Substring Search Patterns in DSA Java - Time & Space Complexity

Time Complexity: Substring Search Patterns

O(n * m)

Understanding Time Complexity

When searching for a smaller string inside a bigger one, we want to know how long it takes as the strings grow.

We ask: How does the search time change when the text or pattern gets longer?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

public int substringSearch(String text, String pattern) {
    int n = text.length();
    int m = pattern.length();
    for (int i = 0; i <= n - m; i++) {
        int j;
        for (j = 0; j < m; j++) {
            if (text.charAt(i + j) != pattern.charAt(j)) {
                break;
            }
        }
        if (j == m) {
            return i; // pattern found
        }
    }
    return -1; // pattern not found
}
    

This code checks each possible position in the text to see if the pattern matches starting there.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Nested loops comparing characters of pattern and text.
• 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 character 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: Operations grow roughly like text length times pattern length.

Final Time Complexity

Time Complexity: O(n * m)

This means the time to search grows proportionally to the product of the text and pattern lengths.

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 compare many characters of the pattern, so pattern length also affects time.

Interview Connect

Understanding this helps you explain why simple substring search can be slow and why smarter methods exist, showing your grasp of algorithm efficiency.

Self-Check

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

Practice

1. What does the indexOf method in Java's String class return when searching for a substring that exists in the string?

easy

A. The starting index of the first occurrence of the substring

B. The length of the substring

C. The number of times the substring appears

D. A boolean indicating if the substring exists

Solution

1. Step 1: Understand the purpose of indexOf

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

2. Step 2: Identify the return value when substring exists

   If the substring is found, indexOf returns the starting index (0-based) of that substring in the string.

3. Final Answer:

   The starting index of the first occurrence of the substring -> Option A

4. Quick Check:

   indexOf returns index, not count or boolean [OK]

Hint: Remember: indexOf returns position, not count or boolean [OK]

Common Mistakes:

• Thinking indexOf returns count of occurrences
• Assuming it returns true/false
• Confusing indexOf with substring length

2. Which of the following is the correct syntax to find the first occurrence of substring "cat" in string str in Java?

easy

A. int pos = str.find("cat");

B. int pos = str.search("cat");

C. int pos = str.indexOf("cat");

D. int pos = str.contains("cat");

Solution

1. Step 1: Recall Java String methods for substring search

   Java's String class uses indexOf to find substring positions.

2. Step 2: Check method names and return types

   find and search are not valid String methods; contains returns boolean, not int.

3. Final Answer:

   int pos = str.indexOf("cat"); -> Option C

4. Quick Check:

   Use indexOf for substring position [OK]

Hint: Use indexOf for position, contains for boolean check [OK]

Common Mistakes:

• Using non-existent methods like find or search
• Using contains which returns boolean, not index
• Confusing method names from other languages

3. What is the output of the following Java code?

String text = "hello world";
int pos = text.indexOf("lo");
System.out.println(pos);

medium

A. -1

B. 3

C. 4

D. 5

Solution

1. Step 1: Locate substring "lo" in "hello world"

   Counting indices: h(0), e(1), l(2), l(3), o(4). The substring "lo" starts at index 3 (l at 3, o at 4).

2. Step 2: Understand indexOf return value

   indexOf returns the starting index of the substring, which is 3 here.

3. Final Answer:

   3 -> Option B

4. Quick Check:

   Substring "lo" starts at index 3 [OK]

Hint: Count string indices from 0 to find substring start [OK]

Common Mistakes:

• Counting from 1 instead of 0
• Returning index of second character in substring
• Assuming -1 means substring found

4. Identify the error in the following Java code snippet for substring search:

String s = "banana";
int index = s.indexOf('an');
System.out.println(index);

medium

A. Using single quotes for substring instead of double quotes

B. Missing semicolon after indexOf call

C. indexOf method does not exist for String

D. Printing index before assignment

Solution

1. Step 1: Check substring argument type in indexOf

   The indexOf method expects a String argument, but single quotes denote a char.

2. Step 2: Identify correct syntax for substring

   Substring literals must be in double quotes, so 'an' should be "an".

3. Final Answer:

   Using single quotes for substring instead of double quotes -> Option A

4. Quick Check:

   Substring literals need double quotes [OK]

Hint: Use double quotes for substrings, single for chars [OK]

Common Mistakes:

• Using single quotes for multi-character substrings
• Confusing char and String types
• Ignoring compiler error messages

5. Given a string text, which code snippet correctly finds all starting indices of substring "ab" in text using indexOf?

hard

A. int index = 0;
while (index < text.length()) {
index = text.indexOf("ab");
System.out.print(index + " ");
index++;
}

B. for (int i = 0; i < text.length(); i++) {
if (text.indexOf("ab") == i) System.out.print(i + " ");
}

C. int index = text.indexOf("ab");
while (index != -1) {
System.out.print(index + " ");
index = text.indexOf("ab");
}

D. int index = 0;
while ((index = text.indexOf("ab", index)) != -1) {
System.out.print(index + " ");
index++;
}

Solution

1. Step 1: Understand how to find multiple occurrences with indexOf

   Use indexOf(substring, fromIndex) to search from a moving start index to find all occurrences.

2. Step 2: Analyze each option's logic

   int index = 0; while ((index = text.indexOf("ab", index)) != -1) { System.out.print(index + " "); index++; } correctly updates index after each found occurrence to avoid infinite loop and find all positions. Others either do not update index properly or repeatedly search from start.

3. Final Answer:

   int index = 0; while ((index = text.indexOf("ab", index)) != -1) { System.out.print(index + " "); index++; } -> Option D

4. Quick Check:

   Use indexOf with start index and increment after each find [OK]

Hint: Increment start index after each found substring to continue search [OK]

Common Mistakes:

• Not updating start index causing infinite loop
• Always searching from start index 0
• Confusing indexOf usage with for-loop index