Edit Distance Problem Levenshtein in DSA C# - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the Edit Distance (Levenshtein distance) between two strings?

It is the minimum number of single-character edits (insertions, deletions, or substitutions) required to change one string into the other.

beginner

Which operations are allowed in calculating Levenshtein distance?

Insertion, deletion, and substitution of a single character.

intermediate

What is the base case when using dynamic programming for Levenshtein distance?

When one string is empty, the distance is the length of the other string (all insertions or deletions).

intermediate

How does the dynamic programming table for Levenshtein distance get filled?

Each cell represents the minimum edit distance between prefixes of the two strings, computed using previous results for smaller prefixes.

intermediate

What is the time complexity of the Levenshtein distance algorithm using dynamic programming?

O(m * n), where m and n are the lengths of the two strings.

Which of the following is NOT an allowed operation in Levenshtein distance?

A Insertion

B Transposition

C Substitution

D Deletion

Levenshtein distance allows insertion, deletion, and substitution, but not transposition (swapping two adjacent characters).

What is the edit distance between "cat" and "cut"?

A 1

B 2

C 3

D 0

Only one substitution ('a' to 'u') is needed.

If one string is empty and the other has length 5, what is the edit distance?

A 0

B 1

C Cannot determine

D 5

All 5 characters must be inserted or deleted.

What does each cell in the DP table represent in Levenshtein distance calculation?

A Number of substitutions only

B Length of the longer string

C Minimum edit distance between prefixes of the two strings

D Number of insertions only

Each cell stores the minimum edit distance between prefixes of the two strings.

What is the space complexity of the classic DP solution for Levenshtein distance?

A O(m * n)

B O(m + n)

C O(1)

D O(n^2)

The DP table requires space proportional to the product of the string lengths.

Explain how dynamic programming helps solve the Edit Distance Problem (Levenshtein distance).

Think about how you compare smaller parts of the strings first.

Describe the three operations used in calculating Levenshtein distance and give a simple example for each.

Consider how to change one word into another by adding, removing, or changing letters.

Practice

1. What does the Levenshtein Edit Distance between two strings measure?

easy

A. The total length of both strings combined

B. The maximum length of common substring between two strings

C. The number of characters that are the same in both strings

D. The minimum number of single-character edits to change one string into another

Solution

1. Step 1: Understand the definition of Levenshtein distance

   It counts the minimum edits (insertions, deletions, substitutions) to convert one string into another.

2. Step 2: Compare with other options

   Options B and C describe different string metrics, not edit distance.

3. Final Answer:

   The minimum number of single-character edits to change one string into another -> Option D

4. Quick Check:

   Edit distance = minimum edits [OK]

Hint: Edit distance counts minimal edits to match strings [OK]

Common Mistakes:

• Confusing edit distance with longest common substring
• Thinking it counts only substitutions
• Assuming it measures string length

2. Which of the following C# method signatures correctly represents a function to compute Levenshtein distance between two strings s1 and s2?

easy

A. private string LevenshteinDistance(string s1, string s2)

B. public void LevenshteinDistance(int s1, int s2)

C. public int LevenshteinDistance(string s1, string s2)

D. public int LevenshteinDistance(int[] s1, int[] s2)

Solution

1. Step 1: Check parameter types for string inputs

   The function should accept two strings, so parameters must be string types.

2. Step 2: Check return type for distance value

   Distance is an integer, so return type should be int, not void or string.

3. Final Answer:

   public int LevenshteinDistance(string s1, string s2) -> Option C

4. Quick Check:

   Function signature matches string inputs and int output [OK]

Hint: Function takes strings, returns int distance [OK]

Common Mistakes:

• Using void return type instead of int
• Using int parameters instead of strings
• Returning string instead of int

3. What is the output of the following C# code snippet?

var s1 = "kitten";
var s2 = "sitting";
Console.WriteLine(LevenshteinDistance(s1, s2));

int LevenshteinDistance(string a, string b) {
    int[,] dp = new int[a.Length + 1, b.Length + 1];
    for (int i = 0; i <= a.Length; i++) dp[i, 0] = i;
    for (int j = 0; j <= b.Length; j++) dp[0, j] = j;
    for (int i = 1; i <= a.Length; i++) {
        for (int j = 1; j <= b.Length; j++) {
            int cost = (a[i - 1] == b[j - 1]) ? 0 : 1;
            dp[i, j] = Math.Min(Math.Min(dp[i - 1, j] + 1, dp[i, j - 1] + 1), dp[i - 1, j - 1] + cost);
        }
    }
    return dp[a.Length, b.Length];
}

medium

A. 3

B. 4

C. 5

D. 6

Solution

1. Step 1: Understand the strings and operations

   "kitten" to "sitting" requires substitutions, insertions, deletions.

2. Step 2: Calculate Levenshtein distance manually

   Changes: k->s (substitution), e->i (substitution), t->t (no change), t->t (no change), e->i (substitution), n->n (no change), add g (insertion) = 3 substitutions + 1 insertion = 4 edits total.

3. Final Answer:

   4 -> Option B

4. Quick Check:

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

Hint: Remember common example: kitten -> sitting = 4 edits [OK]

Common Mistakes:

• Counting only substitutions, ignoring insertions
• Miscounting character positions
• Confusing with Hamming distance

4. Identify the bug in this Levenshtein distance C# code snippet:

int LevenshteinDistance(string a, string b) {
    int[,] dp = new int[a.Length, b.Length];
    for (int i = 0; i < a.Length; i++) dp[i, 0] = i;
    for (int j = 0; j < b.Length; j++) dp[0, j] = j;
    for (int i = 1; i < a.Length; i++) {
        for (int j = 1; j < b.Length; j++) {
            int cost = (a[i] == b[j]) ? 0 : 1;
            dp[i, j] = Math.Min(Math.Min(dp[i - 1, j] + 1, dp[i, j - 1] + 1), dp[i - 1, j - 1] + cost);
        }
    }
    return dp[a.Length - 1, b.Length - 1];
}

medium

A. Array size should be (a.Length+1, b.Length+1) and loops adjusted

B. Cost calculation should use a[i-1] and b[j-1]

C. Return statement should be dp[a.Length, b.Length]

D. No bug, code is correct

Solution

1. Step 1: Check array size and indexing

   dp array must be sized (a.Length+1, b.Length+1) to include empty prefixes; current size misses last row/column.

2. Step 2: Adjust loops to run from 0 to length inclusive

   Loops should run to <= length to fill dp correctly; current loops miss last indices.

3. Final Answer:

   Array size should be (a.Length+1, b.Length+1) and loops adjusted -> Option A

4. Quick Check:

   dp array size must include empty string row/column [OK]

Hint: dp array must be one larger than string lengths [OK]

Common Mistakes:

• Using array size equal to string length only
• Incorrect loop bounds causing index errors
• Ignoring empty string base cases

5. Given two strings s1 = "intention" and s2 = "execution", what is the Levenshtein distance between them? Use the standard operations: insert, delete, substitute.

hard

A. 5

B. 6

C. 4

D. 7

Solution

1. Step 1: Analyze differences between "intention" and "execution"

   Compare characters and count minimal edits: substitutions and insertions/deletions.

2. Step 2: Calculate minimal edits manually

   Edits: i->e (substitution), n->x (substitution), t->c (substitution), insert u, substitute n->o, substitute n->n (no change), total 5 edits.

3. Final Answer:

   5 -> Option A

4. Quick Check:

   Distance("intention", "execution") = 5 [OK]

Hint: Count substitutions and insertions carefully [OK]

Common Mistakes:

• Underestimating substitutions needed
• Ignoring insertions or deletions
• Confusing with Hamming distance