Edit Distance Problem Levenshtein in DSA C# - Time & Space Complexity

Time Complexity: Edit Distance Problem Levenshtein

O(m x n)

Understanding Time Complexity

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

How does the work increase when the input words become bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

public int LevenshteinDistance(string s, string t) {
    int m = s.Length, n = t.Length;
    int[,] dp = new int[m + 1, n + 1];
    for (int i = 0; i <= m; i++) dp[i, 0] = i;
    for (int j = 0; j <= n; j++) dp[0, j] = j;
    for (int i = 1; i <= m; i++) {
        for (int j = 1; j <= n; j++) {
            int cost = s[i - 1] == t[j - 1] ? 0 : 1;
            dp[i, j] = Math.Min(Math.Min(dp[i - 1, j] + 1, dp[i, j - 1] + 1), dp[i - 1, j - 1] + cost);
        }
    }
    return dp[m, n];
}
    

This code calculates the minimum number of edits needed to change one word into another using a grid of subproblems.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Nested loops filling a 2D table of size m by n.
• How many times: The inner loop runs n times for each of the m iterations of the outer loop, so roughly m x n times.

How Execution Grows With Input

As the lengths of the two words increase, the number of operations grows 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 quadruples 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 input words.

Common Mistake

[X] Wrong: "The algorithm runs in linear time because it just loops through the words once."

[OK] Correct: The algorithm uses two nested loops, so it checks every pair of characters, making the work multiply, not just add.

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 Levenshtein Edit Distance between two strings measure?

easy

A. The total length of both strings combined

B. The maximum length of common substring between two strings

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

D. The minimum number of single-character edits to change one string into another

Solution

1. Step 1: Understand the definition of Levenshtein distance

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

2. Step 2: Compare with other options

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

3. Final Answer:

   The minimum number of single-character edits to change one string into another -> Option D

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
• Assuming it measures string length

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

easy

A. private string LevenshteinDistance(string s1, string s2)

B. public void LevenshteinDistance(int s1, int s2)

C. public int LevenshteinDistance(string s1, string s2)

D. public int LevenshteinDistance(int[] s1, int[] s2)

Solution

1. Step 1: Check parameter types for string inputs

   The function should accept two strings, so parameters must be string types.

2. Step 2: Check return type for distance value

   Distance is an integer, so return type should be int, not void or string.

3. Final Answer:

   public int LevenshteinDistance(string s1, string s2) -> Option C

4. Quick Check:

   Function signature matches string inputs and int output [OK]

Hint: Function takes strings, returns int distance [OK]

Common Mistakes:

• Using void return type instead of int
• Using int parameters instead of strings
• Returning string instead of int

3. What is the output of the following C# code snippet?

var s1 = "kitten";
var s2 = "sitting";
Console.WriteLine(LevenshteinDistance(s1, s2));

int LevenshteinDistance(string a, string b) {
    int[,] dp = new int[a.Length + 1, b.Length + 1];
    for (int i = 0; i <= a.Length; i++) dp[i, 0] = i;
    for (int j = 0; j <= b.Length; j++) dp[0, j] = j;
    for (int i = 1; i <= a.Length; i++) {
        for (int j = 1; j <= b.Length; j++) {
            int cost = (a[i - 1] == b[j - 1]) ? 0 : 1;
            dp[i, j] = Math.Min(Math.Min(dp[i - 1, j] + 1, dp[i, j - 1] + 1), dp[i - 1, j - 1] + cost);
        }
    }
    return dp[a.Length, b.Length];
}

medium

A. 3

B. 4

C. 5

D. 6

Solution

1. Step 1: Understand the strings and operations

   "kitten" to "sitting" requires substitutions, insertions, deletions.

2. Step 2: Calculate Levenshtein distance manually

   Changes: k->s (substitution), e->i (substitution), t->t (no change), t->t (no change), e->i (substitution), n->n (no change), add g (insertion) = 3 substitutions + 1 insertion = 4 edits total.

3. Final Answer:

   4 -> Option B

4. Quick Check:

   Distance("kitten", "sitting") = 4 [OK]

Hint: Remember common example: kitten -> sitting = 4 edits [OK]

Common Mistakes:

• Counting only substitutions, ignoring insertions
• Miscounting character positions
• Confusing with Hamming distance

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

int LevenshteinDistance(string a, string b) {
    int[,] dp = new int[a.Length, b.Length];
    for (int i = 0; i < a.Length; i++) dp[i, 0] = i;
    for (int j = 0; j < b.Length; j++) dp[0, j] = j;
    for (int i = 1; i < a.Length; i++) {
        for (int j = 1; j < b.Length; j++) {
            int cost = (a[i] == b[j]) ? 0 : 1;
            dp[i, j] = Math.Min(Math.Min(dp[i - 1, j] + 1, dp[i, j - 1] + 1), dp[i - 1, j - 1] + cost);
        }
    }
    return dp[a.Length - 1, b.Length - 1];
}

medium

A. Array size should be (a.Length+1, b.Length+1) and loops adjusted

B. Cost calculation should use a[i-1] and b[j-1]

C. Return statement should be dp[a.Length, b.Length]

D. No bug, code is correct

Solution

1. Step 1: Check array size and indexing

   dp array must be sized (a.Length+1, b.Length+1) to include empty prefixes; current size misses last row/column.

2. Step 2: Adjust loops to run from 0 to length inclusive

   Loops should run to <= length to fill dp correctly; current loops miss last indices.

3. Final Answer:

   Array size should be (a.Length+1, b.Length+1) and loops adjusted -> Option A

4. Quick Check:

   dp array size must include empty string row/column [OK]

Hint: dp array must be one larger than string lengths [OK]

Common Mistakes:

• Using array size equal to string length only
• Incorrect loop bounds causing index errors
• Ignoring empty string base cases

5. Given two strings s1 = "intention" and s2 = "execution", what is the Levenshtein distance between them? Use the standard operations: insert, delete, substitute.

hard

A. 5

B. 6

C. 4

D. 7

Solution

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

   Compare characters and count minimal edits: substitutions and insertions/deletions.

2. Step 2: Calculate minimal edits manually

   Edits: i->e (substitution), n->x (substitution), t->c (substitution), insert u, substitute n->o, substitute n->n (no change), total 5 edits.

3. Final Answer:

   5 -> Option A

4. Quick Check:

   Distance("intention", "execution") = 5 [OK]

Hint: Count substitutions and insertions carefully [OK]

Common Mistakes:

• Underestimating substitutions needed
• Ignoring insertions or deletions
• Confusing with Hamming distance