = true;
    current.word = word;
}

public List<String> queryPrefix(String prefix) {
    List<String> res = new ArrayList<>();
    TrieNode current = root;
    for (int i = 0; i < prefix.length(); i++) {
        Character key = prefix.charAt(i);
        if (root.next.containsKey(key)) {
            current = root.next.get(key); // mistake -> should have been current
        }
        if (current.isEnd) {
            res.add(current.word);
        }
    }

    dfs(current, res);
    return res;
}

public void dfs(TrieNode root, List<String> res) {
    if (root == null)
        return;

    if (root.isEnd) {
        res.add(root.word);
    }

    for (Map.Entry<Character, TrieNode> entry : root.next.entrySet()) {
        dfs(entry.getValue(), res);
    }
}

}
public class Solution {
public static void main(String[] args) {
Dictionary dict = new Dictionary();
dict.add("abcdef");
dict.add("abcijk");
System.out.println(dict.queryPrefix("abc"));
}
}
// represents a dictionary
// method 1 -> add a word to dictionary (a-z)
// method 2 -> query method list of all the words that match with the given prefix
class TrieNode {
Character c;
HashMap<Character, TrieNode> next;
String word;
boolean isEnd;
public TrieNode(Character c) {
    this.c = c;
    this.next = new HashMap&lt;&gt;();
    this.isEnd = false;
}

}

Verdict : Rejected

Is the job market so brutal that even 2 lines of code cant be incorrect ?