Edit Distance Problem Levenshtein in DSA Python - Time & Space Complexity

Time Complexity: Edit Distance Problem Levenshtein

O(m * n)

Understanding Time Complexity

We want to understand how the time needed to find the minimum edits between two words grows as the words get longer.

The question is: how does the work increase when the input words become bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

def levenshtein_distance(s1: str, s2: str) -> int:
    m, n = len(s1), len(s2)
    dp = [[0] * (n + 1) for _ in range(m + 1)]

    for i in range(m + 1):
        for j in range(n + 1):
            if i == 0:
                dp[i][j] = j
            elif j == 0:
                dp[i][j] = i
            elif 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]

This code calculates the minimum number of edits (insert, delete, replace) to change one word into another using a table.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Nested loops filling a 2D table of size (m+1) x (n+1).
• How many times: Outer loop runs m+1 times, inner loop runs n+1 times for each outer loop.

How Execution Grows With Input

As the length of the two words grows, the number of operations grows roughly by multiplying their lengths.

Input Size (m, n)	Approx. Operations
10, 10	~100
100, 100	~10,000
1000, 1000	~1,000,000

Pattern observation: Doubling the length of both words roughly squares the work needed.

Final Time Complexity

Time Complexity: O(m * n)

This means the time needed grows proportionally to the product of the lengths of the two words.

Common Mistake

[X] Wrong: "The algorithm runs in linear time because it just compares characters."

[OK] Correct: The algorithm compares all pairs of prefixes, so it must fill a table with m times n cells, not just a single pass.

Interview Connect

Understanding this time complexity helps you explain how dynamic programming solves problems efficiently by breaking them into smaller parts.

Self-Check

"What if we used recursion without memoization? How would the time complexity change?"

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