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DSA Goprogramming~10 mins

Word Search in Trie in DSA Go - Interactive Practice

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Practice - 5 Tasks
Answer the questions below
1fill in blank
easy

Complete the code to insert a word into the Trie.

DSA Go
func (t *Trie) Insert(word string) {
    node := t.root
    for _, ch := range word {
        if node.children[ch] == nil {
            node.children[ch] = &TrieNode{children: make(map[rune]*TrieNode)}
        }
        node = node.children[ch]
    }
    node.isWord = [1]
}
Drag options to blanks, or click blank then click option'
Atrue
Bnil
Cfalse
D0
Attempts:
3 left
💡 Hint
Common Mistakes
Setting node.isWord to false instead of true.
Forgetting to mark the end of the word.
2fill in blank
medium

Complete the code to check if a word exists in the Trie.

DSA Go
func (t *Trie) Search(word string) bool {
    node := t.root
    for _, ch := range word {
        if node.children[ch] == nil {
            return [1]
        }
        node = node.children[ch]
    }
    return node.isWord
}
Drag options to blanks, or click blank then click option'
Anil
Btrue
Cfalse
D0
Attempts:
3 left
💡 Hint
Common Mistakes
Returning true when a character is missing.
Not checking if the node marks the end of a word.
3fill in blank
hard

Fix the error in the code to correctly check if a prefix exists in the Trie.

DSA Go
func (t *Trie) StartsWith(prefix string) bool {
    node := t.root
    for _, ch := range prefix {
        if node.children[ch] == [1] {
            return false
        }
        node = node.children[ch]
    }
    return true
}
Drag options to blanks, or click blank then click option'
Anil
Btrue
Cfalse
D0
Attempts:
3 left
💡 Hint
Common Mistakes
Comparing to true or false instead of nil.
Returning true when the prefix path is missing.
4fill in blank
hard

Fill both blanks to complete the recursive helper function for word search in a 2D board using Trie.

DSA Go
func dfs(board [][]byte, node *TrieNode, i, j int, path string, result *map[string]bool) {
    if i < 0 || i >= len(board) || j < 0 || j >= len(board[0]) || board[i][j] == '#' {
        return
    }
    ch := rune(board[i][j])
    nextNode := node.children[[1]]
    if nextNode == nil {
        return
    }
    path += string([2])
    if nextNode.isWord {
        (*result)[path] = true
    }
    board[i][j] = '#'
    dfs(board, nextNode, i+1, j, path, result)
    dfs(board, nextNode, i-1, j, path, result)
    dfs(board, nextNode, i, j+1, path, result)
    dfs(board, nextNode, i, j-1, path, result)
    board[i][j] = byte(ch)
}
Drag options to blanks, or click blank then click option'
Ach
Bi
Cj
Dpath
Attempts:
3 left
💡 Hint
Common Mistakes
Using indices i or j instead of the character.
Using the path string instead of the character.
5fill in blank
hard

Fill all three blanks to complete the function that builds a map of found words from the board using Trie.

DSA Go
func findWords(board [][]byte, words []string) []string {
    trie := NewTrie()
    for _, word := range words {
        trie.Insert([1])
    }
    result := make(map[string]bool)
    for i := 0; i < len(board); i++ {
        for j := 0; j < len(board[0]); j++ {
            dfs(board, trie.root, i, j, [2], &result)
        }
    }
    res := []string{}
    for word := range result {
        res = append(res, [3])
    }
    return res
}
Drag options to blanks, or click blank then click option'
Aword
B""
Cnil
Dboard
Attempts:
3 left
💡 Hint
Common Mistakes
Passing nil or board instead of the word to Insert.
Starting DFS with nil or board instead of empty string.

Practice

(1/5)
1. What is the main purpose of using a Trie data structure in word search problems?
easy
A. To reverse words quickly
B. To sort words alphabetically
C. To store words efficiently for quick prefix-based searches
D. To count the frequency of words

Solution

  1. Step 1: Understand Trie structure

    A Trie stores characters of words in a tree form, sharing prefixes.
  2. Step 2: Identify Trie use in word search

    This structure allows fast prefix checks and word existence queries.
  3. Final Answer:

    To store words efficiently for quick prefix-based searches -> Option C
  4. Quick Check:

    Trie = prefix search efficiency [OK]
Hint: Trie is best for prefix-based word searches [OK]
Common Mistakes:
  • Confusing Trie with sorting structures
  • Thinking Trie counts word frequency
  • Assuming Trie reverses words
2. Which of the following Go code snippets correctly initializes an empty Trie node struct with a children map and end-of-word flag?
easy
A. type TrieNode struct { children map[string]TrieNode; isEnd bool } root := TrieNode{children: map[string]TrieNode{}, isEnd: false}
B. type TrieNode struct { children []TrieNode; isEnd int } root := TrieNode{children: []TrieNode{}, isEnd: 0}
C. type TrieNode struct { children map[rune]TrieNode; isEnd bool } root := &TrieNode{children: make(map[rune]TrieNode), isEnd: false}
D. type TrieNode struct { children map[rune]*TrieNode; isEnd bool } root := &TrieNode{children: make(map[rune]*TrieNode), isEnd: false}

Solution

  1. Step 1: Check struct field types

    Children should be a map from rune to pointer to TrieNode for dynamic nodes.
  2. Step 2: Verify initialization syntax

    Use make(map[rune]*TrieNode) and isEnd as bool with false initial value.
  3. Final Answer:

    type TrieNode struct { children map[rune]*TrieNode; isEnd bool } root := &TrieNode{children: make(map[rune]*TrieNode), isEnd: false} -> Option D
  4. Quick Check:

    Correct map and pointer usage = type TrieNode struct { children map[rune]*TrieNode; isEnd bool } root := &TrieNode{children: make(map[rune]*TrieNode), isEnd: false} [OK]
Hint: Use map[rune]*TrieNode and make() to initialize children [OK]
Common Mistakes:
  • Using slice instead of map for children
  • Using non-pointer TrieNode in map values
  • Incorrect map key type (string instead of rune)
3. Given the following Go code snippet for searching a word in a Trie, what will be the output if the word "cat" is searched after inserting "car" and "cat"?
func (t *TrieNode) Search(word string) bool {
    node := t
    for _, ch := range word {
        if _, ok := node.children[ch]; !ok {
            return false
        }
        node = node.children[ch]
    }
    return node.isEnd
}
medium
A. true
B. false
C. Compilation error
D. Runtime panic

Solution

  1. Step 1: Understand insertion of "car" and "cat"

    Both words share prefix 'ca', nodes created for 'c', 'a', then separate nodes for 'r' and 't' with isEnd true.
  2. Step 2: Trace search for "cat"

    Traverse nodes for 'c', 'a', 't' successfully, last node has isEnd true, so returns true.
  3. Final Answer:

    true -> Option A
  4. Quick Check:

    Word inserted and found = true [OK]
Hint: Search returns true if last node's isEnd is true [OK]
Common Mistakes:
  • Assuming search returns true for prefixes only
  • Confusing isEnd flag usage
  • Thinking search returns true if partial path exists
4. Identify the bug in the following Go code snippet for searching a word in a Trie:
func (t *TrieNode) Search(word string) bool {
    node := t
    for _, ch := range word {
        if node.children[ch] == nil {
            return false
        }
        node = node.children[ch]
    }
    return true
}
medium
A. Returns true even if the word is only a prefix, not a full word
B. Panics if children map is nil
C. Does not check for nil children map before access
D. Incorrect loop variable type causes compilation error

Solution

  1. Step 1: Analyze search return condition

    Function returns true after traversing all characters without checking isEnd flag.
  2. Step 2: Understand consequence of missing isEnd check

    Words that are prefixes but not complete words will incorrectly return true.
  3. Final Answer:

    Returns true even if the word is only a prefix, not a full word -> Option A
  4. Quick Check:

    Missing isEnd check causes false positives [OK]
Hint: Always check isEnd after traversal to confirm full word [OK]
Common Mistakes:
  • Forgetting to check isEnd flag
  • Assuming traversal success means word exists
  • Ignoring nil map checks (not in this code)
5. You have a list of words: ["apple", "app", "application", "bat", "bath", "baton"]. You want to build a Trie and then search for the word "bat". Which of the following statements is true about the search result and the Trie structure after insertion?
hard
A. Searching "bat" returns false because "bat" is a prefix of "bath" and "baton"
B. Searching "bat" returns true; nodes for 'b', 'a', 't' share prefixes with other words
C. Trie stores each word as a separate branch without shared nodes
D. Searching "bat" returns true only if "bat" is inserted last

Solution

  1. Step 1: Insert all words into Trie

    Words share prefixes: "app" and "apple" share 'app', "bat", "bath", "baton" share 'bat'.
  2. Step 2: Search for "bat"

    Since "bat" is inserted, its node's isEnd is true, so search returns true.
  3. Final Answer:

    Searching "bat" returns true; nodes for 'b', 'a', 't' share prefixes with other words -> Option B
  4. Quick Check:

    Shared prefix nodes and isEnd true = Searching "bat" returns true; nodes for 'b', 'a', 't' share prefixes with other words [OK]
Hint: Shared prefixes save space; isEnd marks full words [OK]
Common Mistakes:
  • Thinking prefix words are not found
  • Assuming separate branches for each word
  • Believing insertion order affects search result
Recommended

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10 questions~5 mineasy hard