Substring Search Patterns in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to find the first occurrence of a substring in a string using strstr.

DSA C

char *result = strstr(text, [1]);

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

Apattern

Btext

CNULL

Dresult

Attempts:

3 left

💡 Hint

Common Mistakes

Passing the whole text as the substring to search for.

Using NULL instead of the substring.

Using the result pointer as the substring.

The strstr function searches for the first occurrence of the substring pattern in text.

2fill in blank

medium

Complete the code to compare two characters in the text and pattern during substring search.

DSA C

if (text[i] == [1][j]) {

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

Asubstring

Btext

Cresult

Dpattern

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing text[i] with text[j] instead of pattern[j].

Using result or substring variables incorrectly.

We compare the character at position i in text with the character at position j in pattern to check for a match.

3fill in blank

hard

Fix the error in the loop condition to avoid reading beyond the text length during substring search.

DSA C

for (int i = 0; i <= text_length - [1]; i++) {

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

Atext_length

Bpattern_length

Cresult_length

Dsubstring_length

Attempts:

3 left

💡 Hint

Common Mistakes

Using text_length instead of pattern_length in the subtraction.

Using result_length or substring_length which are undefined.

The loop should run until i reaches text_length - pattern_length to avoid going past the end of the text when checking for the pattern.

4fill in blank

hard

Fill both blanks to correctly check if the substring matches the text segment.

DSA C

if (j == [1]) {
    return [2];
}

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

Apattern_length

Bi

C0

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Returning 0 or 1 instead of the starting index.

Checking j against text_length instead of pattern_length.

When j equals pattern_length, it means the whole pattern matched, so we return the starting index i of the match in the text.

5fill in blank

hard

Fill all three blanks to complete the naive substring search function.

DSA C

int naive_search(char *text, char *pattern) {
    int text_len = strlen(text);
    int pattern_len = strlen(pattern);
    for (int i = 0; i <= text_len - [1]; i++) {
        int j = 0;
        while (j < [2] && text[i + j] == pattern[[3]]) {
            j++;
        }
        if (j == pattern_len) {
            return i;
        }
    }
    return -1;
}

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

Apattern_len

Cj

Di

Attempts:

3 left

💡 Hint

Common Mistakes

Using i instead of j to index pattern characters.

Using text_len instead of pattern_len in loop conditions.

The loop runs until i reaches text_len - pattern_len. The inner while loop checks characters while j is less than pattern_len and characters match. The index j is used to access the pattern characters.

Practice

1. What is the main goal of substring search algorithms in strings?

easy

A. To find if a smaller string exists inside a larger string

B. To sort the characters of a string alphabetically

C. To reverse the entire string

D. To count the total number of characters in a string

Solution

1. Step 1: Understand substring search purpose

   Substring search algorithms check if a smaller string (pattern) appears inside a bigger string (text).

2. Step 2: Compare options with definition

   Only To find if a smaller string exists inside a larger string describes finding a smaller string inside a larger one, which matches substring search.

3. Final Answer:

   To find if a smaller string exists inside a larger string -> Option A

4. Quick Check:

   Substring search = find smaller string inside larger [OK]

Hint: Substring search means finding a smaller string inside a bigger one [OK]

Common Mistakes:

• Confusing substring search with sorting or reversing strings
• Thinking substring search counts characters
• Mixing substring search with string length calculation

2. Which of the following is the correct syntax to check if string pattern is a substring of string text in C?

easy

A. if (text.find(pattern)) { /* found */ }

B. if (text.contains(pattern)) { /* found */ }

C. if (pattern in text) { /* found */ }

D. if (strstr(text, pattern) != NULL) { /* found */ }

Solution

1. Step 1: Recall C substring search function

   In C, strstr() returns a pointer to the first occurrence of pattern in text, or NULL if not found.

2. Step 2: Check each option's syntax

   if (strstr(text, pattern) != NULL) { /* found */ } uses strstr correctly with NULL check. Options A, B, C use syntax from other languages or invalid in C.

3. Final Answer:

   if (strstr(text, pattern) != NULL) { /* found */ } -> Option D

4. Quick Check:

   C substring search uses strstr() [OK]

Hint: Use strstr() in C to find substring, check for NULL [OK]

Common Mistakes:

• Using methods from other languages like contains() or find()
• Not checking if strstr() returns NULL
• Confusing order of arguments in strstr()

3. What will be the output of the following C code snippet?

char text[] = "hello world";
char *pos = strstr(text, "wor");
if (pos != NULL) {
    printf("%ld", pos - text);
} else {
    printf("-1");
}

medium

A. 6

B. 0

C. -1

D. 3

Solution

1. Step 1: Understand strstr() return value

   strstr(text, "wor") returns pointer to first occurrence of "wor" in "hello world", which starts at index 6.

2. Step 2: Calculate pointer difference

   Subtracting text pointer from pos gives index 6, which is printed.

3. Final Answer:

   6 -> Option A

4. Quick Check:

   Substring "wor" starts at index 6 [OK]

Hint: Pointer difference in strstr() gives substring start index [OK]

Common Mistakes:

• Printing pointer value instead of index difference
• Assuming substring starts at 0 or 3 incorrectly
• Not checking if strstr() returns NULL before subtracting

4. Identify the error in the following C code for substring search:

char *find_substring(char *text, char *pattern) {
    for (int i = 0; i <= strlen(text); i++) {
        if (strncmp(&text[i], pattern, strlen(pattern)) == 0) {
            return &text[i];
        }
    }
    return NULL;
}

medium

A. Function should return int index, not pointer

B. strncmp() is used incorrectly; should use strcmp() instead

C. The loop condition should be 'i < strlen(text) - strlen(pattern) + 1' to avoid overflow

D. No error; code works correctly

Solution

1. Step 1: Analyze loop boundary

   The loop runs from 0 to strlen(text), but comparing substrings beyond text length minus pattern length causes overflow.

2. Step 2: Correct loop condition

   Loop should run until i <= strlen(text) - strlen(pattern) to safely compare substrings of pattern length.

3. Final Answer:

   The loop condition should be 'i < strlen(text) - strlen(pattern) + 1' to avoid overflow -> Option C

4. Quick Check:

   Loop boundary must prevent substring overflow [OK]

Hint: Loop till text length minus pattern length to avoid overflow [OK]

Common Mistakes:

• Looping beyond valid substring start positions
• Using strcmp instead of strncmp for partial comparison
• Ignoring return type correctness

5. You want to find all occurrences of substring pat in string txt using C. Which approach correctly finds all start indices of pat in txt?

hard

A. Use strstr() once and return the first found position only

B. Use a loop with strstr() starting from last found position + 1 until strstr() returns NULL

C. Use strcmp() on entire text and pattern directly

D. Use strlen() to compare lengths and decide if substring exists

Solution

1. Step 1: Understand strstr() usage for multiple occurrences

   strstr() returns pointer to first occurrence. To find all, repeatedly call strstr() starting after last found position.

2. Step 2: Evaluate options

   Use a loop with strstr() starting from last found position + 1 until strstr() returns NULL correctly describes looping with strstr() advancing pointer. Use strstr() once and return the first found position only finds only first occurrence. Options C and D do not find all occurrences.

3. Final Answer:

   Use a loop with strstr() starting from last found position + 1 until strstr() returns NULL -> Option B

4. Quick Check:

   Loop with strstr() to find all occurrences [OK]

Hint: Loop with strstr() advancing pointer to find all matches [OK]

Common Mistakes:

• Using strstr() only once for all matches
• Using strcmp() which compares whole strings only
• Relying on length checks without searching