Substring Search Patterns in DSA Javascript - Time & Space Complexity

Time Complexity: Substring Search Patterns

O(n * m)

Understanding Time Complexity

When searching for a smaller string inside a bigger one, it is important to know how the time needed grows as the strings get longer.

We want to find out how the search time changes with input size.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function containsSubstring(text, pattern) {
  for (let i = 0; i <= text.length - pattern.length; i++) {
    let j = 0;
    while (j < pattern.length && text[i + j] === pattern[j]) {
      j++;
    }
    if (j === pattern.length) {
      return true;
    }
  }
  return false;
}
    

This code checks if the pattern string exists inside the text string by comparing characters one by one.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Nested loops where the outer loop moves through the text and the inner loop compares characters of the pattern.
• How many times: Outer loop runs about (text length - pattern length + 1) times; inner loop runs up to pattern length times for each outer loop iteration.

How Execution Grows With Input

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

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

Pattern observation: The operations grow roughly by the product of text length and pattern length.

Final Time Complexity

Time Complexity: O(n * m)

This means the time to search grows proportionally to the size of the text times the size of the pattern.

Common Mistake

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

[OK] Correct: Because for each position in the text, we may need to check all characters of the pattern, so pattern length also affects time.

Interview Connect

Understanding how substring search time grows helps you explain and improve string matching in real applications, showing your grasp of algorithm efficiency.

Self-Check

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

Practice

1. What does the indexOf method do when used on a string in JavaScript?

easy

A. Converts the string to uppercase

B. Replaces a substring with another string

C. Splits the string into an array of substrings

D. Finds the position of the first occurrence of a substring

Solution

1. Step 1: Understand the purpose of indexOf

   The indexOf method searches a string for a specified substring and returns the position of its first occurrence.

2. Step 2: Confirm behavior on strings

   If the substring is found, it returns the zero-based index; if not found, it returns -1.

3. Final Answer:

   Finds the position of the first occurrence of a substring -> Option D

4. Quick Check:

   indexOf = position of substring [OK]

Hint: Remember: indexOf returns position or -1 if not found [OK]

Common Mistakes:

• Confusing indexOf with replace method
• Thinking indexOf splits the string
• Assuming indexOf changes the string case

2. Which of the following is the correct syntax to check if the substring 'cat' exists in the string text using includes method?

easy

A. text.include('cat')

B. includes(text, 'cat')

C. text.includes('cat')

D. text.contains('cat')

Solution

1. Step 1: Recall the correct method name

   The correct method to check substring presence is includes, called on the string object.

2. Step 2: Verify syntax

   The syntax is string.includes(substring), so text.includes('cat') is correct.

3. Final Answer:

   text.includes('cat') -> Option C

4. Quick Check:

   Correct method call syntax = text.includes('cat') [OK]

Hint: Use string.includes(substring) to check presence [OK]

Common Mistakes:

• Using wrong method name like include or contains
• Calling includes as a standalone function
• Passing arguments in wrong order

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

const text = 'hello world';
const pos = text.indexOf('o');
console.log(pos);

medium

A. 4

B. 5

C. -1

D. 6

Solution

1. Step 1: Identify the string and substring

   The string is 'hello world' and the substring searched is 'o'.

2. Step 2: Find the first occurrence of 'o'

   Counting from zero: h(0) e(1) l(2) l(3) o(4) (space 5) w(6) o(7). The first 'o' is at index 4.

3. Step 3: Verify the output

   The code uses indexOf('o') which returns the first occurrence index, which is 4.

4. Final Answer:

   4 -> Option A

5. Quick Check:

   First 'o' index = 4 [OK]

Hint: Count string index from zero to find first substring position [OK]

Common Mistakes:

• Counting index starting from 1 instead of 0
• Confusing second 'o' position with first
• Assuming indexOf returns -1 if substring exists

4. The following code is intended to check if the substring 'test' exists in str. What is the error?

const str = 'this is a test';
if (str.indexOf == 'test') {
  console.log('Found');
} else {
  console.log('Not Found');
}

medium

A. The variable str is not declared properly

B. indexOf is used without parentheses, so it compares function to string

C. The if condition should use === instead of ==

D. The substring 'test' is misspelled

Solution

1. Step 1: Analyze the use of indexOf

   The code uses str.indexOf without parentheses, so it refers to the function itself, not the result of calling it.

2. Step 2: Understand the comparison

   Comparing a function to a string ('test') is always false, so the condition is incorrect.

3. Final Answer:

   indexOf is used without parentheses, so it compares function to string -> Option B

4. Quick Check:

   Missing parentheses on indexOf causes error [OK]

Hint: Always call indexOf with parentheses and substring argument [OK]

Common Mistakes:

• Forgetting parentheses on indexOf
• Using wrong comparison operator
• Assuming indexOf returns boolean directly

5. You want to find all starting positions of the substring 'ana' in the string 'banana'. Which approach correctly finds all positions?

hard

A. Use a loop with indexOf and update the start position after each find

B. Use includes repeatedly until it returns false

C. Use lastIndexOf to find all positions from the end

D. Use replace to remove substring and count occurrences

Solution

1. Step 1: Understand the problem of multiple occurrences

   To find all positions, you must repeatedly search starting after the last found position.

2. Step 2: Use a loop with indexOf and update start

   Use indexOf('ana', start) in a loop, updating start to position + 1 each time until no more found.

3. Final Answer:

   Use a loop with indexOf and update the start position after each find -> Option A

4. Quick Check:

   Loop + indexOf with updated start = all positions [OK]

Hint: Loop with indexOf(start) to find all substring positions [OK]

Common Mistakes:

• Using includes which returns only true/false
• Using lastIndexOf which finds only last occurrence
• Using replace which changes string and loses positions