Edit Distance Problem Levenshtein in DSA Typescript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

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

DSA Typescript

const dp: number[][] = Array(m + 1).fill(null).map(() => Array(n + 1).fill([1]));

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

A1

Bnull

C0

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing with 1 causes incorrect base cases.

Using null or -1 leads to runtime errors.

The DP table is initialized with 0 because the base case for edit distance starts with zero cost for empty strings.

2fill in blank

medium

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

DSA Typescript

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

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

A+

B*

C-

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using multiplication or division instead of addition.

Using subtraction causes negative values.

We add 1 for each insertion to transform an empty string to the first j characters.

3fill in blank

hard

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

DSA Typescript

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 '!=' reverses the logic causing wrong results.

Using '>=' or '<=' is incorrect for character comparison.

We check if characters are equal to avoid extra cost; use '==' for equality.

4fill in blank

hard

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

DSA Typescript

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

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

A[i - 1][j - 1]

B[i][j - 1]

C[i - 1][j]

D[i][j]

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing indices causes wrong minimum calculation.

Using dp[i][j] inside Math.min causes infinite recursion.

The three options are delete (dp[i-1][j]), insert (dp[i][j-1]), and replace (dp[i-1][j-1]). Here, blanks fill the replace case.

5fill in blank

hard

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

DSA Typescript

function levenshtein([1]: string, [2]: string): number {
  const m = [1].length;
  const n = [2].length;
  // ... DP initialization and computation ...
  return dp[m][n];
}

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

Aword1

Bword2

Cs1

Ds2

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variable names causes reference errors.

Mismatching parameter names and length variables.

The function takes two strings named word1 and word2. We use word1.length and word2.length to get lengths.

Practice

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

easy

A. The maximum length of common substring between two strings

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

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

D. The total length of both strings combined

Solution

1. Step 1: Understand the definition of Levenshtein distance

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

2. Step 2: Compare with other options

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

3. Final Answer:

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

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, not insertions/deletions
• Assuming it measures similarity score instead of edit count

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

easy

A. function levenshtein(s1: string, s2: string): number

B. function levenshtein(s1: number, s2: number): string

C. function levenshtein(s1: string[], s2: string[]): boolean

D. function levenshtein(s1: any, s2: any): void

Solution

1. Step 1: Identify correct parameter and return types

   Levenshtein distance takes two strings and returns a number representing the edit distance.

2. Step 2: Check each option's signature

   function levenshtein(s1: string, s2: string): number matches the expected input and output types. Others mismatch types or return void.

3. Final Answer:

   function levenshtein(s1: string, s2: string): number -> Option A

4. Quick Check:

   Input strings, output number [OK]

Hint: Input strings, output number for edit distance [OK]

Common Mistakes:

• Using wrong parameter types like arrays or numbers
• Returning wrong types like boolean or void
• Using any type without specifying string inputs

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

function levenshtein(s1: string, s2: string): number {
  const dp: number[][] = [];
  for (let i = 0; i <= s1.length; i++) {
    dp[i] = [i];
  }
  for (let j = 0; j <= s2.length; j++) {
    dp[0][j] = j;
  }
  for (let i = 1; i <= s1.length; i++) {
    for (let j = 1; j <= s2.length; 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], dp[i][j - 1], dp[i - 1][j - 1]) + 1;
      }
    }
  }
  return dp[s1.length][s2.length];
}

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

medium

A. 3

B. 4

C. 5

D. 6

Solution

1. Step 1: Initialize DP table for 'kitten' and 'sitting'

   DP table rows = 7, columns = 8. Base cases set for empty prefixes.

2. Step 2: Fill DP table comparing characters and calculating minimum edits

   Calculations show minimum edits needed to convert 'kitten' to 'sitting' is 3 (substitute 'k'->'s', 'e'->'i', insert 'g').

3. Final Answer:

   3 -> Option A

4. Quick Check:

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

Hint: Trace DP table for small strings to find edit count [OK]

Common Mistakes:

• Counting only substitutions, ignoring insertions/deletions
• Off-by-one errors in indexing strings
• Misreading the DP initialization steps

4. Identify the bug in this TypeScript implementation of Levenshtein distance:

function levenshtein(s1: string, s2: string): number {
  const dp: number[][] = [];
  for (let i = 0; i < s1.length; i++) {
    dp[i] = [i];
  }
  for (let j = 0; j < s2.length; j++) {
    dp[0][j] = j;
  }
  for (let i = 1; i <= s1.length; i++) {
    for (let j = 1; j <= s2.length; 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], dp[i][j - 1], dp[i - 1][j - 1]) + 1;
      }
    }
  }
  return dp[s1.length][s2.length];
}

medium

A. The function should return dp[0][0] instead of dp[s1.length][s2.length]

B. The dp array should be initialized with zeros instead of indices

C. The comparison should be s1[i] === s2[j], not s1[i - 1] === s2[j - 1]

D. The loops initializing dp should run up to <= length, not < length

Solution

1. Step 1: Check dp initialization loops

   Loops run from 0 to less than length, but dp needs size length+1 to include empty prefix cases.

2. Step 2: Correct loop ranges to include dp[s1.length] and dp[s2.length]

   Change loops to run up to <= length to properly initialize dp edges.

3. Final Answer:

   The loops initializing dp should run up to <= length, not < length -> Option D

4. Quick Check:

   DP size must cover full string lengths [OK]

Hint: Initialize DP arrays with length+1 for empty string cases [OK]

Common Mistakes:

• Using < length instead of <= length in loops
• Incorrect string index comparisons causing off-by-one errors
• Returning wrong dp cell ignoring full string lengths

5. Given two strings s1 = 'intention' and s2 = 'execution', what is the minimum edit distance using Levenshtein algorithm? Also, explain why this distance is correct.

hard

A. 6

B. 5

C. 4

D. 7

Solution

1. Step 1: Analyze transformations from 'intention' to 'execution'

   Operations include substituting 'i'->'e', 'n'->'x', deleting 't', substituting 'n'->'c', substituting 't'->'u'. Counting these gives 5 edits.

2. Step 2: Confirm minimal edits with DP approach

   DP table confirms minimum edits needed is 5, as fewer edits cannot convert all differing characters.

3. Final Answer:

   5 -> Option B

4. Quick Check:

   Minimum edits = 5 for 'intention' to 'execution' [OK]

Hint: Count substitutions, insertions, deletions carefully [OK]

Common Mistakes:

• Underestimating edits by ignoring deletions
• Counting only substitutions
• Misaligning characters causing wrong counts