Edit Distance Problem Levenshtein in DSA C# - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the 2D array for dynamic programming.

DSA C#

int[,] dp = new int[[1], [2]];

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

Aword1.Length + 1

Bword1.Length

Cword2.Length

Dword2.Length - 1

Eword2.Length + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using word1.Length instead of word1.Length + 1

Using word2.Length - 1 which causes index errors

The DP table size must be (length + 1) to include empty prefixes.

2fill in blank

medium

Complete the code to fill the first row of the DP table.

DSA C#

for (int j = 0; j < dp.GetLength(1); j++) dp[0, j] = [1];

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

Aj

B0

C1

Dj + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Setting all to 0 which ignores insertions

Using j + 1 which is off by one

The first row represents converting empty string to prefixes of word2, so cost equals j.

3fill in blank

hard

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

DSA C#

if (word1[i - 1] [1] word2[j - 1]) dp[i, j] = dp[i - 1, j - 1];

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

A!=

B<

C>

D==

Attempts:

3 left

💡 Hint

Common Mistakes

Using != which causes wrong cost calculation

Using > or < which are invalid for char comparison here

We check if characters are equal to avoid extra cost.

4fill in blank

hard

Fill both blanks to compute the minimum edit distance for dp[i, j].

DSA C#

dp[i, j] = 1 + Math.Min(dp[i - 1, j], dp[i, [1]]);
dp[i, j] = Math.Min(dp[i, j], 1 + dp[[2], j - 1]);

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

Aj - 1

Bi - 1

Cj + 1

Di + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping i and j indices

Using j+1 or i+1 which are out of bounds

We consider deletion (dp[i-1,j]), insertion (dp[i,j-1]), and substitution (dp[i-1,j-1]) costs.

5fill in blank

hard

Fill all three blanks to complete the Levenshtein distance function.

DSA C#

public int LevenshteinDistance(string word1, string word2) {
    int m = word1.Length;
    int n = word2.Length;
    int[,] dp = new int[m + 1, n + 1];

    for (int i = 0; i <= m; i++) dp[i, 0] = i;
    for (int j = 0; j <= n; j++) dp[0, j] = j;

    for (int i = 1; i <= m; i++) {
        for (int j = 1; j <= n; j++) {
            if (word1[i - 1] [1] word2[j - 1]) {
                dp[i, j] = dp[i - 1, j - 1];
            } else {
                dp[i, j] = 1 + Math.Min(dp[i - 1, j], Math.Min(dp[i, [2]], dp[[3], j - 1]));
            }
        }
    }
    return dp[m, n];
}

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

A==

Bj - 1

Ci - 1

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '!=' instead of '==' in character comparison

Swapping indices in Math.Min calls

The function compares characters, then calculates minimum edits considering insertion, deletion, and substitution.

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