Edit Distance Problem Levenshtein in DSA Rust - Build from Scratch

Edit Distance Problem Levenshtein

📖 Scenario: You are building a simple text comparison tool that helps find how different two words are. This is useful in spell checkers and search suggestions.

🎯 Goal: Build a program that calculates the Levenshtein edit distance between two words. This distance tells how many single-letter changes (insertions, deletions, or substitutions) are needed to change one word into the other.

📋 What You'll Learn

Create two string variables with exact values

Create a 2D vector to hold distances

Use nested loops to fill the distance matrix using the Levenshtein formula

Print the final edit distance value

💡 Why This Matters

🌍 Real World

Spell checkers, search engines, and auto-correct features use edit distance to suggest correct words.

💼 Career

Understanding dynamic programming and string algorithms is essential for software engineers working on text processing, search, and data cleaning.

1

Create the two words to compare

Create two string variables called word1 and word2 with the exact values "kitten" and "sitting" respectively.

DSA Rust

# Create two string variables word1 and word2 with values "kitten" and "sitting"
# Your code here

Hint

Use let to create variables and assign the exact strings.

2

Initialize the distance matrix

Create a 2D vector called dist with dimensions (word1.len() + 1) by (word2.len() + 1) filled with zeros. Then fill the first row with values from 0 to word2.len() and the first column with values from 0 to word1.len().

DSA Rust

let word1 = "kitten";
let word2 = "sitting";
// Initialize dist matrix with zeros and fill first row and column
# Your code here

Hint

Use vec![vec![0; cols]; rows] to create the 2D vector. Use for loops with 0..=length to fill edges.

3

Fill the distance matrix with Levenshtein logic

Use nested for loops with variables i from 1 to word1.len() and j from 1 to word2.len(). Inside, check if word1.chars().nth(i - 1).unwrap() equals word2.chars().nth(j - 1).unwrap(). If equal, set dist[i][j] = dist[i - 1][j - 1]. Otherwise, set dist[i][j] to 1 plus the minimum of dist[i - 1][j], dist[i][j - 1], and dist[i - 1][j - 1].

DSA Rust

let word1 = "kitten";
let word2 = "sitting";

let mut dist = vec![vec![0; word2.len() + 1]; word1.len() + 1];

for i in 0..=word1.len() {
    dist[i][0] = i;
}
for j in 0..=word2.len() {
    dist[0][j] = j;
}

// Fill dist matrix with Levenshtein distance logic
# Your code here

Hint

Use nested loops and compare characters at positions i-1 and j-1. Use std::cmp::min to find the smallest cost.

4

Print the final edit distance

Print the value at dist[word1.len()][word2.len()] which is the Levenshtein edit distance between word1 and word2.

DSA Rust

let word1 = "kitten";
let word2 = "sitting";

let mut dist = vec![vec![0; word2.len() + 1]; word1.len() + 1];

for i in 0..=word1.len() {
    dist[i][0] = i;
}
for j in 0..=word2.len() {
    dist[0][j] = j;
}

for i in 1..=word1.len() {
    for j in 1..=word2.len() {
        let cost = if word1.chars().nth(i - 1).unwrap() == word2.chars().nth(j - 1).unwrap() { 0 } else { 1 };
        dist[i][j] = std::cmp::min(
            std::cmp::min(dist[i - 1][j] + 1, dist[i][j - 1] + 1),
            dist[i - 1][j - 1] + cost
        );
    }
}

// Print the final edit distance
# Your code here

Hint

Use println!("{}", dist[word1.len()][word2.len()]) to print the final distance.

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