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

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of prefix count function with simple input

What is the output of the following JavaScript code that counts words starting with a given prefix?

DSA Javascript

function countWordsWithPrefix(words, prefix) {
  let count = 0;
  for (const word of words) {
    if (word.startsWith(prefix)) {
      count++;
    }
  }
  return count;
}

const words = ['apple', 'app', 'application', 'banana', 'apex'];
console.log(countWordsWithPrefix(words, 'app'));

A 2

B 3

C 4

D 1

💡 Hint

Count how many words start exactly with 'app'.

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

🧠 Conceptual

intermediate

Understanding prefix matching in strings

Which of the following statements correctly describes how to check if a word starts with a given prefix in JavaScript?

A Use word.indexOf(prefix) === word.length - prefix.length to check prefix.

B Use word.includes(prefix) to check if the prefix is anywhere in the word.

C Use word.endsWith(prefix) to check if the word starts with the prefix.

D Use word.startsWith(prefix) to check if the word begins with the prefix.

💡 Hint

Think about the method that checks the start of a string.

The startsWith() method returns true if the string begins with the specified prefix.

🔧 Debug

advanced

Identify the error in prefix counting code

What error will the following code produce when run, and why?

DSA Javascript

function countPrefix(words, prefix) {
  let count = 0;
  for (let i = 0; i <= words.length; i++) {
    if (words[i].startsWith(prefix)) {
      count++;
    }
  }
  return count;
}

console.log(countPrefix(['cat', 'car', 'dog'], 'ca'));

A SyntaxError due to missing semicolon.

B Returns 0 because prefix is not found.

C TypeError: Cannot read property 'startsWith' of undefined because loop goes out of bounds.

D Works correctly and returns 2.

💡 Hint

Check the loop condition and array indexing.

The loop uses i <= words.length which causes words[words.length] to be undefined, so calling startsWith on undefined throws a TypeError.

🚀 Application

advanced

Count words with prefix ignoring case

Which code snippet correctly counts words starting with a prefix ignoring case differences?

A words.filter(word => word.toLowerCase().startsWith(prefix.toLowerCase())).length

B words.filter(word => word.startsWith(prefix)).length

C words.filter(word => word.includes(prefix.toLowerCase())).length

D words.filter(word => word.toUpperCase() === prefix.toUpperCase()).length

💡 Hint

Convert both word and prefix to the same case before checking.

To ignore case, convert both strings to lower (or upper) case before using startsWith.

🧠 Conceptual

expert

Time complexity of counting words with prefix

What is the time complexity of counting how many words start with a given prefix in a list of n words, each of average length m?

A O(n * m) because each word is checked character by character for the prefix.

B O(n^2) because nested loops are needed.

C O(m) because only prefix length matters.

D O(n + m) because prefix check is constant time.

💡 Hint

Consider how many words and how many characters per word are checked.

Each of the n words is checked for prefix of length up to m, so total is O(n * m).

Practice

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

easy

A. Counts how many words contain the prefix anywhere

B. Counts the total number of words in the list

C. Counts how many words end with the given prefix

D. Counts how many words start with the given prefix

Solution

1. Step 1: Understand the function purpose

   The function checks each word to see if it starts with the given prefix.

2. Step 2: Count matching words

   It counts only those words that begin exactly with the prefix, ignoring others.

3. Final Answer:

   Counts how many words start with the given prefix -> Option D

4. Quick Check:

   Starts with prefix = Count [OK]

Hint: Look for words starting exactly with prefix [OK]

Common Mistakes:

• Counting words containing prefix anywhere
• Counting words ending with prefix
• Counting total words instead of filtered ones

2. Which of the following is the correct syntax to check if a string word starts with a prefix prefix in JavaScript?

easy

A. word.beginWith(prefix)

B. word.hasPrefix(prefix)

C. word.startsWith(prefix)

D. word.starts(prefix)

Solution

1. Step 1: Recall JavaScript string methods

   JavaScript strings have a method called startsWith() to check prefix.

2. Step 2: Match method name

   Only startsWith() is the correct method; others do not exist.

3. Final Answer:

   word.startsWith(prefix) -> Option C

4. Quick Check:

   Correct method name = startsWith [OK]

Hint: Use startsWith() method for prefix check [OK]

Common Mistakes:

• Using non-existent methods like hasPrefix
• Using incorrect method names like beginWith
• Confusing startsWith with other string methods

3. What is the output of the following code?

const words = ['apple', 'app', 'application', 'banana'];
const prefix = 'app';
function countWordsWithPrefix(words, prefix) {
  let count = 0;
  for (const word of words) {
    if (word.startsWith(prefix)) {
      count++;
    }
  }
  return count;
}
console.log(countWordsWithPrefix(words, prefix));

medium

A. 3

B. 2

C. 1

D. 4

Solution

1. Step 1: Check each word against prefix 'app'

   'apple', 'app', and 'application' all start with 'app'. 'banana' does not.

2. Step 2: Count matching words

   There are 3 words starting with 'app', so count is 3.

3. Final Answer:

   3 -> Option A

4. Quick Check:

   Words starting with 'app' = 3 [OK]

Hint: Count words starting exactly with prefix [OK]

Common Mistakes:

• Counting words containing prefix anywhere
• Missing one of the matching words
• Counting all words regardless of prefix

4. The following code is intended to count words starting with a prefix, but it has a bug. What is the bug?

function countWordsWithPrefix(words, prefix) {
  let count = 0;
  for (let i = 0; i <= words.length; i++) {
    if (words[i].startsWith(prefix)) {
      count++;
    }
  }
  return count;
}

medium

A. startsWith method is used incorrectly

B. Loop goes out of bounds causing an error

C. count variable is not initialized

D. Prefix is not passed as a parameter

Solution

1. Step 1: Analyze loop condition

   The loop runs with i from 0 to words.length inclusive, but array indices go from 0 to words.length - 1.

2. Step 2: Identify out of bounds access

   At i = words.length, words[i] is undefined, so calling startsWith causes an error.

3. Final Answer:

   Loop goes out of bounds causing an error -> Option B

4. Quick Check:

   Loop index must be less than length [OK]

Hint: Use i < words.length, not <= [OK]

Common Mistakes:

• Using <= instead of < in loop condition
• Not checking array bounds before access
• Assuming startsWith works on undefined

5. Given the array words = ['Test', 'testing', 'Tester', 'tent', 'team'] and prefix 'te', which code snippet correctly counts words starting with the prefix ignoring case sensitivity?

hard

A. words.filter(w => w.toLowerCase().startsWith(prefix.toLowerCase())).length

B. words.filter(w => w.startsWith(prefix)).length

C. words.filter(w => w.includes(prefix)).length

D. words.filter(w => w.endsWith(prefix)).length

Solution

1. Step 1: Understand case sensitivity issue

   Direct startsWith is case sensitive, so 'Test' and 'test' differ.

2. Step 2: Use toLowerCase() to normalize

   Convert both word and prefix to lowercase before checking startsWith to ignore case.

3. Final Answer:

   words.filter(w => w.toLowerCase().startsWith(prefix.toLowerCase())).length -> Option A

4. Quick Check:

   Normalize case before prefix check [OK]

Hint: Convert both strings to lowercase before startsWith [OK]

Common Mistakes:

• Ignoring case sensitivity
• Using includes instead of startsWith
• Using endsWith instead of startsWith