Edit Distance Problem Levenshtein in DSA Javascript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

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

DSA Javascript

const dp = Array(m + 1).fill().map(() => Array([1]).fill(0));

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

An + 1

Bm + 1

Cn

Dm

Attempts:

3 left

💡 Hint

Common Mistakes

Using m+1 for both dimensions, which causes incorrect table size.

Using n instead of n+1, missing the base case column.

The dp array should have dimensions (m+1) x (n+1) to cover all prefixes of both strings.

2fill in blank

medium

Complete the code to fill the base case for the first column of the dp table.

DSA Javascript

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

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

Ai

B0

Cm

Dn

Attempts:

3 left

💡 Hint

Common Mistakes

Setting dp[i][0] to 0, which ignores the cost of deletions.

Using n or m instead of i, which is incorrect indexing.

The first column represents converting the first i characters of word1 to an empty string, so dp[i][0] = i.

3fill in blank

hard

Fix the error in the condition that checks if characters are equal.

DSA Javascript

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 reverses the logic.

Using '>=' or '<=' which are not meaningful for character equality.

We check if characters are equal using '=='. If equal, no operation is needed, so dp[i][j] = dp[i-1][j-1].

4fill in blank

hard

Fill both blanks to compute the minimum edit distance considering insert, delete, and replace operations.

DSA Javascript

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

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Adp[i][j - 1]

Bdp[i - 1][j - 1]

Cdp[i][j + 1]

Ddp[i + 1][j]

Attempts:

3 left

💡 Hint

Common Mistakes

Using invalid indices like dp[i][j+1] or dp[i+1][j] which are out of bounds.

Mixing up insert and delete indices.

The three operations are delete (dp[i-1][j]), insert (dp[i][j-1]), and replace (dp[i-1][j-1]). We take the minimum plus one.

5fill in blank

hard

Fill all three blanks to complete the function that returns the edit distance between word1 and word2.

DSA Javascript

function editDistance(word1, word2) {
  const m = word1.length;
  const n = word2.length;
  const dp = Array(m + 1).fill().map(() => Array([1]).fill(0));
  for (let i = 0; i <= m; i++) dp[i][0] = i;
  for (let j = 0; j <= n; j++) dp[0][j] = j;
  for (let i = 1; i <= m; i++) {
    for (let j = 1; j <= n; j++) {
      if (word1[i - 1] [2] word2[j - 1]) {
        dp[i][j] = dp[i - 1][j - 1];
      } else {
        dp[i][j] = 1 + Math.min(dp[i - 1][j], dp[i][j - 1], dp[i - 1][j - 1]);
      }
    }
  }
  return dp[[3]][n];
}

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

An + 1

B==

Cm

Dm + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using m+1 for columns instead of n+1.

Using '!=' instead of '==' for character comparison.

Returning dp[m+1][m+1] which is out of bounds.

The dp array has n+1 columns, character equality uses '==', and the final answer is dp[m][n].

Practice

1. What does the Levenshtein distance between two strings measure?

easy

A. The length difference between the two strings

B. The number of matching characters in the same positions

C. The minimum number of single-character edits needed to change one string into the other

D. The number of vowels in both strings combined

Solution

1. Step 1: Understand the definition of Levenshtein distance

   It measures how many single-character edits (insertions, deletions, or substitutions) are needed to convert one string into another.

2. Step 2: Compare with other options

   Options B, C, and D do not describe edit distance but other unrelated string properties.

3. Final Answer:

   The minimum number of single-character edits needed to change one string into the other -> Option C

4. Quick Check:

   Levenshtein distance = minimum edits [OK]

Hint: Think of edits needed to transform one word into another [OK]

Common Mistakes:

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

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

easy

A. function levenshtein(s1, s2) { /* code */ }

B. function levenshtein(s1, s2, s3) { /* code */ }

C. function levenshtein() { /* code */ }

D. function levenshtein(s1) { /* code */ }

Solution

1. Step 1: Identify required parameters for Levenshtein function

   The function needs exactly two strings to compare, so it should accept two parameters.

2. Step 2: Check options for correct parameter count

   function levenshtein(s1, s2) { /* code */ } has two parameters (s1, s2), which is correct. Options A, B, and C have incorrect parameter counts.

3. Final Answer:

   function levenshtein(s1, s2) { /* code */ } -> Option A

4. Quick Check:

   Two strings needed as input [OK]

Hint: Levenshtein compares two strings, so function needs two parameters [OK]

Common Mistakes:

• Using wrong number of parameters
• Omitting parameters entirely
• Adding unnecessary extra parameters

3. What is the output of the following JavaScript code snippet?

function min(a, b, c) {
  return Math.min(a, b, c);
}

function levenshtein(s1, s2) {
  const len1 = s1.length;
  const len2 = s2.length;
  const dp = Array(len1 + 1).fill(null).map(() => Array(len2 + 1).fill(0));

  for (let i = 0; i <= len1; i++) dp[i][0] = i;
  for (let j = 0; j <= len2; j++) dp[0][j] = j;

  for (let i = 1; i <= len1; i++) {
    for (let j = 1; j <= len2; j++) {
      if (s1[i - 1] === s2[j - 1]) {
        dp[i][j] = dp[i - 1][j - 1];
      } else {
        dp[i][j] = min(dp[i - 1][j] + 1, dp[i][j - 1] + 1, dp[i - 1][j - 1] + 1);
      }
    }
  }
  return dp[len1][len2];
}

console.log(levenshtein('kitten', 'sitting'));

medium

A. 3

B. 4

C. 5

D. 6

Solution

1. Step 1: Understand the Levenshtein distance calculation for 'kitten' and 'sitting'

   Transform 'kitten' to 'sitting' requires 3 substitutions and 1 insertion: k->s, i->i (same), t->t (same), t->t (same), e->i, n->n (same), plus insert 'g'. Total edits = 3.

2. Step 2: Trace the code logic

   The code correctly implements the dynamic programming approach to calculate the minimum edits, returning 3.

3. Final Answer:

   3 -> Option A

4. Quick Check:

   Levenshtein('kitten', 'sitting') = 3 [OK]

Hint: Count edits: substitutions + insertions + deletions [OK]

Common Mistakes:

• Counting only substitutions
• Ignoring insertions or deletions
• Misindexing string characters

4. Identify the bug in the following Levenshtein distance function and select the fix:

function levenshtein(s1, s2) {
  const len1 = s1.length;
  const len2 = s2.length;
  const dp = Array(len1).fill(null).map(() => Array(len2).fill(0));

  for (let i = 0; i <= len1; i++) dp[i][0] = i;
  for (let j = 0; j <= len2; j++) dp[0][j] = j;

  for (let i = 1; i <= len1; i++) {
    for (let j = 1; j <= len2; j++) {
      if (s1[i - 1] === s2[j - 1]) {
        dp[i][j] = dp[i - 1][j - 1];
      } else {
        dp[i][j] = Math.min(dp[i - 1][j] + 1, dp[i][j - 1] + 1, dp[i - 1][j - 1] + 1);
      }
    }
  }
  return dp[len1][len2];
}

medium

A. Change loops to run from 0 to len1-1 and len2-1

B. Replace Math.min with Math.max in dp calculation

C. Remove the initialization loops for dp first row and column

D. Change dp array initialization to size (len1+1) x (len2+1)

Solution

1. Step 1: Analyze dp array initialization

   The dp array is initialized with size len1 x len2, but the loops access indices up to len1 and len2, causing out-of-bounds errors.

2. Step 2: Fix dp size to (len1+1) x (len2+1)

   Allocating dp with one extra row and column prevents index errors and allows proper initialization of base cases.

3. Final Answer:

   Change dp array initialization to size (len1+1) x (len2+1) -> Option D

4. Quick Check:

   dp size must match loop bounds [OK]

Hint: dp size must be string length + 1 for indexing [OK]

Common Mistakes:

• Not allocating dp with extra row/column
• Using wrong loop ranges causing errors
• Replacing min with max breaks logic

5. Given two strings s1 = 'intention' and s2 = 'execution', what is the Levenshtein distance between them? Use the concept of minimum edits (insertions, deletions, substitutions) to find the answer.

hard

A. 4

B. 6

C. 5

D. 7

Solution

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

   Transform 'intention' to 'execution' involves substitutions and insertions: i->e, n->x, t->c, e->u, n->t, plus insert 'o'. Total edits count to 6.

2. Step 2: Count minimum edits carefully

   Counting substitutions and insertions needed gives 6 as the minimum number of edits.

3. Final Answer:

   6 -> Option B

4. Quick Check:

   Minimum edits between 'intention' and 'execution' = 6 [OK]

Hint: Count each letter change and insertion carefully [OK]

Common Mistakes:

• Underestimating substitutions needed
• Ignoring insertions or deletions
• Confusing with string length difference