Substring Search Patterns in DSA Python - 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.

def substring_search(text: str, pattern: str) -> int:
    for i in range(len(text) - len(pattern) + 1):
        match = True
        for j in range(len(pattern)):
            if text[i + j] != pattern[j]:
                match = False
                break
        if match:
            return i
    return -1

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: Checking each character of the pattern against the text.
• How many times: For each possible start in the text, it checks up to the length of the pattern.

How Execution Grows With Input

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

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

Pattern observation: The work grows roughly by n times m, so doubling text or pattern length roughly doubles the work.

Final Time Complexity

Time Complexity: O(n * m)

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

Common Mistake

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

[OK] Correct: Because for each position in the text, we may need to check every character of the pattern, so pattern length matters too.

Interview Connect

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

Self-Check

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

Practice

1. What does the in operator check when used like 'abc' in 'abcdef' in Python?

easy

A. If the substring 'abc' exists anywhere inside the string 'abcdef'

B. If the string 'abcdef' starts with 'abc'

C. If the string 'abcdef' ends with 'abc'

D. If the strings 'abc' and 'abcdef' are exactly equal

Solution

1. Step 1: Understand the in operator for strings

   The in operator checks if the left string is found anywhere inside the right string.

2. Step 2: Apply to given strings

   Since 'abc' is at the start of 'abcdef', it is found inside it.

3. Final Answer:

   If the substring 'abc' exists anywhere inside the string 'abcdef' -> Option A

4. Quick Check:

   'abc' in 'abcdef' = True [OK]

Hint: Remember: 'in' checks anywhere inside the string [OK]

Common Mistakes:

• Thinking 'in' only checks start of string
• Confusing 'in' with equality
• Assuming 'in' checks only end of string

2. Which of the following is the correct syntax to find the index of substring 'cat' in string s = 'concatenate' using Python?

easy

A. s.indexOf('cat')

B. find(s, 'cat')

C. s.find('cat')

D. index(s, 'cat')

Solution

1. Step 1: Recall Python string method for substring index

   Python strings have a method called find() to locate substring index.

2. Step 2: Check method usage

   The correct syntax is s.find('cat'). Other options are invalid in Python.

3. Final Answer:

   s.find('cat') -> Option C

4. Quick Check:

   Use string.find(substring) syntax [OK]

Hint: Use string.find(substring) to get index [OK]

Common Mistakes:

• Using non-Python syntax like indexOf
• Calling find as a standalone function
• Confusing method name or order of arguments

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

text = 'abracadabra'
pos = text.find('ra')
print(pos)

medium

A. 2

B. 1

C. 3

D. -1

Solution

1. Step 1: Understand find() method behavior

   The find() method returns the lowest index where the substring is found or -1 if not found.

2. Step 2: Locate substring 'ra' in 'abracadabra'

   The substring 'ra' first appears starting at index 1 (abracadabra).

3. Final Answer:

   1 -> Option B

4. Quick Check:

   First 'ra' at index 1 [OK]

Hint: Remember: find() returns first occurrence index [OK]

Common Mistakes:

• Assuming find() returns all occurrences
• Confusing index with count
• Mistaking find() return -1 for found substring

4. Identify the error in the following code snippet that tries to check if substring 'dog' is in string s:

s = 'hotdog'
if s.contains('dog'):
    print('Found')
else:
    print('Not Found')

medium

A. The method contains does not exist for strings in Python

B. The substring 'dog' is misspelled

C. The string s is not defined

D. The print statements are incorrect

Solution

1. Step 1: Check string method availability in Python

   Python strings do not have a contains() method; membership is checked with in.

2. Step 2: Correct way to check substring presence

   Use 'dog' in s instead of s.contains('dog').

3. Final Answer:

   The method contains does not exist for strings in Python -> Option A

4. Quick Check:

   Use in operator, not contains() [OK]

Hint: Use in for substring check, not contains() [OK]

Common Mistakes:

• Using Java-like contains() method in Python
• Misspelling substring or variable
• Assuming print syntax is wrong

5. Given a string s = 'banana', which code snippet correctly finds all starting indices of substring 'ana' in s?

hard

A. indices = [s.find('ana')]

B. indices = [i for i in range(len(s)) if 'ana' in s[i]]

C. indices = [i for i in range(len(s)) if s.find('ana') == i]

D. indices = [i for i in range(len(s)) if s[i:i+3] == 'ana']

Solution

1. Step 1: Understand substring search for all occurrences

   To find all starting indices, check each position if substring matches.

2. Step 2: Analyze each option

   indices = [i for i in range(len(s)) if s[i:i+3] == 'ana'] uses slicing and compares substring at each index correctly. indices = [s.find('ana')] finds only first occurrence. indices = [i for i in range(len(s)) if s.find('ana') == i] incorrectly compares find() result to index. indices = [i for i in range(len(s)) if 'ana' in s[i]] incorrectly checks membership of substring in single character.

3. Final Answer:

   indices = [i for i in range(len(s)) if s[i:i+3] == 'ana'] -> Option D

4. Quick Check:

   Slice and compare at each index for all matches [OK]

Hint: Use slicing and loop to find all substring start indices [OK]

Common Mistakes:

• Using find() only returns first match
• Comparing find() result incorrectly
• Checking substring in single character