Why Edit Distance Problem Levenshtein in DSA Typescript?

The Big Idea

Discover how a simple table can solve the tricky problem of measuring text differences perfectly!

The Scenario

Imagine you have two long lists of words from different books, and you want to find out how different they are by counting how many changes it takes to turn one list into the other.

Doing this by hand means checking every word, letter by letter, and guessing how many changes are needed. This is like trying to fix a long messy text without any tools.

The Problem

Manually comparing two strings or texts is slow and tiring. You might miss small differences or count changes wrongly.

It's easy to get confused about whether to add, remove, or change letters, especially when the texts are long.

This leads to mistakes and wastes a lot of time.

The Solution

The Edit Distance Problem, solved by the Levenshtein algorithm, gives a clear way to count the smallest number of changes needed to turn one string into another.

It uses a simple table to keep track of changes step-by-step, so you never lose track or guess wrong.

This method is fast, reliable, and works even for very long texts.

Before vs After

✗ Before

function manualCompare(str1: string, str2: string): number {
  // Guess changes by checking letters one by one
  let changes = 0;
  for (let i = 0; i < Math.min(str1.length, str2.length); i++) {
    if (str1[i] !== str2[i]) changes++;
  }
  changes += Math.abs(str1.length - str2.length);
  return changes;
}

✓ After

function levenshteinDistance(str1: string, str2: string): number {
  const dp: number[][] = [];
  for (let i = 0; i <= str1.length; i++) {
    dp[i] = [];
    dp[i][0] = i;
  }
  for (let j = 0; j <= str2.length; j++) {
    dp[0][j] = j;
  }
  for (let i = 1; i <= str1.length; i++) {
    for (let j = 1; j <= str2.length; j++) {
      if (str1[i - 1] === str2[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[str1.length][str2.length];
}

What It Enables

This lets you quickly and accurately measure how similar or different two texts are, enabling spell checkers, DNA analysis, and more.

Real Life Example

Spell checkers use this to suggest the closest correct word when you mistype something.

Key Takeaways

Manual comparison is slow and error-prone.

Levenshtein algorithm uses a table to count minimum changes.

This method is fast, reliable, and widely useful.

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