Count Words with Given Prefix in DSA Python - Cheat Sheet & Quick Revision

Recall & Review

beginner

What does 'Count Words with Given Prefix' mean?

It means finding how many words in a list start with a specific set of letters called a prefix.

beginner

Why is a prefix important in searching words?

A prefix helps to quickly find words that begin with certain letters without checking the whole word.

beginner

How can you count words with a given prefix using a simple loop?

Check each word in the list. If the word starts with the prefix, add 1 to the count. At the end, the count shows how many words matched.

beginner

What Python string method helps to check if a word starts with a prefix?

The method is called startswith(). It returns True if the word starts with the prefix, otherwise False.

intermediate

What is a more efficient data structure than a list for prefix counting in many queries?

A Trie (prefix tree) is more efficient because it stores words by their prefixes, allowing fast counting of words with a given prefix.

Which Python method checks if a string starts with a specific prefix?

A startswith()

B endswith()

C contains()

D find()

startswith() returns True if the string begins with the given prefix.

What does the prefix in 'Count Words with Given Prefix' refer to?

A The ending letters of a word

B The starting letters of a word

C Any letters inside a word

D The middle letters of a word

A prefix is the starting part of a word.

If you have a list of words and want to count how many start with 'pre', what is the simplest approach?

A Sort the list and pick the first word

B Count words that end with 'pre'

C Use a loop and check each word with startswith('pre')

D Use a dictionary to store word lengths

Looping and checking each word with startswith() is the direct way.

Which data structure is best for fast prefix queries when many queries are needed?

A Trie

B List

C Stack

D Queue

Trie stores words by prefixes, making prefix queries efficient.

What is the output of counting words with prefix 'a' in ['apple', 'banana', 'apricot', 'cherry']?

A 1

B 3

C 4

D 2

'apple' and 'apricot' start with 'a', so count is 2.

Explain how to count words with a given prefix using a simple Python approach.

Think about checking each word one by one.

Describe why a Trie is useful for counting words with a given prefix when there are many queries.

Imagine a tree where each branch is a letter.

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