Longest Word in Dictionary Using Trie in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to insert a word into the Trie.

DSA C++

void insert(string word) {
    TrieNode* node = root;
    for (char c : word) {
        int index = c - 'a';
        if (node->children[[1]] == nullptr) {
            node->children[index] = new TrieNode();
        }
        node = node->children[index];
    }
    node->isEnd = true;
}

Drag options to blanks, or click blank then click option'

Aindex

Bc

Cword

Dnode

Attempts:

3 left

💡 Hint

Common Mistakes

Using the character 'c' directly as an index causes errors.

Using the whole word instead of the character index.

We calculate the index of the character in the children array using 'index'. So, the blank should be filled with 'index'.

2fill in blank

medium

Complete the code to check if a node marks the end of a word.

DSA C++

bool isEndOfWord(TrieNode* node) {
    return node->[1];
}

Drag options to blanks, or click blank then click option'

AwordEnd

BisWord

Cend

DisEnd

Attempts:

3 left

💡 Hint

Common Mistakes

Using incorrect variable names like 'isWord' or 'wordEnd'.

The TrieNode uses 'isEnd' boolean to mark the end of a word.

3fill in blank

hard

Fix the error in the DFS function to find the longest word.

DSA C++

void dfs(TrieNode* node, string& current, string& longest) {
    if (!node || !node->[1]) return;
    if (current.length() > longest.length()) {
        longest = current;
    }
    for (int i = 0; i < 26; i++) {
        if (node->children[i] && node->children[i]->isEnd) {
            current.push_back('a' + i);
            dfs(node->children[i], current, longest);
            current.pop_back();
        }
    }
}

Drag options to blanks, or click blank then click option'

AisWord

BisEnd

Cend

DwordEnd

Attempts:

3 left

💡 Hint

Common Mistakes

Checking the wrong variable for word end.

Not stopping DFS when node is null.

The DFS should stop if the current node is not the end of a word, so we check 'isEnd'.

4fill in blank

hard

Fill both blanks to build the longest word using DFS traversal.

DSA C++

void dfs(TrieNode* node, string& current, string& longest) {
    if (!node || !node->[1]) return;
    if (current.length() > longest.length() || (current.length() == longest.length() && current < longest)) {
        longest = current;
    }
    for (int i = 0; i < 26; i++) {
        if (node->children[i] && node->children[i]->[2]) {
            current.push_back('a' + i);
            dfs(node->children[i], current, longest);
            current.pop_back();
        }
    }
}

Drag options to blanks, or click blank then click option'

AisEnd

BisWord

Cend

DwordEnd

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variable names for the two blanks.

Using variables that do not exist in the TrieNode.

Both blanks check if the node marks the end of a word using 'isEnd'.

5fill in blank

hard

Fill all three blanks to complete the function that finds the longest word in the dictionary.

DSA C++

string longestWord(vector<string>& words) {
    Trie trie;
    for (auto& word : words) {
        trie.[1](word);
    }
    string longest = "";
    string current = "";
    trie.[2](trie.root, current, longest);
    return longest;
}

class Trie {
public:
    TrieNode* root;
    Trie() {
        root = new TrieNode();
    }
    void [3](string word) {
        TrieNode* node = root;
        for (char c : word) {
            int index = c - 'a';
            if (!node->children[index]) {
                node->children[index] = new TrieNode();
            }
            node = node->children[index];
        }
        node->isEnd = true;
    }
    void dfs(TrieNode* node, string& current, string& longest) {
        if (!node || !node->isEnd) return;
        if (current.length() > longest.length() || (current.length() == longest.length() && current < longest)) {
            longest = current;
        }
        for (int i = 0; i < 26; i++) {
            if (node->children[i] && node->children[i]->isEnd) {
                current.push_back('a' + i);
                dfs(node->children[i], current, longest);
                current.pop_back();
            }
        }
    }
};

Drag options to blanks, or click blank then click option'

Ainsert

Bdfs

CinsertWord

Dsearch

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'search' instead of 'insert' to add words.

Confusing function names for DFS.

We insert words using 'insert', start DFS with 'dfs', and the insert function is named 'insert'.

Practice

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

easy

A. To sort words alphabetically

B. To reverse the words in the dictionary

C. To count the total number of words

D. To efficiently store and search words by their prefixes

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 prefix-based storage.

3. Final Answer:

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

4. Quick Check:

   Trie = prefix search [OK]

Hint: Trie stores prefixes for fast word checks [OK]

Common Mistakes:

• Thinking Trie sorts words alphabetically
• Confusing Trie with simple arrays
• Assuming Trie counts words only

2. Which of the following is the correct way to declare a Trie node class in C++ for lowercase English letters?

easy

A. class TrieNode { public: vector<TrieNode> children; bool isEnd; TrieNode() { isEnd = false; } };

B. class TrieNode { public: TrieNode* children[26]; bool isEnd; TrieNode() { for(int i=0; i<26; i++) children[i] = nullptr; isEnd = false; } };

C. class TrieNode { public: TrieNode* children[52]; bool isEnd; TrieNode() { for(int i=0; i<52; i++) children[i] = nullptr; isEnd = false; } };

D. class TrieNode { public: int children[26]; bool isEnd; TrieNode() { for(int i=0; i<26; i++) children[i] = 0; isEnd = true; } };

Solution

1. Step 1: Check children type and size

   Children should be pointers to TrieNode, size 26 for lowercase letters.

2. Step 2: Check initialization

   Initialize all children to nullptr and isEnd to false.

3. Final Answer:

   class TrieNode { public: TrieNode* children[26]; bool isEnd; TrieNode() { for(int i=0; i<26; i++) children[i] = nullptr; isEnd = false; } }; -> Option B

4. Quick Check:

   Pointer array size 26, initialized null [OK]

Hint: Use pointer array size 26 for lowercase letters [OK]

Common Mistakes:

• Using int array instead of pointers
• Wrong array size (52 instead of 26)
• Not initializing children to nullptr

3. What will be the output of the following code snippet that inserts words into a Trie and finds the longest word where all prefixes exist? vector<string> words = {"a", "ap", "app", "appl", "apple", "apply"}; // After building Trie and searching longest word cout << longestWord(words) << endl;

medium

A. "apple"

B. "apply"

C. "appl"

D. "app"

Solution

1. Step 1: Insert all words into Trie

   Words "a", "ap", "app", "appl", "apple", "apply" are inserted.

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

   "apple" and "apply" both have all prefixes present, but "apple" comes first lexicographically.

3. Final Answer:

   "apple" -> Option A

4. Quick Check:

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

Hint: Check all prefixes exist; choose longest lex smallest [OK]

Common Mistakes:

• Choosing "apply" ignoring lex order
• Choosing shorter words like "app"
• Not verifying all prefixes exist

4. Identify the error in this Trie insertion code snippet for lowercase English letters: void insert(TrieNode* root, string word) { TrieNode* node = root; for (char c : word) { int index = c - 'a'; if (!node->children[index]) node->children[index] = new TrieNode(); node = node->children[index]; } node->isEnd = true; }

medium

A. No error; code is correct

B. Incorrect calculation of index from character

C. Missing initialization of root node before insertion

D. Not marking the end of word after insertion

Solution

1. Step 1: Check index calculation

   Index is correctly calculated as c - 'a' for lowercase letters.

2. Step 2: Verify insertion and end marking

   Children nodes are created if null, and isEnd is set true after insertion.

3. Final Answer:

   No error; code is correct -> Option A

4. Quick Check:

   Insertion logic correct [OK]

Hint: Check index and isEnd marking carefully [OK]

Common Mistakes:

• Miscomputing index for uppercase letters
• Forgetting to mark isEnd true
• Not initializing root before use (outside snippet)

5. Given a list of words, how would you modify the Trie-based approach to find the longest word where every prefix is also a valid word, but now the dictionary may contain uppercase and lowercase letters mixed? Choose the best approach.

hard

A. Ignore case and treat uppercase and lowercase as same during insertion

B. Convert all words to lowercase before inserting into Trie and searching

C. Use a Trie with 52 children per node to handle both cases distinctly

D. Use a hash map instead of Trie for faster prefix checks

Solution

1. Step 1: Understand case sensitivity

   Uppercase and lowercase letters differ; need to distinguish them in Trie.

2. Step 2: Adjust Trie structure

   Use 52 children per node (26 lowercase + 26 uppercase) to store all letters distinctly.

3. Final Answer:

   Use a Trie with 52 children per node to handle both cases distinctly -> Option C

4. Quick Check:

   Handle case by expanding Trie size [OK]

Hint: Expand Trie children to 52 for mixed case [OK]

Common Mistakes:

• Converting all to lowercase loses case info
• Ignoring case causes wrong matches
• Using hash map loses prefix tree benefits