String Pattern Matching Naive in DSA Rust - Time & Space Complexity

Time Complexity: String Pattern Matching Naive

O(n * m)

Understanding Time Complexity

We want to understand how long it takes to find a pattern inside a text using a simple method.

How does the time needed grow when the text or pattern gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

fn naive_pattern_search(text: &str, pattern: &str) -> Vec {
    let n = text.len();
    let m = pattern.len();
    let mut result = Vec::new();
    if m > n {
        return result;
    }
    for i in 0..=n - m {
        let mut j = 0;
        while j < m && text.as_bytes()[i + j] == pattern.as_bytes()[j] {
            j += 1;
        }
        if j == m {
            result.push(i);
        }
    }
    result
}
    

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

Identify Repeating Operations

• Primary operation: The outer loop runs over the text positions, and the inner loop compares characters of the pattern with the text.
• How many times: The outer loop runs about n - m + 1 times, and for each, the inner loop can run up to m times.

How Execution Grows With Input

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

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

Pattern observation: The total work grows roughly by multiplying the text length by the pattern length.

Final Time Complexity

Time Complexity: O(n * m)

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

Common Mistake

[X] Wrong: "The search takes only O(n) time because it looks through the text once."

[OK] Correct: The inner loop compares each character of the pattern for every position, so the work multiplies by the pattern length.

Interview Connect

Understanding this simple search helps build a foundation for more efficient methods and shows you can analyze nested loops clearly.

Self-Check

"What if we stopped checking as soon as one character didn't match? How would that affect the time complexity in the best case?"

Practice

1. What is the main idea behind the naive string pattern matching algorithm?

easy

A. Check every possible position in the text for the pattern

B. Use a hash function to find the pattern quickly

C. Build a prefix table to skip unnecessary checks

D. Sort the text and pattern before matching

Solution

1. Step 1: Understand naive matching approach

   The naive algorithm tries to match the pattern starting at every position in the text.

2. Step 2: Compare with other methods

   Other methods like hashing or prefix tables are not part of naive matching.

3. Final Answer:

   Check every possible position in the text for the pattern -> Option A

4. Quick Check:

   Naive = Check all positions [OK]

Hint: Naive means try all positions one by one [OK]

Common Mistakes:

• Confusing naive with hashing or prefix methods
• Thinking naive skips positions
• Assuming sorting is involved

2. Which Rust code snippet correctly starts the naive pattern matching loop over text and pattern?

easy

A. for i in 0..text.len() + pattern.len() { /* check pattern */ }

B. for i in 0..=text.len() - pattern.len() { /* check pattern */ }

C. for i in 0..text.len() { /* check pattern */ }

D. for i in 0..pattern.len() { /* check pattern */ }

Solution

1. Step 1: Determine valid start positions

   The pattern can only start at positions where it fits fully in the text, so from 0 to text length minus pattern length.

2. Step 2: Check Rust range syntax

   Using 0..=text.len() - pattern.len() includes the last valid index.

3. Final Answer:

   for i in 0..=text.len() - pattern.len() { /* check pattern */ } -> Option B

4. Quick Check:

   Loop range = 0 to text.len()-pattern.len() [OK]

Hint: Loop till text length minus pattern length inclusive [OK]

Common Mistakes:

• Looping beyond valid start positions causing panic
• Using pattern length as loop limit
• Off-by-one errors in range

3. What will be the output of this Rust code using naive pattern matching?

let text = "abracadabra";
let pattern = "abra";
let mut positions = Vec::new();
for i in 0..=text.len() - pattern.len() {
    if &text[i..i+pattern.len()] == pattern {
        positions.push(i);
    }
}
println!("{:?}", positions);

medium

A. [3, 7]

B. [0, 3, 7]

C. [7]

D. [0, 7]

Solution

1. Step 1: Identify pattern matches in text

   The pattern "abra" appears at index 0 ("abra") and index 7 ("abra") in "abracadabra".

2. Step 2: Confirm positions collected

   Positions vector collects indices 0 and 7 where pattern matches exactly.

3. Final Answer:

   [0, 7] -> Option D

4. Quick Check:

   Matches at 0 and 7 only [OK]

Hint: Check substring slices equal to pattern [OK]

Common Mistakes:

• Including partial matches like at index 3
• Off-by-one errors in slicing
• Missing second occurrence

4. Identify the bug in this naive pattern matching Rust code:

let text = "hello";
let pattern = "ll";
for i in 0..text.len() - pattern.len() {
    if &text[i..i+pattern.len()] == pattern {
        println!("Found at {}", i);
    }
}

medium

A. Range should be 0..=text.len() - pattern.len() to include last position

B. Pattern length is used incorrectly in slicing

C. Text and pattern should be converted to chars before slicing

D. println! macro syntax is wrong

Solution

1. Step 1: Check loop range correctness

   The loop uses 0..text.len() - pattern.len(), which excludes the last valid start index.

2. Step 2: Fix range to include last index

   Using 0..=text.len() - pattern.len() includes the last position where pattern can start.

3. Final Answer:

   Range should be 0..=text.len() - pattern.len() to include last position -> Option A

4. Quick Check:

   Use inclusive range for last valid index [OK]

Hint: Use inclusive range to cover all valid start indices [OK]

Common Mistakes:

• Using exclusive range causing missed matches
• Incorrect slicing indices
• Unnecessary conversion to chars

5. Given a text and a pattern, how can you modify the naive pattern matching to count overlapping occurrences of the pattern in Rust?
Example: text = "aaaa", pattern = "aa" should count 3 occurrences.

hard

A. Increment index by pattern length after each match

B. Use a hash map to store counts

C. Increment index by 1 after each check instead of pattern length

D. Reverse the text and pattern before matching

Solution

1. Step 1: Understand overlapping matches

   Overlapping means after finding a match, next check starts just one position ahead, not skipping full pattern length.

2. Step 2: Adjust loop increment

   Increment index by 1 each time to allow overlapping matches to be found.

3. Final Answer:

   Increment index by 1 after each check instead of pattern length -> Option C

4. Quick Check:

   Step by 1 to count overlaps [OK]

Hint: Move index by 1 to catch overlaps [OK]

Common Mistakes:

• Skipping by pattern length missing overlaps
• Using hash map unnecessarily
• Reversing strings does not help