String Pattern Matching Naive in DSA Rust - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to get the length of the pattern string.

DSA Rust

let pattern_len = pattern.[1]();

Drag options to blanks, or click blank then click option'

Alength

Blen

Csize

Dcount

Attempts:

3 left

💡 Hint

Common Mistakes

Using length() which does not exist for &str

Using size() or count() which are invalid methods here

The correct method to get the length of a string slice in Rust is len().

2fill in blank

medium

Complete the code to get the length of the text string.

DSA Rust

let text_len = text.[1]();

Drag options to blanks, or click blank then click option'

Alength

Bsize

Clen

Dcount

Attempts:

3 left

💡 Hint

Common Mistakes

Using length() which is not a method on &str

Using size() or count() which are invalid here

The .len() method returns the length of the string slice in Rust.

3fill in blank

hard

Fix the error in the loop to iterate over possible start indices for matching.

DSA Rust

for i in 0..=text_len [1] pattern_len {

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A-

B+

C*

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using + causes index out of bounds

Using * or / are invalid for index range

We iterate from 0 up to text_len - pattern_len inclusive to avoid overflow.

4fill in blank

hard

Fill the blank to compare characters at current indices in text and pattern.

DSA Rust

if text[i[1]j] == pattern[j] {

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A+

B-

C*

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using - causes negative indices

Using * or / are invalid for indexing

We add j to i to get the correct character index in text, and use j as is for pattern.

5fill in blank

hard

Fill all three blanks to complete the naive string matching function.

DSA Rust

fn naive_search(text: &str, pattern: &str) -> Vec<usize> {
    let mut result = Vec::new();
    let text_len = text.[1]();
    let pattern_len = pattern.[2]();
    for i in 0..=text_len[3]pattern_len {
        let mut j = 0;
        while j < pattern_len && text[i + j] == pattern[j] {
            j += 1;
        }
        if j == pattern_len {
            result.push(i);
        }
    }
    result
}

Drag options to blanks, or click blank then click option'

Alen

Blength

C-

D+

Attempts:

3 left

💡 Hint

Common Mistakes

Using length instead of len()

Using + instead of - in loop range

Not using .len() for lengths

Use .len() to get lengths and subtract pattern_len from text_len for the loop range.

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