Edit Distance Problem Levenshtein in DSA Python - Build from Scratch

Edit Distance Problem Levenshtein

📖 Scenario: Imagine you are building a spell checker that suggests corrections for misspelled words. To do this, you need to find how different two words are by counting the minimum number of changes needed to turn one word into the other. This is called the Levenshtein edit distance.

🎯 Goal: You will build a program that calculates the Levenshtein edit distance between two given words. This distance counts how many single-letter changes, insertions, or deletions are needed to change one word into the other.

📋 What You'll Learn

Create two string variables with exact given words

Create a 2D list (matrix) to store distances

Fill the matrix using the Levenshtein algorithm

Print the final edit distance value

💡 Why This Matters

🌍 Real World

Spell checkers, auto-correct features, DNA sequence analysis, and natural language processing use edit distance to find how similar two strings are.

💼 Career

Understanding edit distance is important for software engineers working on text processing, search engines, and bioinformatics.

1

Create the two words to compare

Create two string variables called word1 and word2 with the exact values "kitten" and "sitting" respectively.

DSA Python

# Create the two words to compare
# Your code here

Hint

Use simple assignment to create word1 and word2 with the exact words.

2

Create the distance matrix

Create a 2D list called distance with dimensions len(word1) + 1 rows and len(word2) + 1 columns. Initialize it with zeros. Then fill the first row with values from 0 to len(word2) and the first column with values from 0 to len(word1).

DSA Python

word1 = "kitten"
word2 = "sitting"
# Create the distance matrix and initialize first row and column
# Your code here

Hint

Use list comprehension to create the matrix. Then use two for loops to fill the first row and column with increasing numbers.

3

Fill the matrix with Levenshtein distances

Use nested for loops with variables i from 1 to len(word1) and j from 1 to len(word2). For each position, if word1[i-1] equals word2[j-1], set distance[i][j] to distance[i-1][j-1]. Otherwise, set distance[i][j] to 1 plus the minimum of distance[i-1][j], distance[i][j-1], and distance[i-1][j-1].

DSA Python

word1 = "kitten"
word2 = "sitting"
distance = [[0] * (len(word2) + 1) for _ in range(len(word1) + 1)]
for i in range(len(word1) + 1):
    distance[i][0] = i
for j in range(len(word2) + 1):
    distance[0][j] = j
# Fill the matrix with Levenshtein distances
# Your code here

Hint

Use two nested loops and compare characters at i-1 and j-1. Update the matrix cell with the correct minimum cost.

4

Print the final edit distance

Print the value at distance[len(word1)][len(word2)] which is the Levenshtein edit distance between word1 and word2.

DSA Python

word1 = "kitten"
word2 = "sitting"
distance = [[0] * (len(word2) + 1) for _ in range(len(word1) + 1)]
for i in range(len(word1) + 1):
    distance[i][0] = i
for j in range(len(word2) + 1):
    distance[0][j] = j
for i in range(1, len(word1) + 1):
    for j in range(1, len(word2) + 1):
        if word1[i - 1] == word2[j - 1]:
            distance[i][j] = distance[i - 1][j - 1]
        else:
            distance[i][j] = 1 + min(distance[i - 1][j], distance[i][j - 1], distance[i - 1][j - 1])
# Print the final edit distance
# Your code here

Hint

The last cell of the matrix holds the answer. Use print(distance[len(word1)][len(word2)]).

Practice

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

easy

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

B. The number of common characters between two strings

C. The length of the longest common substring

D. The number of characters in the longer string

Solution

1. Step 1: Understand the definition of Edit Distance

   Edit Distance counts how many changes (insert, delete, replace) are needed to convert one string to another.

2. Step 2: Compare options with definition

   Only The minimum number of single-character edits to change one string into the other correctly describes this minimum number of single-character edits.

3. Final Answer:

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

4. Quick Check:

   Edit Distance = minimum edits [OK]

Hint: Edit Distance counts minimal edits to match strings [OK]

Common Mistakes:

• Confusing with longest common substring
• Thinking it counts common characters
• Assuming it measures string length

2. Which of the following Python code snippets correctly initializes a 2D list for dynamic programming of size (m+1) x (n+1)?

easy

A. dp = [0] * (m+1) * (n+1)

B. dp = [[0] * (n+1) for _ in range(m+1)]

C. dp = [[0] * m for _ in range(n)]

D. dp = [[0] * (m+1)] * (n+1)

Solution

1. Step 1: Understand 2D list initialization in Python

   Using list comprehension with separate inner lists avoids shared references.

2. Step 2: Analyze each option

   dp = [[0] * (n+1) for _ in range(m+1)] correctly creates independent rows. dp = [[0] * (m+1)] * (n+1) creates shallow copies causing bugs. Options A and C have wrong dimensions or types.

3. Final Answer:

   dp = [[0] * (n+1) for _ in range(m+1)] -> Option B

4. Quick Check:

   Proper 2D list init = dp = [[0] * (n+1) for _ in range(m+1)] [OK]

Hint: Use list comprehension for 2D lists to avoid shared rows [OK]

Common Mistakes:

• Using multiplication for 2D lists causing shared references
• Mixing up m and n dimensions
• Creating 1D list instead of 2D

3. What is the output of the following Python code snippet that calculates the Levenshtein distance between 'kitten' and 'sitting'?

def levenshtein(s1, s2):
    m, n = len(s1), len(s2)
    dp = [[0] * (n+1) for _ in range(m+1)]
    for i in range(m+1):
        dp[i][0] = i
    for j in range(n+1):
        dp[0][j] = j
    for i in range(1, m+1):
        for j in range(1, n+1):
            if s1[i-1] == s2[j-1]:
                dp[i][j] = dp[i-1][j-1]
            else:
                dp[i][j] = 1 + min(dp[i-1][j], dp[i][j-1], dp[i-1][j-1])
    return dp[m][n]

print(levenshtein('kitten', 'sitting'))

medium

A. 4

B. 5

C. 3

D. 6

Solution

1. Step 1: Understand the Levenshtein distance calculation

   Compare 'kitten' and 'sitting' stepwise counting minimal edits: substitute 'k'->'s' (kitten -> sitten), substitute 'e'->'i' (sitten -> sittin), insert 'g' at end (sittin -> sitting). Total 3 edits.

2. Step 2: Verify with code logic

   The code correctly implements the DP approach and returns dp[m][n] which is 3 for these inputs.

3. Final Answer:

   3 -> Option C

4. Quick Check:

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

Hint: Count minimal edits: replace, insert, delete [OK]

Common Mistakes:

• Counting only replacements, ignoring insertions
• Miscounting characters to change
• Confusing with substring length

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

def levenshtein_bug(s1, s2):
    m, n = len(s1), len(s2)
    dp = [[0] * (n+1)] * (m+1)
    for i in range(m+1):
        dp[i][0] = i
    for j in range(n+1):
        dp[0][j] = j
    for i in range(1, m+1):
        for j in range(1, n+1):
            if s1[i-1] == s2[j-1]:
                dp[i][j] = dp[i-1][j-1]
            else:
                dp[i][j] = 1 + min(dp[i-1][j], dp[i][j-1], dp[i-1][j-1])
    return dp[m][n]

medium

A. Using dp = [[0] * (n+1)] * (m+1) causes all rows to reference the same list

B. The loops should start from 0 instead of 1

C. The condition s1[i-1] == s2[j-1] should be s1[i] == s2[j]

D. The return statement should be dp[0][0]

Solution

1. Step 1: Analyze dp initialization

   Using multiplication to create dp causes all rows to point to the same list, so updating one row updates all rows.

2. Step 2: Understand impact on algorithm

   This shared reference breaks the DP table updates, causing incorrect results.

3. Final Answer:

   Using dp = [[0] * (n+1)] * (m+1) causes all rows to reference the same list -> Option A

4. Quick Check:

   Shared row references cause bugs [OK]

Hint: Avoid list multiplication for 2D lists; use list comprehension [OK]

Common Mistakes:

• Using wrong dp initialization
• Incorrect loop ranges
• Wrong indexing in condition

5. Given two strings s1 = 'intention' and s2 = 'execution', what is the minimum edit distance using Levenshtein algorithm? Consider insert, delete, and replace operations each cost 1.

hard

A. 7

B. 5

C. 4

D. 6

Solution

1. Step 1: Compare strings and count minimal edits

   Transform 'intention' to 'execution' requires the minimum number of single-character edits (insertions, deletions, substitutions) of 6.

2. Step 2: Confirm with Levenshtein DP logic

   Dynamic programming confirms minimum edits as 6 for these strings.

3. Final Answer:

   6 -> Option D

4. Quick Check:

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

Hint: Count replacements and insertions carefully for longer strings [OK]

Common Mistakes:

• Underestimating number of replacements
• Ignoring insertions
• Confusing with Hamming distance