Longest Word in Dictionary Using Trie in DSA Java - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is a Trie data structure?

A Trie is a tree-like data structure that stores a dynamic set of strings, where each node represents a character. It is used for efficient retrieval of words, especially prefixes.

intermediate

How does a Trie help in finding the longest word in a dictionary where all prefixes are also words?

By inserting all words into the Trie, we can traverse nodes only if the node marks the end of a valid word. This ensures all prefixes exist. The longest path following this rule gives the longest word.

beginner

In the context of the longest word problem, why do we check if a Trie node is an end of a word before continuing traversal?

Because the problem requires that all prefixes of the word must be valid words. Checking the end-of-word flag ensures the prefix exists in the dictionary.

beginner

What is the time complexity of inserting a word of length n into a Trie?

The time complexity is O(n), where n is the length of the word, because we insert one character per level in the Trie.

intermediate

What is the main advantage of using a Trie over sorting when finding the longest word with all prefixes present?

Trie allows checking prefix existence efficiently during traversal without repeated string operations, making it faster and more memory efficient for large dictionaries.

What does each node in a Trie typically represent?

A An integer value

B A whole word

C A number of children

D A character

Each node in a Trie represents a single character of a word.

Why do we only continue traversal in the Trie if the current node marks the end of a word?

A To sort the words alphabetically

B To ensure all prefixes are valid words

C To count the number of children

D To save memory

This ensures that every prefix of the word exists in the dictionary.

What is the main goal of the 'Longest Word in Dictionary Using Trie' problem?

A Count the total number of words

B Find the shortest word in the dictionary

C Find the longest word where all prefixes are also words

D Sort words by length

The problem asks for the longest word such that every prefix is also a valid word.

What is the time complexity to insert all words into a Trie if total characters are M?

A O(M)

B O(M log M)

C O(M^2)

D O(log M)

Inserting all words takes O(M) time where M is the total number of characters across all words.

Which traversal method is commonly used to find the longest word in a Trie with prefix constraints?

A Depth-First Search (DFS)

B Breadth-First Search (BFS)

C Level Order Traversal

D Inorder Traversal

DFS is used to explore all paths and find the longest valid word.

Explain how a Trie can be used to find the longest word in a dictionary where all prefixes are also words.

Think about how prefix validation works with Trie nodes.

Describe the steps to insert a word into a Trie and how to check if a prefix exists.

Focus on node creation and marking end of words.

Practice

1. What is the main purpose of using a Trie data structure when finding the longest word in a dictionary?

easy

A. To efficiently store and search words by their prefixes

B. To sort words alphabetically

C. To count the total number of words

D. To reverse the words quickly

Solution

1. Step 1: Understand Trie usage

   A Trie stores words by their prefixes, allowing quick prefix checks.

2. Step 2: Relate to longest word search

   Finding the longest word with all prefixes present is easier with Trie prefix checks.

3. Final Answer:

   To efficiently store and search words by their prefixes -> Option A

4. Quick Check:

   Trie = prefix search [OK]

Hint: Trie is best for prefix-based word searches [OK]

Common Mistakes:

• Confusing Trie with sorting structures
• Thinking Trie reverses words
• Assuming Trie counts words only

2. Which of the following is the correct way to insert a word into a Trie node in Java?

easy

A. node.children[word[i] - 'a'] = new TrieNode();

B. node.children[word.charAt(i) - 'a'] = new TrieNode();

C. node.children[word.charAt(i)] = new TrieNode();

D. node.children[word.charAt(i) + 'a'] = new TrieNode();

Solution

1. Step 1: Understand character indexing

   Characters are converted to index by subtracting 'a' to map 'a' to 0, 'b' to 1, etc.

2. Step 2: Check Java syntax for charAt

   Java uses word.charAt(i) to get character at index i.

3. Final Answer:

   node.children[word.charAt(i) - 'a'] = new TrieNode(); -> Option B

4. Quick Check:

   Correct char indexing = node.children[word.charAt(i) - 'a'] = new TrieNode(); [OK]

Hint: Use charAt(i) - 'a' for array index in Java [OK]

Common Mistakes:

• Using word[i] instead of charAt(i)
• Adding 'a' instead of subtracting
• Using char directly without index conversion

3. What will be the output of the following Java code snippet that uses a Trie to find the longest word?

String[] words = {"a", "app", "ap", "apply", "apple"};
LongestWordInDictionaryUsingTrie trie = new LongestWordInDictionaryUsingTrie();
System.out.println(trie.longestWord(words));

medium

A. "app"

B. "apply"

C. "apple"

D. "ap"

Solution

1. Step 1: Insert all words into Trie

   Words inserted: "a", "app", "ap", "apply", "apple".

2. Step 2: Check longest word with all prefixes present

   "app" is longest where all prefixes "a", "ap", "app" exist. "apple" and "apply" prefixes "appl" missing.

3. Final Answer:

   "app" -> Option A

4. Quick Check:

   Longest word with all prefixes = "app" [OK]

Hint: Longest word must have all prefixes present [OK]

Common Mistakes:

• Choosing longest word ignoring prefix condition
• Confusing "apple" and "apply" as valid
• Ignoring shorter prefix words

4. Identify the error in this Java method snippet for searching the longest word in a Trie:

public String longestWord(String[] words) {
  TrieNode root = new TrieNode();
  for (String w : words) insert(root, w);
  String result = "";
  dfs(root, new StringBuilder(), result);
  return result;
}

private void dfs(TrieNode node, StringBuilder path, String result) {
  if (node == null) return;
  if (!node.isWord && path.length() != 0) return;
  if (path.length() > result.length()) result = path.toString();
  for (int i = 0; i < 26; i++) {
    if (node.children[i] != null) {
      path.append((char)(i + 'a'));
      dfs(node.children[i], path, result);
      path.deleteCharAt(path.length() - 1);
    }
  }
}

medium

A. The insert method is missing

B. The dfs method does not check for null children

C. The result string is not updated correctly inside dfs due to pass-by-value

D. The path StringBuilder is not cleared after recursion

Solution

1. Step 1: Analyze result parameter usage

   String is immutable and passed by value, so updating result inside dfs does not affect outer variable.

2. Step 2: Understand effect on final output

   Result remains empty because changes inside dfs do not persist outside recursion.

3. Final Answer:

   The result string is not updated correctly inside dfs due to pass-by-value -> Option C

4. Quick Check:

   String immutability causes result update failure [OK]

Hint: Use a mutable object or class field for result [OK]

Common Mistakes:

• Assuming String updates persist in recursion
• Ignoring missing insert method
• Overlooking null checks in children

5. Given a list of words, how would you modify the Trie-based algorithm to return the lexicographically smallest longest word when multiple longest words exist?

hard

A. Store all words in a list and pick the smallest after Trie traversal

B. Sort the input words first and insert into Trie in reverse order

C. Use a max heap to store all longest words and pop the smallest lexicographically

D. Traverse children in alphabetical order during DFS and update result only if new word is longer or lex smaller

Solution

1. Step 1: Traverse children in alphabetical order

   DFS visits children from 'a' to 'z' ensuring lex order traversal.

2. Step 2: Update result conditionally

   Update result only if new word is longer or same length but lex smaller.

3. Step 3: Ensure final result is lex smallest longest word

   This approach naturally picks lex smallest among longest words.

4. Final Answer:

   Traverse children in alphabetical order during DFS and update result only if new word is longer or lex smaller -> Option D

5. Quick Check:

   Lex order DFS + conditional update = Traverse children in alphabetical order during DFS and update result only if new word is longer or lex smaller [OK]

Hint: DFS children from 'a' to 'z' and update result carefully [OK]

Common Mistakes:

• Sorting input but not controlling DFS order
• Using data structures that add unnecessary complexity
• Ignoring lex order during traversal