Word Search in Trie in DSA Java - Time & Space Complexity

Time Complexity: Word Search in Trie

O(n)

Understanding Time Complexity

We want to understand how long it takes to find a word in a Trie data structure.

Specifically, how the search time changes as the word length grows.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

public boolean search(String word) {
    TrieNode current = root;
    for (int i = 0; i < word.length(); i++) {
        char ch = word.charAt(i);
        if (!current.children.containsKey(ch)) {
            return false;
        }
        current = current.children.get(ch);
    }
    return current.isEndOfWord;
}
    

This code checks if a given word exists in the Trie by following each character step-by-step.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Looping through each character of the word.
• How many times: Exactly once per character in the word (word length times).

How Execution Grows With Input

As the word gets longer, the search checks more characters one by one.

Input Size (n)	Approx. Operations
10	About 10 character checks
100	About 100 character checks
1000	About 1000 character checks

Pattern observation: The number of operations grows directly with the word length.

Final Time Complexity

Time Complexity: O(n)

This means the search time grows linearly with the length of the word you are looking for.

Common Mistake

[X] Wrong: "Searching a word in a Trie takes constant time because it's a tree."

[OK] Correct: The search depends on the word length, so it grows linearly, not constant time.

Interview Connect

Knowing how Trie search time depends on word length helps you explain efficient prefix searches clearly.

Self-Check

"What if the Trie stored words with only lowercase letters, and we used an array instead of a map for children? How would the time complexity change?"

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