Edit Distance Problem Levenshtein in DSA Rust - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the DP matrix with zeros.

DSA Rust

let mut dp = vec![vec![[1]; n + 1]; m + 1];

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

A0

B1

CNone

Dvec![]

Attempts:

3 left

💡 Hint

Common Mistakes

Using 1 instead of 0 causes incorrect base initialization.

Using None or empty vectors causes type errors.

The DP matrix should be initialized with zeros to represent base cases before filling in edit distances.

2fill in blank

medium

Complete the code to fill the first row of the DP matrix with increasing values.

DSA Rust

for j in 0..=n {
    dp[0][j] = [1];
}

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

Aj

B0

Cm

Dn

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning 0 to all cells causes wrong distances.

Using m or n instead of j ignores the loop index.

The first row represents converting empty string to prefix of second string, so distance equals j.

3fill in blank

hard

Fix the error in the condition to check if characters are equal.

DSA Rust

if s1.chars().nth(i - 1) [1] s2.chars().nth(j - 1) {
    dp[i][j] = dp[i - 1][j - 1];
} else {
    dp[i][j] = 1 + std::cmp::min(
        dp[i - 1][j - 1],
        std::cmp::min(dp[i][j - 1], dp[i - 1][j])
    );
}

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

A>

B!=

C=

D==

Attempts:

3 left

💡 Hint

Common Mistakes

Using assignment '=' causes syntax error.

Using '!=' reverses logic and gives wrong results.

We check if characters are equal using '=='. If equal, no edit needed, so copy previous value.

4fill in blank

hard

Fill both blanks to correctly compute the minimum edit distance.

DSA Rust

dp[i][j] = 1 + std::cmp::min(dp[i - 1][j - 1], std::cmp::min(dp[[1]][j], dp[i][[2]]));

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

Ai - 1

Bi + 1

Cj - 1

Dj + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using i+1 or j+1 causes index out of bounds errors.

Mixing row and column indices incorrectly.

We compare substitution (i-1,j-1), deletion (i-1,j), and insertion (i,j-1) costs to find minimum.

5fill in blank

hard

Fill all three blanks to complete the function returning the edit distance.

DSA Rust

fn levenshtein(s1: &str, s2: &str) -> usize {
    let m = s1.len();
    let n = s2.len();
    let mut dp = vec![vec![0; n + 1]; m + 1];

    for i in 0..=m {
        dp[i][0] = i;
    }
    for j in 0..=n {
        dp[0][j] = j;
    }

    for i in 1..=m {
        for j in 1..=n {
            if s1.chars().nth(i - 1) [1] s2.chars().nth(j - 1) {
                dp[i][j] = dp[i - 1][j - 1];
            } else {
                dp[i][j] = 1 + std::cmp::min(
                    dp[i - 1][j - 1],
                    std::cmp::min(dp[i - 1][j], dp[i][j - 1])
                );
            }
        }
    }
    dp[[2]][[3]]
}

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

A==

B!=

Cm

Dn

Attempts:

3 left

💡 Hint

Common Mistakes

Using '!=' reverses logic and causes wrong results.

Returning dp[0][0] or dp[n][m] gives incorrect answer.

Use '==' to compare characters. The final answer is at dp[m][n], the full lengths of both strings.

Practice

1. What does the Levenshtein distance between two strings measure?

easy

A. The minimum number of single-character edits to change one string into the other

B. The maximum number of matching characters between two strings

C. The length difference between two strings

D. The number of common substrings between two strings

Solution

1. Step 1: Understand the definition of Levenshtein distance

   It counts the smallest number of edits (insertions, deletions, substitutions) needed to convert one string into another.

2. Step 2: Compare options with the definition

   Only The minimum number of single-character edits to change one string into the other correctly describes this minimum edit count concept.

3. Final Answer:

   The minimum number of single-character edits to change one string into the other -> Option A

4. Quick Check:

   Levenshtein distance = minimum edits [OK]

Hint: Think of edits needed to transform one word to another [OK]

Common Mistakes:

• Confusing with longest common subsequence
• Thinking it measures string length difference only
• Assuming it counts matching characters

2. Which Rust code snippet correctly initializes a 2D vector for Levenshtein distance with dimensions (m+1) x (n+1)?

easy

A. let dp = vec![vec![0; m+1]; n+1];

B. let mut dp = vec![vec![0; n+1]; m+1];

C. let mut dp = vec![vec![0; m]; n];

D. let dp = vec![vec![0; n]; m];

Solution

1. Step 1: Identify correct dimensions for DP table

   The DP table for Levenshtein distance has (m+1) rows and (n+1) columns, where m and n are string lengths.

2. Step 2: Match Rust vector initialization syntax

   let mut dp = vec![vec![0; n+1]; m+1]; correctly creates a mutable 2D vector with m+1 rows and n+1 columns filled with zeros.

3. Final Answer:

   let mut dp = vec![vec![0; n+1]; m+1]; -> Option B

4. Quick Check:

   Rows = m+1, Columns = n+1, mutable vector [OK]

Hint: Rows = m+1, columns = n+1 for DP matrix [OK]

Common Mistakes:

• Swapping m and n in dimensions
• Forgetting to add 1 to lengths
• Not making vector mutable

3. What is the output of the following Rust code snippet calculating Levenshtein distance between "kitten" and "sitting"?

fn main() {
    let s1 = "kitten";
    let s2 = "sitting";
    let m = s1.len();
    let n = s2.len();
    let mut dp = vec![vec![0; n + 1]; m + 1];

    for i in 0..=m {
        dp[i][0] = i;
    }
    for j in 0..=n {
        dp[0][j] = j;
    }

    for i in 1..=m {
        for j in 1..=n {
            if s1.as_bytes()[i - 1] == s2.as_bytes()[j - 1] {
                dp[i][j] = dp[i - 1][j - 1];
            } else {
                dp[i][j] = 1 + std::cmp::min(
                    dp[i - 1][j - 1],
                    std::cmp::min(dp[i][j - 1], dp[i - 1][j]),
                );
            }
        }
    }
    println!("{}", dp[m][n]);
}

medium

A. 4

B. 3

C. 5

D. 6

Solution

1. Step 1: Understand the strings and DP setup

   Strings are "kitten" (length 6) and "sitting" (length 7). The DP table is correctly initialized and filled using Levenshtein logic.

2. Step 2: Calculate Levenshtein distance manually

   Transform "kitten" to "sitting" requires 3 substitutions and 1 insertion, total 4 edits.

3. Final Answer:

   4 -> Option A

4. Quick Check:

   Levenshtein("kitten", "sitting") = 4 [OK]

Hint: Count edits: kitten -> sitten -> sittin -> sitting [OK]

Common Mistakes:

• Counting only substitutions, ignoring insertions
• Misindexing string characters
• Off-by-one errors in loops

4. Identify the bug in this Rust code snippet for Levenshtein distance:

for i in 1..m {
    for j in 1..n {
        if s1.as_bytes()[i - 1] == s2.as_bytes()[j - 1] {
            dp[i][j] = dp[i - 1][j - 1];
        } else {
            dp[i][j] = 1 + std::cmp::min(
                dp[i - 1][j - 1],
                std::cmp::min(dp[i][j - 1], dp[i - 1][j]),
            );
        }
    }
}

medium

A. Not initializing dp[0][j] and dp[i][0] before loops

B. Indexing s1 and s2 with i and j instead of i-1 and j-1

C. Loop ranges should be 1..=m and 1..=n to include last indices

D. Using std::cmp::max instead of std::cmp::min in calculation

Solution

1. Step 1: Check loop ranges for DP filling

   Loops use 1..m and 1..n which exclude the last index (m and n). They should be inclusive (1..=m and 1..=n) to cover all characters.

2. Step 2: Verify other code parts

   Indexing with i-1 and j-1 is correct, min is used properly, and initialization is assumed done outside snippet.

3. Final Answer:

   Loop ranges should be 1..=m and 1..=n to include last indices -> Option C

4. Quick Check:

   Inclusive loops needed for full DP table [OK]

Hint: Use inclusive ranges 1..=m and 1..=n for full coverage [OK]

Common Mistakes:

• Using exclusive ranges missing last character
• Incorrect indexing causing panic
• Skipping DP initialization

5. Given two strings s1 = "intention" and s2 = "execution", what is the Levenshtein distance computed by the Rust algorithm? Use the standard operations: insert, delete, substitute with cost 1 each.

hard

A. 7

B. 8

C. 5

D. 6

Solution

1. Step 1: Analyze the strings and possible edits

   "intention" (i n t e n t i o n) to "execution" (e x e c u t i o n). Minimal edits: substitute i->e, n->x, t->e, e->c, n->u, plus one insertion or deletion needed, total 6 edits.

2. Step 2: Confirm with Levenshtein algorithm logic

   The DP computes the minimum cost as 6.

3. Final Answer:

   6 -> Option D

4. Quick Check:

   Levenshtein("intention", "execution") = 6 [OK]

Hint: Count minimal edits stepwise between strings [OK]

Common Mistakes:

• Ignoring insertions or deletions
• Miscounting substitutions
• Assuming equal length means zero edits