Substring Search Patterns in DSA C - Time & Space Complexity

Time Complexity: Substring Search Patterns

O(n * m)

Understanding Time Complexity

We want to understand how long it takes to find a smaller string inside a bigger string.

How does the time needed grow when the strings get longer?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

int substring_search(const char *text, const char *pattern) {
    int n = strlen(text);
    int m = strlen(pattern);
    for (int i = 0; i <= n - m; i++) {
        int j = 0;
        while (j < m && text[i + j] == pattern[j]) {
            j++;
        }
        if (j == m) {
            return i; // pattern found at index i
        }
    }
    return -1; // pattern not found
}
    

This code looks for the first place where the pattern appears inside the text.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The outer loop runs from start to end of text minus pattern length.
• Nested operation: The inner while loop compares characters of pattern and text one by one.
• How many times: Outer loop runs about n - m + 1 times; inner loop runs up to m times per outer loop.

How Execution Grows With Input

As the text and pattern get longer, the number of comparisons grows roughly by multiplying their lengths.

Input Size (n)	Approx. Operations
10 (text), 3 (pattern)	About 8 * 3 = 24 comparisons
100 (text), 5 (pattern)	About 96 * 5 = 480 comparisons
1000 (text), 10 (pattern)	About 991 * 10 = 9,910 comparisons

Pattern observation: The work grows roughly by the product of text length and pattern length.

Final Time Complexity

Time Complexity: O(n * m)

This means the time needed grows roughly by multiplying the length of the text and the pattern.

Common Mistake

[X] Wrong: "The search always takes time proportional to the text length only."

[OK] Correct: Because each position in the text may require checking many characters of the pattern, so pattern length matters too.

Interview Connect

Understanding how substring search time grows helps you explain and improve searching methods in real problems.

Self-Check

"What if we used a smarter algorithm like KMP that avoids re-checking characters? How would the time complexity change?"

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