Count Words with Given Prefix in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to return the length of the prefix string.

DSA C

int prefixLength(char *prefix) {
    return [1];
}

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

Astrlen(prefix)

Bsizeof(prefix)

Cprefix.length

Dlength(prefix)

Attempts:

3 left

💡 Hint

Common Mistakes

Using sizeof on a pointer instead of strlen

Using non-existent string length properties

The strlen function returns the length of a string in C.

2fill in blank

medium

Complete the code to compare the prefix with the start of the word.

DSA C

int startsWith(char *word, char *prefix) {
    int len = strlen(prefix);
    return strncmp(word, prefix, [1]) == 0;
}

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

Astrlen(word)

Blen

Csizeof(prefix)

Dstrlen(prefix) + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using length of word instead of prefix

Adding 1 unnecessarily

We compare only the first len characters to check if word starts with prefix.

3fill in blank

hard

Fix the error in the loop that counts words starting with the prefix.

DSA C

int countWordsWithPrefix(char **words, int wordsSize, char *prefix) {
    int count = 0;
    for (int i = 0; i < [1]; i++) {
        if (startsWith(words[i], prefix)) {
            count++;
        }
    }
    return count;
}

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

Aprefix

Bsizeof(words)

Cstrlen(words)

DwordsSize

Attempts:

3 left

💡 Hint

Common Mistakes

Using prefix length instead of wordsSize

Using sizeof on pointer which gives wrong size

The loop should run from 0 to wordsSize to cover all words.

4fill in blank

hard

Fill both blanks to create a function that counts words starting with the prefix.

DSA C

int countWordsWithPrefix(char **words, int wordsSize, char *prefix) {
    int count = 0;
    for (int [1] = 0; [2] < wordsSize; [2]++) {
        if (startsWith(words[i], prefix)) {
            count++;
        }
    }
    return count;
}

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

Ai

Bj

Ck

Dindex

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variable names in loop initialization and condition

The loop variable should be consistently named i in initialization and condition.

5fill in blank

hard

Fill all three blanks to implement the full function counting words with the prefix.

DSA C

int countWordsWithPrefix(char **words, int wordsSize, char *prefix) {
    int count = 0;
    int prefixLen = [1](prefix);
    for (int i = 0; i < [2]; i++) {
        if (strncmp(words[i], prefix, [3]) == 0) {
            count++;
        }
    }
    return count;
}

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

Astrlen

BwordsSize

CprefixLen

Dsizeof

Attempts:

3 left

💡 Hint

Common Mistakes

Using sizeof instead of strlen

Using wrong loop limit

Comparing wrong number of characters

We get the prefix length with strlen, loop over wordsSize, and compare first prefixLen characters.

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 given prefix

B. Count the total number of characters in all words

C. Find the longest word in the list

D. Sort the words alphabetically

Solution

1. Step 1: Understand the problem statement

   The problem asks to find how many words start with a specific prefix string.

2. Step 2: Identify the key operation

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

3. Final Answer:

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

4. Quick Check:

   Count words starting with prefix = Count how many words in a list start with a given prefix [OK]

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

Common Mistakes:

• Counting total characters instead of words
• Looking for words containing prefix anywhere
• Sorting words instead of counting

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

easy

A. strcmp(word, prefix) == 0

B. strncmp(word, prefix, strlen(prefix)) == 0

C. strstr(word, prefix) != NULL

D. strlen(word) == strlen(prefix)

Solution

1. Step 1: Understand string comparison functions

   strncmp compares first n characters; here n = length of prefix.

2. Step 2: Check which code confirms prefix match

   strncmp(word, prefix, strlen(prefix)) == 0 means word starts with prefix exactly.

3. Final Answer:

   strncmp(word, prefix, strlen(prefix)) == 0 -> Option B

4. Quick Check:

   strncmp with prefix length = 0 means prefix match [OK]

Hint: Use strncmp with prefix length to check start [OK]

Common Mistakes:

• Using strcmp which requires full string match
• Using strstr without checking start position
• Comparing lengths only

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

char *words[] = {"apple", "app", "application", "banana"};
char prefix[] = "app";
int count = 0;
for (int i = 0; i < 4; i++) {
  if (strncmp(words[i], prefix, strlen(prefix)) == 0) {
    count++;
  }
}
printf("%d\n", count);

medium

A. 4

B. 2

C. 1

D. 3

Solution

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

   "apple" starts with "app" (yes), "app" (yes), "application" (yes), "banana" (no).

2. Step 2: Count matching words

   Three words match the prefix, so count = 3.

3. Final Answer:

   3 -> Option D

4. Quick Check:

   Words starting with "app" = 3 [OK]

Hint: Count words where strncmp returns 0 for prefix length [OK]

Common Mistakes:

• Counting words that contain prefix anywhere
• Off-by-one errors in loop
• Incorrect prefix length in strncmp

4. Find the bug in this C code that counts words starting with prefix prefix:

int countWords(char *words[], int n, char *prefix) {
  int count = 0;
  for (int i = 0; i <= n; i++) {
    if (strncmp(words[i], prefix, strlen(prefix)) == 0) {
      count++;
    }
  }
  return count;
}

medium

A. Initialize count to 1 instead of 0

B. Use strcmp instead of strncmp

C. Loop condition should be i < n, not i <= n

D. strlen(prefix) should be replaced with strlen(words[i])

Solution

1. Step 1: Analyze loop boundary

   Loop runs from i = 0 to i <= n, which accesses words[n] (out of bounds).

2. Step 2: Correct loop condition

   Change to i < n to avoid accessing beyond array size.

3. Final Answer:

   Loop condition should be i < n, not i <= n -> Option C

4. Quick Check:

   Array index must be less than size [OK]

Hint: Use i < n for array loops to avoid overflow [OK]

Common Mistakes:

• Using <= instead of < in loops
• Confusing strcmp and strncmp usage
• Wrong length argument in strncmp

5. You have a list of words and want to count how many start with prefix pre. Which approach is best to optimize counting if you must do many prefix queries efficiently?

hard

A. Build a trie (prefix tree) from words and count nodes matching prefix

B. Use strncmp on each word for every query

C. Sort words and use binary search for prefix range

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

Solution

1. Step 1: Understand multiple prefix queries

   Checking each word with strncmp repeatedly is slow for many queries.

2. Step 2: Choose data structure for fast prefix counting

   Trie stores words by prefixes, allowing quick count of words with given prefix.

3. Step 3: Compare alternatives

   Binary search helps but trie is more direct for prefix counts; hash set doesn't support prefix queries well.

4. Final Answer:

   Build a trie (prefix tree) from words and count nodes matching prefix -> Option A

5. Quick Check:

   Trie = fast prefix count [OK]

Hint: Use trie for fast multiple prefix queries [OK]

Common Mistakes:

• Using linear search for many queries
• Ignoring trie benefits for prefix problems
• Using hash set which doesn't support prefix search