Count Words with Given Prefix in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

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

DSA C++

bool startsWith(const string& word, const string& prefix) {
    return word.[1](0, prefix.size()) == prefix;
}

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

Areplace

Bfind

Ccompare

Dsubstr

Attempts:

3 left

💡 Hint

Common Mistakes

Using find instead of substr returns the position, not the substring.

Using compare returns an int, not a string.

The substr function extracts a substring from the word starting at index 0 with length equal to prefix size, which is then compared to the prefix.

2fill in blank

medium

Complete the code to count how many words start with the given prefix.

DSA C++

int countPrefix(const vector<string>& words, const string& prefix) {
    int count = 0;
    for (const auto& word : words) {
        if (word.size() >= prefix.size() && word.[1](0, prefix.size()) == prefix) {
            count++;
        }
    }
    return count;
}

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Areplace

Bfind

Csubstr

Dcompare

Attempts:

3 left

💡 Hint

Common Mistakes

Using find returns position, which is not suitable for prefix check.

Using compare returns int, not string.

Using substr to extract the prefix part of each word allows comparison with the given prefix to count matches.

3fill in blank

hard

Fix the error in the code to correctly count words starting with the prefix.

DSA C++

int countPrefix(const vector<string>& words, const string& prefix) {
    int count = 0;
    for (const auto& word : words) {
        if (word.size() >= prefix.size() && word.[1](0, prefix.size(), prefix) == 0) {
            count++;
        }
    }
    return count;
}

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Acompare

Bsubstr

Cfind

Dreplace

Attempts:

3 left

💡 Hint

Common Mistakes

Using find returns position, not 0 for match.

Using substr returns string, cannot be compared to 0.

The compare function can compare the prefix with the start of the word and returns 0 if they match.

4fill in blank

hard

Fill both blanks to create a map of prefix counts for given words.

DSA C++

unordered_map<string, int> prefixCount(const vector<string>& words, int prefixLen) {
    unordered_map<string, int> counts;
    for (const auto& word : words) {
        if (word.size() >= [1]) {
            string prefix = word.[2](0, prefixLen);
            counts[prefix]++;
        }
    }
    return counts;
}

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AprefixLen

Bword.size()

Csubstr

Dfind

Attempts:

3 left

💡 Hint

Common Mistakes

Using find instead of substr to extract prefix.

Checking word.size() < prefixLen instead of >=.

Check if word length is at least prefixLen, then extract prefix using substr from index 0 with length prefixLen.

5fill in blank

hard

Fill all three blanks to return the count of words starting with prefix using a loop and condition.

DSA C++

int countWordsWithPrefix(const vector<string>& words, const string& prefix) {
    int count = 0;
    for (const auto& [1] : words) {
        if ([2].[3](0, prefix.size()) == prefix) {
            count++;
        }
    }
    return count;
}

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

Aword

Bwords

Csubstr

Dfind

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'words' as loop variable instead of 'word'.

Using find instead of substr for prefix extraction.

Use 'word' as loop variable, then call word.substr(0, prefix.size()) to compare with prefix.

Practice

1. What does the problem "Count Words with Given Prefix" ask you to do?

easy

A. Count how many words in a list start with a specific prefix

B. Count the total number of words in a list

C. Find the longest word in a list

D. Replace the prefix in all words with another string

Solution

1. Step 1: Understand the problem statement

   The problem asks to find how many words start with a given prefix from a list of words.

2. Step 2: Identify the key operation

   We need to check each word and count those that begin with the prefix.

3. Final Answer:

   Count how many words in a list start with a specific prefix -> Option A

4. Quick Check:

   Counting words with prefix = Count how many words in a list start with a specific prefix [OK]

Hint: Focus on words starting with prefix, not total words [OK]

Common Mistakes:

• Counting all words instead of only those with prefix
• Looking for longest word instead of counting
• Replacing prefix instead of counting

2. Which of the following C++ code snippets correctly checks if a string word starts with prefix pref?

easy

A. if (word.compare(pref) == 1) { /* starts with prefix */ }

B. if (word == pref) { /* starts with prefix */ }

C. if (word.substr(1, pref.size()) == pref) { /* starts with prefix */ }

D. if (word.find(pref) == 0) { /* starts with prefix */ }

Solution

1. Step 1: Understand string::find usage

   Using word.find(pref) == 0 checks if pref is found at the start (index 0) of word.

2. Step 2: Check other options

   if (word == pref) { /* starts with prefix */ } checks equality, not prefix. if (word.substr(1, pref.size()) == pref) { /* starts with prefix */ } checks substring starting at index 1, which is wrong. if (word.compare(pref) == 1) { /* starts with prefix */ } compares strings but result 1 means greater, not prefix.

3. Final Answer:

   if (word.find(pref) == 0) { /* starts with prefix */ } -> Option D

4. Quick Check:

   find(pref) == 0 means prefix match [OK]

Hint: Use find(pref) == 0 to check prefix in C++ strings [OK]

Common Mistakes:

• Using equality instead of prefix check
• Checking substring from wrong index
• Misunderstanding compare() return values

3. What is the output of the following C++ code?

std::vector words = {"apple", "app", "application", "banana"};
std::string prefix = "app";
int count = 0;
for (const auto& w : words) {
    if (w.find(prefix) == 0) {
        count++;
    }
}
std::cout << count << std::endl;

medium

A. 3

B. 2

C. 1

D. 4

Solution

1. Step 1: Check each word for prefix "app"

   "apple" starts with "app" (count=1), "app" starts with "app" (count=2), "application" starts with "app" (count=3), "banana" does not.

2. Step 2: Count total matches

   Total words starting with "app" are 3.

3. Final Answer:

   3 -> Option A

4. Quick Check:

   Words starting with "app" = 3 [OK]

Hint: Count words where find(prefix) == 0 inside loop [OK]

Common Mistakes:

• Counting words that contain prefix anywhere
• Off-by-one errors in counting
• Ignoring words that exactly equal prefix

4. Identify the error in this C++ code snippet that tries to count words starting with prefix pref:

int count = 0;
for (auto w : words) {
    if (w.find(pref)) {
        count++;
    }
}

medium

A. The count variable should be initialized to -1

B. The loop variable should be a reference: auto& w

C. The condition should check if w.find(pref) == 0, not just w.find(pref)

D. The find function should be replaced with substr

Solution

1. Step 1: Understand find() return value

   find() returns position or npos. Using if(w.find(pref)) treats 0 as false, so words starting at index 0 are skipped.

2. Step 2: Correct condition for prefix check

   We must check if w.find(pref) == 0 to confirm prefix at start.

3. Final Answer:

   The condition should check if w.find(pref) == 0, not just w.find(pref) -> Option C

4. Quick Check:

   Check find() == 0 for prefix [OK]

Hint: Always compare find() result to 0 for prefix check [OK]

Common Mistakes:

• Using if(w.find(pref)) which fails for prefix at start
• Not using reference in loop (less critical here)
• Initializing count incorrectly

5. Given a list of words and a prefix, how would you efficiently count words starting with the prefix if the list is very large and queried multiple times?
Choose the best approach.

hard

A. Check each word with find(prefix) == 0 every time a query comes

B. Use a Trie (prefix tree) to store words and count prefix matches quickly

C. Sort the list and use binary search to find prefix range each time

D. Store words in a hash set and check prefix membership

Solution

1. Step 1: Understand repeated prefix queries

   Checking each word every time is slow for large lists and many queries.

2. Step 2: Use Trie data structure

   A Trie stores words by prefixes, allowing fast counting of words with given prefix in O(length of prefix) time.

3. Step 3: Evaluate other options

   Binary search after sorting helps but is less efficient for many queries. Hash set doesn't support prefix queries efficiently.

4. Final Answer:

   Use a Trie (prefix tree) to store words and count prefix matches quickly -> Option B

5. Quick Check:

   Trie = fast prefix counting [OK]

Hint: Use Trie for fast repeated prefix counts [OK]

Common Mistakes:

• Checking all words each query (slow)
• Using hash set which doesn't support prefix queries
• Ignoring Trie advantages for prefix problems