Count Words with Given Prefix in DSA Python - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check if a word starts with the given prefix.

DSA Python

if word.[1](prefix):
    count += 1

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

Astartswith

Bendswith

Cfind

Dcontains

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'endswith' instead of 'startswith'.

Using 'find' which returns an index, not a boolean.

The startswith method checks if the word begins with the prefix.

2fill in blank

medium

Complete the code to loop through each word in the list.

DSA Python

for [1] in words:
    if word.startswith(prefix):
        count += 1

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

Aprefix

Bitem

Cword

Dcount

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'prefix' which is the prefix string, not the loop variable.

Using 'count' which is a number, not a word.

We use word as the variable to represent each element in the list words.

3fill in blank

hard

Fix the error in the return statement to return the count.

DSA Python

return [1]

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

Aprefix

Bcount

Cword

Dwords

Attempts:

3 left

💡 Hint

Common Mistakes

Returning the list of words instead of the count.

Returning the prefix string instead of the count.

The function should return the variable count which holds the number of words with the prefix.

4fill in blank

hard

Complete the code to create a dictionary comprehension counting word lengths for words with prefix.

DSA Python

{word: len(word) for word in words if word.[1](prefix)}

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

Bstartswith

Cendswith

Dcontains

Attempts:

3 left

💡 Hint

Common Mistakes

Adding extra characters after word in the key.

Using 'endswith' or 'contains' instead of 'startswith'.

The dictionary comprehension uses the word itself as the key (no extra characters), so blank 1 is empty. The condition uses startswith to check the prefix.

5fill in blank

hard

Fill all three blanks to create a dictionary of uppercase words with their lengths if they start with prefix.

DSA Python

{word[1]: [2](word) for word in words if word.[3](prefix)}

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

A.upper()

Blen

Cstartswith

Dlower

Attempts:

3 left

💡 Hint

Common Mistakes

Using lower instead of upper.

Using endswith instead of startswith.

The key is the uppercase version of the word (word.upper()), the value is the length (len(word)), and the condition checks if the word starts with the prefix (startswith).

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