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]);

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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]) {

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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++) {

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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];
}

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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;
}

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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.