Edit Distance Problem Levenshtein in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the base case for the first string length in the edit distance matrix.

DSA C

for (int i = 0; i <= m; i++) {
    dp[i][0] = [1];
}

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

An

B0

Ci

Dm

Attempts:

3 left

💡 Hint

Common Mistakes

Setting dp[i][0] to 0 instead of i

Using n instead of i

Using m instead of i

The first column of the matrix represents converting the first i characters of string1 to an empty string, which requires i deletions.

2fill in blank

medium

Complete the code to initialize the base case for the first string length in the edit distance matrix for the second string.

DSA C

for (int j = 0; j <= n; j++) {
    dp[0][j] = [1];
}

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

Am

Bj

C0

Dn

Attempts:

3 left

💡 Hint

Common Mistakes

Setting dp[0][j] to 0 instead of j

Using m instead of j

Using n instead of j

The first row of the matrix represents converting an empty string to the first j characters of string2, which requires j insertions.

3fill in blank

hard

Fix the error in the condition that checks if characters at positions i-1 and j-1 are the same.

DSA C

if (s1[i - 1] [1] s2[j - 1]) {
    dp[i][j] = dp[i - 1][j - 1];
} else {
    dp[i][j] = 1 + min(dp[i - 1][j], min(dp[i][j - 1], 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 '=' instead of '=='

Using '!=' which reverses the logic

Using '<' which is not correct for equality check

The condition must check if the characters are equal using '=='. Using '=' assigns a value and causes an error.

4fill in blank

hard

Fill both blanks to complete the min function calls to find the minimum edit distance operation cost.

DSA C

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

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

A-

B+

C*

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Using '+' or '*' instead of '-'

Using '/' which is invalid for indices

Using incorrect numbers for indices

The last term compares dp[i-1][j-1], so the blanks must be '-' and '1' respectively.

5fill in blank

hard

Fill all three blanks to complete the function signature and return statement for the Levenshtein distance function.

DSA C

[1] levenshtein_distance([2] s1, [3] s2) {
    int m = strlen(s1);
    int n = strlen(s2);
    int dp[m + 1][n + 1];
    // initialization and computation code here
    return dp[m][n];
}

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

Aint

Bchar*

Cconst char*

Dvoid

Attempts:

3 left

💡 Hint

Common Mistakes

Using void as return type

Using char* without const for input strings

Using int for input strings

The function returns an int distance and takes two constant character pointers as input strings.

Practice

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

easy

A. The minimum number of insertions, deletions, or substitutions to convert one string into another

B. The number of matching characters in the same positions

C. The length difference between two strings

D. The maximum number of characters that differ between two strings

Solution

1. Step 1: Understand the definition of Edit Distance

   Edit Distance counts the smallest number of changes needed to turn one string into another.

2. Step 2: Identify allowed operations

   These operations include insertion, deletion, and substitution of characters.

3. Final Answer:

   The minimum number of insertions, deletions, or substitutions to convert one string into another -> Option A

4. Quick Check:

   Edit Distance = minimum edits needed [OK]

Hint: Think of edits needed to match strings exactly [OK]

Common Mistakes:

• Confusing edit distance with matching characters count
• Thinking it only counts substitutions
• Assuming it measures length difference only

2. Which of the following is the correct way to initialize the first row of the DP table for Levenshtein distance in C?

easy

A. for (int j = 0; j <= n; j++) dp[0][j] = j;

B. for (int i = 0; i <= m; i++) dp[i][0] = i;

C. for (int j = 1; j <= n; j++) dp[0][j] = 0;

D. for (int i = 1; i <= m; i++) dp[0][i] = i;

Solution

1. Step 1: Identify the DP table dimensions

   The DP table has dimensions (m+1) x (n+1), where m and n are string lengths.

2. Step 2: Initialize first row correctly

   The first row dp[0][j] represents converting empty string to prefix of second string, so dp[0][j] = j.

3. Final Answer:

   for (int j = 0; j <= n; j++) dp[0][j] = j; -> Option A

4. Quick Check:

   First row = 0 to n [OK]

Hint: First row = indices of second string length [OK]

Common Mistakes:

• Mixing row and column indices
• Starting loop from 1 instead of 0
• Initializing first column instead of first row

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

int min(int a, int b, int c) {
    int m = a < b ? a : b;
    return m < c ? m : c;
}

int levenshtein(char *s, char *t) {
    int m = strlen(s), n = strlen(t);
    int dp[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++) {
            int cost = (s[i-1] == t[j-1]) ? 0 : 1;
            dp[i][j] = min(dp[i-1][j] + 1, dp[i][j-1] + 1, dp[i-1][j-1] + cost);
        }
    }
    return dp[m][n];
}

printf("%d", levenshtein("kitten", "sitting"));

medium

A. 3

B. 4

C. 5

D. 6

Solution

1. Step 1: Understand the strings and operations

   "kitten" to "sitting" requires substitutions and insertions: k->s, e->i, add g.

2. Step 2: Calculate minimum edits

   Minimum edits are 3 substitutions and 1 insertion = 4 total edits.

3. Final Answer:

   4 -> Option B

4. Quick Check:

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

Hint: Recall classic example: kitten -> sitting = 4 edits [OK]

Common Mistakes:

• Counting only substitutions, missing insertions
• Confusing character positions
• Off-by-one errors in indexing

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

int levenshtein(char *s, char *t) {
    int m = strlen(s), n = strlen(t);
    int dp[m][n];
    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++) {
            int cost = (s[i-1] == t[j-1]) ? 0 : 1;
            dp[i][j] = min(dp[i-1][j] + 1, dp[i][j-1] + 1, dp[i-1][j-1] + cost);
        }
    }
    return dp[m][n];
}

medium

A. Loops should start from 0, not 1

B. The cost calculation is incorrect

C. dp array size should be (m+1) x (n+1), not m x n

D. Return statement should be dp[m-1][n-1]

Solution

1. Step 1: Check dp array declaration

   dp is declared as dp[m][n], but indices go up to m and n inclusive, so size is insufficient.

2. Step 2: Understand indexing needs

   We need dp of size (m+1) x (n+1) to store base cases and full computation.

3. Final Answer:

   dp array size should be (m+1) x (n+1), not m x n -> Option C

4. Quick Check:

   dp size = (m+1)*(n+1) [OK]

Hint: DP table size must cover all indices including zero [OK]

Common Mistakes:

• Declaring dp too small causing out-of-bounds errors
• Misunderstanding loop ranges
• Incorrect return index

5. Given two strings "intention" and "execution", what is the minimum edit distance using Levenshtein algorithm? Consider insertions, deletions, and substitutions only.

hard

A. 7

B. 8

C. 5

D. 6

Solution

1. Step 1: Analyze string lengths and differences

   "intention" and "execution" differ in several characters and positions.

2. Step 2: Calculate minimum edits stepwise

   Edits include substitutions (i->e, n->x, t->e, e->c, n->u, etc.) totaling 6 changes.

3. Final Answer:

   6 -> Option D

4. Quick Check:

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

Hint: Count substitutions and insertions carefully [OK]

Common Mistakes:

• Underestimating substitutions needed
• Ignoring insertions or deletions
• Confusing similar characters as matches