Edit Distance Problem Levenshtein in DSA Rust - Time & Space Complexity

Time Complexity: Edit Distance Problem Levenshtein

O(m * n)

Understanding Time Complexity

We want to understand how the time needed to find the edit distance between two words grows as the words get longer.

The question is: How does the number of steps change when the input words become bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

fn levenshtein_distance(s1: &str, s2: &str) -> usize {
    let len1 = s1.len();
    let len2 = s2.len();
    let mut dp = vec![vec![0; len2 + 1]; len1 + 1];

    for i in 0..=len1 {
        dp[i][0] = i;
    }
    for j in 0..=len2 {
        dp[0][j] = j;
    }

    for i in 1..=len1 {
        for j in 1..=len2 {
            dp[i][j] = if s1.as_bytes()[i - 1] == s2.as_bytes()[j - 1] {
                dp[i - 1][j - 1]
            } else {
                1 + std::cmp::min(dp[i - 1][j - 1], std::cmp::min(dp[i][j - 1], dp[i - 1][j]))
            };
        }
    }
    dp[len1][len2]
}

This code calculates the minimum number of edits needed to change one word into another using a table to store results.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Nested loops filling a 2D table of size (length of first word + 1) by (length of second word + 1).
• How many times: The inner loop runs for each character of the second word, and the outer loop runs for each character of the first word, so total steps are roughly the product of the two lengths.

How Execution Grows With Input

As the words get longer, the number of operations grows by multiplying their lengths.

Input Size (n)	Approx. Operations
10	About 100 steps
100	About 10,000 steps
1000	About 1,000,000 steps

Pattern observation: Doubling the length roughly quadruples the work, showing a square growth pattern.

Final Time Complexity

Time Complexity: O(m * n)

This means the time needed grows proportionally to the product of the lengths of the two words.

Common Mistake

[X] Wrong: "The algorithm only looks at one word at a time, so it should be O(n)."

[OK] Correct: The algorithm compares every character of the first word with every character of the second, so it must check all pairs, making it grow with both lengths multiplied.

Interview Connect

Understanding this time complexity helps you explain how dynamic programming solves problems efficiently by avoiding repeated work, a key skill in many coding challenges.

Self-Check

"What if we used recursion without storing results? How would the time complexity change?"

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