Count Words with Given Prefix in DSA Python - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of prefix count after insertions

What is the output of the code after inserting words and counting the prefix 'app'?

DSA Python

class TrieNode:
    def __init__(self):
        self.children = {}
        self.count = 0

class Trie:
    def __init__(self):
        self.root = TrieNode()

    def insert(self, word):
        node = self.root
        for ch in word:
            if ch not in node.children:
                node.children[ch] = TrieNode()
            node = node.children[ch]
            node.count += 1

    def count_prefix(self, prefix):
        node = self.root
        for ch in prefix:
            if ch not in node.children:
                return 0
            node = node.children[ch]
        return node.count

trie = Trie()
words = ['apple', 'app', 'application', 'apt', 'banana']
for w in words:
    trie.insert(w)

print(trie.count_prefix('app'))

A 4

B 3

C 2

D 1

💡 Hint

Count how many words start with 'app' in the list.

The words starting with 'app' are 'apple', 'app', and 'application', so the count is 3.

❓ Predict Output

intermediate

Count prefix with no matching words

What is the output of counting prefix 'bat' after inserting given words?

DSA Python

class TrieNode:
    def __init__(self):
        self.children = {}
        self.count = 0

class Trie:
    def __init__(self):
        self.root = TrieNode()

    def insert(self, word):
        node = self.root
        for ch in word:
            if ch not in node.children:
                node.children[ch] = TrieNode()
            node = node.children[ch]
            node.count += 1

    def count_prefix(self, prefix):
        node = self.root
        for ch in prefix:
            if ch not in node.children:
                return 0
            node = node.children[ch]
        return node.count

trie = Trie()
words = ['cat', 'car', 'cart', 'dog']
for w in words:
    trie.insert(w)

print(trie.count_prefix('bat'))

A 0

B 1

C 2

D 3

💡 Hint

Check if any word starts with 'bat'.

No word in the list starts with 'bat', so the count is 0.

❓ Predict Output

advanced

Count prefix after multiple insertions with overlapping prefixes

What is the output after inserting words and counting prefix 'a'?

DSA Python

class TrieNode:
    def __init__(self):
        self.children = {}
        self.count = 0

class Trie:
    def __init__(self):
        self.root = TrieNode()

    def insert(self, word):
        node = self.root
        for ch in word:
            if ch not in node.children:
                node.children[ch] = TrieNode()
            node = node.children[ch]
            node.count += 1

    def count_prefix(self, prefix):
        node = self.root
        for ch in prefix:
            if ch not in node.children:
                return 0
            node = node.children[ch]
        return node.count

trie = Trie()
words = ['a', 'ab', 'abc', 'abcd', 'abcde']
for w in words:
    trie.insert(w)

print(trie.count_prefix('a'))

A 4

B 0

C 5

D 1

💡 Hint

Count all words starting with 'a'.

All 5 words start with 'a', so the count is 5.

❓ Predict Output

advanced

Count prefix after inserting words with shared and unique prefixes

What is the output of counting prefix 'ban' after inserting the words?

DSA Python

class TrieNode:
    def __init__(self):
        self.children = {}
        self.count = 0

class Trie:
    def __init__(self):
        self.root = TrieNode()

    def insert(self, word):
        node = self.root
        for ch in word:
            if ch not in node.children:
                node.children[ch] = TrieNode()
            node = node.children[ch]
            node.count += 1

    def count_prefix(self, prefix):
        node = self.root
        for ch in prefix:
            if ch not in node.children:
                return 0
            node = node.children[ch]
        return node.count

trie = Trie()
words = ['banana', 'band', 'bandana', 'ban', 'banner', 'bat']
for w in words:
    trie.insert(w)

print(trie.count_prefix('ban'))

A 5

B 6

C 3

D 4

💡 Hint

Count words starting with 'ban' only.

The words starting with 'ban' are 'banana', 'band', 'bandana', 'ban', and 'banner' totaling 5.

🧠 Conceptual

expert

Understanding Trie node count increments

In a Trie implementation for counting words with a given prefix, why is the count incremented at each node during insertion?

A To store the length of the word inserted.

B To count the total number of children nodes under that node.

C To mark the end of a word at that node.

D To keep track of how many words pass through that node, enabling quick prefix counts.

💡 Hint

Think about how prefix counts are retrieved efficiently.

Incrementing count at each node during insertion tracks how many words share that prefix, allowing quick retrieval of prefix counts.

Practice

1. What does the function countWordsStartingWith(words, prefix) do when given a list of words and a prefix?

easy

A. Counts how many words contain the prefix anywhere inside

B. Returns the first word that starts with the prefix

C. Counts how many words in the list start with the given prefix

D. Removes all words that start with the prefix from the list

Solution

1. Step 1: Understand the function purpose

   The function is designed to count words that begin with a specific prefix, not just contain it anywhere.

2. Step 2: Compare options with function behavior

   Only Counts how many words in the list start with the given prefix correctly describes counting words starting with the prefix. Others describe different behaviors.

3. Final Answer:

   Counts how many words in the list start with the given prefix -> Option C

4. Quick Check:

   Counting words starting with prefix = Counts how many words in the list start with the given prefix [OK]

Hint: Focus on 'start with' meaning prefix at word beginning [OK]

Common Mistakes:

• Confusing 'start with' and 'contain' prefix
• Thinking function returns words instead of count
• Assuming it removes words instead of counting

2. Which of the following Python code snippets correctly counts words starting with a prefix p in a list words?

easy

A. sum(word.startswith(p) for word in words)

B. sum(word.startswith(p) == 'True' for word in words)

C. sum(word.startswith(p) is 'True' for word in words)

D. sum(word.startswith(p) = True for word in words)

Solution

1. Step 1: Check syntax correctness

   sum(word.startswith(p) = True for word in words) uses '=' inside sum which is invalid syntax for comparison.

2. Step 2: Understand boolean summation

   Options B and C incorrectly compare boolean to string 'True' (always False). Only sum(word.startswith(p) for word in words) correctly sums booleans directly (True counts as 1).

3. Final Answer:

   sum(word.startswith(p) for word in words) -> Option A

4. Quick Check:

   Use sum with boolean directly for counting [OK]

Hint: Use sum with boolean expressions directly, no need for == or is [OK]

Common Mistakes:

• Using '=' instead of '==' for comparison
• Comparing boolean to string 'True'
• Confusing syntax causing errors

3. What is the output of the following code?

words = ['apple', 'app', 'application', 'banana', 'apply']
prefix = 'app'
count = sum(word.startswith(prefix) for word in words)
print(count)

medium

A. 2

B. 3

C. 5

D. 4

Solution

1. Step 1: Identify words starting with 'app'

   Words starting with 'app' are 'apple', 'app', 'application', and 'apply' -- total 4 words.

2. Step 2: Calculate sum of True values

   Each word starting with prefix counts as 1, so sum is 4.

3. Final Answer:

   4 -> Option D

4. Quick Check:

   Count words starting with 'app' = 4 [OK]

Hint: Count words starting exactly with prefix, not just containing it [OK]

Common Mistakes:

• Counting words that contain prefix anywhere
• Missing one word that starts with prefix
• Confusing prefix length with word length

4. Find the error in this code that tries to count words starting with prefix 'pre':

def count_prefix(words, prefix):
    count = 0
    for word in words:
        if word.startswith == prefix:
            count += 1
    return count

words = ['prefix', 'preach', 'prevent', 'post']
prefix = 'pre'
print(count_prefix(words, prefix))

medium

A. Using 'word.startswith == prefix' instead of 'word.startswith(prefix)'

B. Missing return statement in function

C. Incorrect variable name 'count' used

D. Loop should be 'for prefix in words' instead of 'for word in words'

Solution

1. Step 1: Identify incorrect method usage

   The code uses 'word.startswith == prefix' which compares method to string, incorrect syntax.

2. Step 2: Correct method call

   It should be 'word.startswith(prefix)' to check if word starts with prefix.

3. Final Answer:

   Using 'word.startswith == prefix' instead of 'word.startswith(prefix)' -> Option A

4. Quick Check:

   Method call needs parentheses with argument [OK]

Hint: Remember to call methods with parentheses and arguments [OK]

Common Mistakes:

• Comparing method itself instead of calling it
• Forgetting parentheses after method name
• Mixing variable names in loop

5. Given a list of words, how would you count words starting with any prefix from a list of prefixes prefixes? Choose the correct code snippet.

words = ['cat', 'car', 'dog', 'cart', 'cow']
prefixes = ['ca', 'do', 'co']

hard

A. sum(word.startswith(prefixes[0]) or word.startswith(prefixes[1]) for word in words)

B. sum(any(word.startswith(p) for p in prefixes) for word in words)

C. sum(word.startswith(prefixes) for word in words)

D. sum(word.startswith(p) and p in prefixes for word in words)

Solution

1. Step 1: Understand requirement for multiple prefixes

   We need to check if each word starts with any prefix in the prefixes list.

2. Step 2: Analyze options for correct logic

   sum(any(word.startswith(p) for p in prefixes) for word in words) uses any() to correctly check if word starts with any prefix. sum(word.startswith(prefixes[0]) or word.startswith(prefixes[1]) for word in words) only checks first two prefixes (misses 'cow' with 'co'). sum(word.startswith(prefixes) for word in words) passes list to startswith (TypeError). sum(word.startswith(p) and p in prefixes for word in words) uses unbound p and incorrect and logic.

3. Final Answer:

   sum(any(word.startswith(p) for p in prefixes) for word in words) -> Option B

4. Quick Check:

   Use any() to check multiple prefixes per word [OK]

Hint: Use any() inside sum() to check multiple prefixes per word [OK]

Common Mistakes:

• Passing list directly to startswith instead of iterating
• Using 'and' instead of 'or' logic
• Hardcoding prefix checks instead of looping