Word Search in Trie in DSA Java - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Trie Word Search

What is the output of the following Java code that inserts words into a Trie and searches for a specific word?

DSA Java

class TrieNode {
    TrieNode[] children = new TrieNode[26];
    boolean isEnd = false;
}

class Trie {
    TrieNode root = new TrieNode();

    void insert(String word) {
        TrieNode node = root;
        for (char c : word.toCharArray()) {
            int idx = c - 'a';
            if (node.children[idx] == null) {
                node.children[idx] = new TrieNode();
            }
            node = node.children[idx];
        }
        node.isEnd = true;
    }

    boolean search(String word) {
        TrieNode node = root;
        for (char c : word.toCharArray()) {
            int idx = c - 'a';
            if (node.children[idx] == null) {
                return false;
            }
            node = node.children[idx];
        }
        return node.isEnd;
    }
}

public class Main {
    public static void main(String[] args) {
        Trie trie = new Trie();
        trie.insert("apple");
        trie.insert("app");
        trie.insert("application");
        System.out.println(trie.search("app"));
        System.out.println(trie.search("appl"));
    }
}

A true\nfalse

B true\ntrue

C false\nfalse

D false\ntrue

💡 Hint

Remember that search returns true only if the word is fully inserted and marked as an end.

The word "app" was inserted and marked as an end, so search("app") returns true. The prefix "appl" is not marked as an end, so search("appl") returns false.

🧠 Conceptual

intermediate

Trie Node Children Count

After inserting the words ["cat", "car", "cart", "dog"] into an empty Trie, how many children does the root node have?

A 1

B 3

C 4

D 2

💡 Hint

Count distinct first letters of inserted words.

Words start with 'c' and 'd', so root has children at indices for 'c' and 'd', total 2 children.

🔧 Debug

advanced

Identify the Error in Trie Search

What error will the following Java code produce when searching for a word not inserted in the Trie?

DSA Java

class TrieNode {
    TrieNode[] children = new TrieNode[26];
    boolean isEnd = false;
}

class Trie {
    TrieNode root = new TrieNode();

    void insert(String word) {
        TrieNode node = root;
        for (char c : word.toCharArray()) {
            int idx = c - 'a';
            if (node.children[idx] == null) {
                node.children[idx] = new TrieNode();
            }
            node = node.children[idx];
        }
        node.isEnd = true;
    }

    boolean search(String word) {
        TrieNode node = root;
        for (char c : word.toCharArray()) {
            int idx = c - 'a';
            node = node.children[idx];
        }
        return node != null && node.isEnd;
    }
}

public class Main {
    public static void main(String[] args) {
        Trie trie = new Trie();
        trie.insert("hello");
        System.out.println(trie.search("hey"));
    }
}

A No error, returns false

B ArrayIndexOutOfBoundsException

C NullPointerException

D Compilation error

💡 Hint

Check if the code verifies children before accessing.

The search method does not check if node.children[idx] is null before assigning node = node.children[idx]. For the word "hey", the path diverges at the second character ('e'), and at the third character ('y'), node.children[idx] is null, so node becomes null and the next iteration or access causes a NullPointerException.

📝 Syntax

advanced

Correct Syntax for Trie Insert Method

Which option shows the correct syntax for inserting a word into a Trie in Java?

A

for (char c : word.toCharArray()) {
    int idx = c - 'a';
    if (node.children[idx] == null) {
        node.children[idx] = new TrieNode();
    }
    node = node.children[idx];
}
node.isEnd = true;

B

for (char c in word) {
    int idx = c - 'a';
    if (node.children[idx] == null) {
        node.children[idx] = new TrieNode();
    }
    node = node.children[idx];
}
node.isEnd = true;

C

for (int i = 0; i < word.length(); i++) {
    int idx = word.charAt(i) - 'a';
    if (node.children[idx] == null) {
        node.children[idx] = new TrieNode();
    }
    node = node.children[idx];
}
node.isEnd = true;

D

for (char c : word.toCharArray()) {
    int idx = c - 'a';
    if (node.children[idx] != null) {
        node.children[idx] = new TrieNode();
    }
    node = node.children[idx];
}
node.isEnd = true;

💡 Hint

Java uses for-each with colon, not 'in'. Also check condition for creating new nodes.

Option A uses correct for-each syntax and creates new TrieNode only if child is null. Option A uses invalid 'in' keyword. Option A is valid syntax but uses index-based loop (also correct but not the preferred here). Option A incorrectly creates new node if child is NOT null.

🚀 Application

expert

Number of Words with Given Prefix in Trie

Given a Trie with words inserted: ["test", "team", "teach", "tent", "toast"], what is the number of words in the Trie that start with the prefix "te"?

A 3

B 4

C 2

D 5

💡 Hint

Count all words starting with 'te' from the list.

Words starting with 'te' are "test", "team", "teach", and "tent" — total 4 words.

Practice

1. What is the main purpose of using a Trie data structure in word search problems?

easy

A. To perform arithmetic operations on words

B. To sort words alphabetically

C. To store words in a linked list

D. To efficiently store and search words by their prefixes

Solution

1. Step 1: Understand Trie structure

   A Trie stores characters of words in a tree-like structure where each node represents a character.

2. Step 2: Identify Trie usage in word search

   It allows quick prefix-based search, making it efficient to find words starting with given letters.

3. Final Answer:

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

4. Quick Check:

   Trie = prefix search [OK]

Hint: Trie is all about prefix-based fast word search [OK]

Common Mistakes:

• Confusing Trie with sorting data structures
• Thinking Trie stores words as linked lists
• Assuming Trie performs arithmetic on words

2. Which of the following is the correct way to declare a Trie node class in Java?

easy

A. class TrieNode { char val; TrieNode[] children = new TrieNode[26]; boolean isWord; }

B. class TrieNode { int val; TrieNode children; boolean isWord; }

C. class TrieNode { String val; TrieNode[] children = new TrieNode[26]; }

D. class TrieNode { char val; List<TrieNode> children; boolean isWord; }

Solution

1. Step 1: Check data types for Trie node

   Each node stores a character, an array of 26 children for letters a-z, and a boolean to mark word end.

2. Step 2: Validate options

   class TrieNode { char val; TrieNode[] children = new TrieNode[26]; boolean isWord; } matches this structure exactly with char val, TrieNode[] children of size 26, and boolean isWord.

3. Final Answer:

   class TrieNode { char val; TrieNode[] children = new TrieNode[26]; boolean isWord; } -> Option A

4. Quick Check:

   Correct TrieNode declaration = class TrieNode { char val; TrieNode[] children = new TrieNode[26]; boolean isWord; } [OK]

Hint: Trie nodes use char + array of 26 children + boolean flag [OK]

Common Mistakes:

• Using int or String instead of char for node value
• Using single child instead of array for children
• Using List instead of fixed-size array for children

3. What will be the output of the following Java code snippet that inserts words into a Trie and searches for "cat"?

class TrieNode {
  TrieNode[] children = new TrieNode[26];
  boolean isWord = false;
}

class Trie {
  TrieNode root = new TrieNode();

  void insert(String word) {
    TrieNode node = root;
    for (char c : word.toCharArray()) {
      int idx = c - 'a';
      if (node.children[idx] == null) node.children[idx] = new TrieNode();
      node = node.children[idx];
    }
    node.isWord = true;
  }

  boolean search(String word) {
    TrieNode node = root;
    for (char c : word.toCharArray()) {
      int idx = c - 'a';
      if (node.children[idx] == null) return false;
      node = node.children[idx];
    }
    return node.isWord;
  }
}

public class Main {
  public static void main(String[] args) {
    Trie trie = new Trie();
    trie.insert("cat");
    trie.insert("car");
    System.out.println(trie.search("cat"));
    System.out.println(trie.search("cap"));
  }
}

medium

A. false
false

B. true
false

C. true
true

D. false
true

Solution

1. Step 1: Insert words "cat" and "car" into Trie

   Both words are inserted character by character, marking the last node's isWord as true.

2. Step 2: Search for "cat" and "cap"

   "cat" exists with isWord true, so search returns true. "cap" path is missing 'p' node, so returns false.

3. Final Answer:

   true false -> Option B

4. Quick Check:

   Inserted words found = true, missing word = false [OK]

Hint: Search returns true only if full word path and isWord true [OK]

Common Mistakes:

• Assuming partial prefix returns true
• Not marking isWord correctly on insert
• Confusing character indexing in children array

4. Identify the bug in the following Trie search method and choose the correct fix:

boolean search(String word) {
  TrieNode node = root;
  for (int i = 0; i <= word.length(); i++) {
    int idx = word.charAt(i) - 'a';
    if (node.children[idx] == null) return false;
    node = node.children[idx];
  }
  return node.isWord;
}

medium

A. Initialize idx before loop

B. Return true inside the loop

C. Change loop condition to i < word.length()

D. Remove return statement inside loop

Solution

1. Step 1: Analyze loop boundary

   The loop runs from i = 0 to i <= word.length(), which causes index out of bounds at last iteration.

2. Step 2: Fix loop condition

   Changing condition to i < word.length() prevents accessing invalid index and correctly iterates over characters.

3. Final Answer:

   Change loop condition to i < word.length() -> Option C

4. Quick Check:

   Loop boundary fix prevents index error [OK]

Hint: Loop over string chars with i < length, not <= length [OK]

Common Mistakes:

• Using <= instead of < in loop condition
• Trying to fix by moving idx initialization
• Removing return inside loop causing wrong logic

5. Given a 2D board of characters and a list of words, which approach best uses a Trie to efficiently find all words present on the board? A sample board:

[
  ['o','a','a','n'],
  ['e','t','a','e'],
  ['i','h','k','r'],
  ['i','f','l','v']
]

Words: ["oath", "pea", "eat", "rain"]

hard

A. Build a Trie from words, then DFS on board checking prefixes in Trie

B. Check each word by scanning entire board without Trie

C. Use a hash set for words and scan board rows only

D. Sort words and board rows, then binary search each word

Solution

1. Step 1: Build Trie from given words

   Insert all words into a Trie to enable prefix checking during search.

2. Step 2: Perform DFS on board cells

   From each cell, explore neighbors recursively, checking if current path matches any prefix in Trie to prune search.

3. Step 3: Collect words found when Trie node marks word end

   This approach efficiently finds all words without redundant searches.

4. Final Answer:

   Build a Trie from words, then DFS on board checking prefixes in Trie -> Option A

5. Quick Check:

   Trie + DFS = efficient word search on board [OK]

Hint: Use Trie for prefix pruning during DFS on board [OK]

Common Mistakes:

• Searching words one by one scanning whole board
• Ignoring prefix pruning causing slow search
• Using sorting and binary search which doesn't fit 2D board