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 change when the strings get longer?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

bool containsSubstring(const std::string &text, const std::string &pattern) {
    int n = text.size();
    int m = pattern.size();
    for (int i = 0; i <= n - m; i++) {
        int j = 0;
        while (j < m && text[i + j] == pattern[j]) {
            j++;
        }
        if (j == m) return true;
    }
    return false;
}
    

This code checks if the pattern appears anywhere inside the text by comparing characters one by one.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The outer loop moves through each possible start position in the text.
• How many times: Up to n - m + 1 times, where n is text length and m is pattern length.
• Secondary operation: The inner while loop compares characters of pattern and text.
• How many times: Up to m times for each outer loop iteration in the worst case.
• Dominant operation: The nested character comparisons inside the loops.

How Execution Grows With Input

As the text and pattern get longer, the number of character comparisons grows quickly.

Input Size (n)	Approx. Operations
10 (text), 3 (pattern)	Up to 8 * 3 = 24 comparisons
100 (text), 10 (pattern)	Up to 91 * 10 = 910 comparisons
1000 (text), 50 (pattern)	Up to 951 * 50 = 47,550 comparisons

Pattern observation: The operations grow roughly by multiplying text length by pattern length.

Final Time Complexity

Time Complexity: O(n * m)

This means the time needed grows proportionally to the product of text length and pattern length.

Common Mistake

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

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

Interview Connect

Understanding this helps you explain why simple substring search can be slow and why smarter algorithms exist.

Self-Check

"What if we used a more advanced algorithm like KMP? How would the time complexity change?"

Practice

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

easy

A. To count the number of vowels in a string

B. To sort the characters of a string alphabetically

C. To reverse the entire string

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

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 instead of finding patterns

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) != std::string::npos) { /* found */ }

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

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

D. if (text.search(pattern) == true) { /* found */ }

Solution

1. Step 1: Recall C++ substring search syntax

   In C++, std::string::find returns the position of the substring or std::string::npos if not found.

2. Step 2: Analyze each option

   if (text.find(pattern) != std::string::npos) { /* found */ } uses text.find(pattern) != std::string::npos, which is correct. Others use invalid or non-existent methods.

3. Final Answer:

   if (text.find(pattern) != std::string::npos) { /* found */ } -> Option A

4. Quick Check:

   Use find() and compare with npos [OK]

Hint: Use text.find(pattern) != npos to check substring [OK]

Common Mistakes:

• Using methods like contains or indexOf which don't exist in std::string
• Comparing find() result directly to true or false

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

std::string text = "hello world";
std::string pattern = "world";
size_t pos = text.find(pattern);
if (pos != std::string::npos) {
    std::cout << pos << std::endl;
} else {
    std::cout << -1 << std::endl;
}

medium

A. 5

B. -1

C. 6

D. 0

Solution

1. Step 1: Find substring position in text

   The substring "world" starts at index 6 in "hello world" (counting from 0).

2. Step 2: Check condition and output

   Since find returns 6 (not npos), the code prints 6.

3. Final Answer:

   6 -> Option C

4. Quick Check:

   "world" starts at index 6 in "hello world" [OK]

Hint: Remember string indices start at 0 in C++ [OK]

Common Mistakes:

• Counting index starting at 1 instead of 0
• Printing -1 when substring is actually found

4. Identify the error in this substring search code snippet:

std::string text = "example";
std::string pattern = "amp";
if (text.find(pattern) == -1) {
    std::cout << "Not found" << std::endl;
} else {
    std::cout << "Found" << std::endl;
}

medium

A. Comparing find() result to -1 instead of std::string::npos

B. Using std::cout incorrectly

C. Pattern string is empty

D. Missing semicolon after if statement

Solution

1. Step 1: Understand find() return value

   std::string::find returns std::string::npos if substring not found, not -1.

2. Step 2: Identify incorrect comparison

   Comparing to -1 is wrong; must compare to std::string::npos for correctness.

3. Final Answer:

   Comparing find() result to -1 instead of std::string::npos -> Option A

4. Quick Check:

   Use npos, not -1, to check find() failure [OK]

Hint: Always compare find() result with std::string::npos [OK]

Common Mistakes:

• Using -1 instead of std::string::npos for not found
• Assuming find() returns boolean

5. Given a large text and multiple patterns, which substring search approach is best to efficiently find all pattern occurrences?

// Patterns: ["he", "she", "his", "hers"]
// Text: "ahishers"

hard

A. Use Rabin-Karp algorithm with hashing for each pattern

B. Build an Aho-Corasick automaton for all patterns and search once

C. Use naive search for each pattern separately

D. Use std::string::find in a loop for each pattern

Solution

1. Step 1: Understand multiple pattern search problem

   Searching many patterns separately is inefficient; better to search all at once.

2. Step 2: Identify best algorithm for multiple patterns

   Aho-Corasick builds a combined automaton to find all patterns in one pass efficiently.

3. Step 3: Compare other options

   Naive and Rabin-Karp per pattern or repeated find calls are slower for many patterns.

4. Final Answer:

   Build an Aho-Corasick automaton for all patterns and search once -> Option B

5. Quick Check:

   Multiple patterns = Aho-Corasick automaton [OK]

Hint: Use Aho-Corasick for multiple pattern substring search [OK]

Common Mistakes:

• Using naive search for each pattern separately
• Applying single-pattern algorithms repeatedly
• Ignoring combined automaton approach