Count Words with Given Prefix in DSA C - Time & Space Complexity

Time Complexity: Count Words with Given Prefix

O(n * m)

Understanding Time Complexity

We want to understand how the time needed to count words with a certain prefix grows as the input changes.

Specifically, how does the work increase when we have more words or longer prefixes?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

This code counts how many words in the array start with the given prefix.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Loop through all words and compare prefix characters.
• How many times: The loop runs once for each word (n times), and each comparison checks up to prefix length characters.

How Execution Grows With Input

As the number of words (n) grows, the code checks each word once. For each word, it compares characters up to the prefix length.

Input Size (n)	Approx. Operations
10	About 10 x prefix length comparisons
100	About 100 x prefix length comparisons
1000	About 1000 x prefix length comparisons

Pattern observation: The work grows linearly with the number of words and linearly with the prefix length.

Final Time Complexity

Time Complexity: O(n * m)

This means the time grows proportionally with the number of words (n) and the length of the prefix (m).

Common Mistake

[X] Wrong: "The time depends only on the number of words, not the prefix length."

[OK] Correct: Each word comparison checks characters up to the prefix length, so longer prefixes mean more work per word.

Interview Connect

Understanding how input size and prefix length affect performance helps you explain your code clearly and choose better approaches if needed.

Self-Check

"What if the words were stored in a trie data structure? How would the time complexity change?"

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