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

Time Complexity: Edit Distance Problem Levenshtein

O(3^{min(m,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 number of steps change when the input words become bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

int min(int a, int b, int c) {
    int m = a < b ? a : b;
    return m < c ? m : c;
}

int editDistance(char *s1, char *s2, int m, int n) {
    if (m == 0) return n;
    if (n == 0) return m;

    if (s1[m - 1] == s2[n - 1])
        return editDistance(s1, s2, m - 1, n - 1);

    return 1 + min(
        editDistance(s1, s2, m, n - 1),    // Insert
        editDistance(s1, s2, m - 1, n),    // Remove
        editDistance(s1, s2, m - 1, n - 1) // Replace
    );
}
    

This code finds the minimum number of edits (insert, remove, replace) to change one string into another using recursion.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive calls that explore all combinations of substring lengths.
• How many times: Each call branches into three more calls unless base case is reached, leading to many repeated calls.

How Execution Grows With Input

Each character in the first word can match with each character in the second word, and the recursion tries all possibilities.

Input Size (m, n)	Approx. Operations
5, 5	Hundreds of calls
10, 10	Thousands of calls
20, 20	Millions of calls

Pattern observation: The number of calls grows very fast, roughly tripling with each increase in input size.

Final Time Complexity

Time Complexity: O(3^{min(m,n)})

This means the time needed grows very fast as the lengths of the two strings increase, because the recursion explores many overlapping cases.

Common Mistake

[X] Wrong: "The recursive solution runs quickly because it only checks each character once."

[OK] Correct: The recursion repeats many calls for the same substrings, causing exponential growth in steps.

Interview Connect

Understanding this complexity helps you explain why a simple recursive solution is slow and why dynamic programming is needed to improve it.

Self-Check

"What if we add memoization to store results of subproblems? 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 insertions, deletions, or substitutions to convert one string into another

B. The number of matching characters in the same positions

C. The length difference between two strings

D. The maximum number of characters that differ between two strings

Solution

1. Step 1: Understand the definition of Edit Distance

   Edit Distance counts the smallest number of changes needed to turn one string into another.

2. Step 2: Identify allowed operations

   These operations include insertion, deletion, and substitution of characters.

3. Final Answer:

   The minimum number of insertions, deletions, or substitutions to convert one string into another -> Option A

4. Quick Check:

   Edit Distance = minimum edits needed [OK]

Hint: Think of edits needed to match strings exactly [OK]

Common Mistakes:

• Confusing edit distance with matching characters count
• Thinking it only counts substitutions
• Assuming it measures length difference only

2. Which of the following is the correct way to initialize the first row of the DP table for Levenshtein distance in C?

easy

A. for (int j = 0; j <= n; j++) dp[0][j] = j;

B. for (int i = 0; i <= m; i++) dp[i][0] = i;

C. for (int j = 1; j <= n; j++) dp[0][j] = 0;

D. for (int i = 1; i <= m; i++) dp[0][i] = i;

Solution

1. Step 1: Identify the DP table dimensions

   The DP table has dimensions (m+1) x (n+1), where m and n are string lengths.

2. Step 2: Initialize first row correctly

   The first row dp[0][j] represents converting empty string to prefix of second string, so dp[0][j] = j.

3. Final Answer:

   for (int j = 0; j <= n; j++) dp[0][j] = j; -> Option A

4. Quick Check:

   First row = 0 to n [OK]

Hint: First row = indices of second string length [OK]

Common Mistakes:

• Mixing row and column indices
• Starting loop from 1 instead of 0
• Initializing first column instead of first row

3. What is the output of the following C code snippet computing Levenshtein distance between "kitten" and "sitting"?

int min(int a, int b, int c) {
    int m = a < b ? a : b;
    return m < c ? m : c;
}

int levenshtein(char *s, char *t) {
    int m = strlen(s), n = strlen(t);
    int dp[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] = min(dp[i-1][j] + 1, dp[i][j-1] + 1, dp[i-1][j-1] + cost);
        }
    }
    return dp[m][n];
}

printf("%d", levenshtein("kitten", "sitting"));

medium

A. 3

B. 4

C. 5

D. 6

Solution

1. Step 1: Understand the strings and operations

   "kitten" to "sitting" requires substitutions and insertions: k->s, e->i, add g.

2. Step 2: Calculate minimum edits

   Minimum edits are 3 substitutions and 1 insertion = 4 total edits.

3. Final Answer:

   4 -> Option B

4. Quick Check:

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

Hint: Recall classic example: kitten -> sitting = 4 edits [OK]

Common Mistakes:

• Counting only substitutions, missing insertions
• Confusing character positions
• Off-by-one errors in indexing

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

int levenshtein(char *s, char *t) {
    int m = strlen(s), n = strlen(t);
    int dp[m][n];
    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] = min(dp[i-1][j] + 1, dp[i][j-1] + 1, dp[i-1][j-1] + cost);
        }
    }
    return dp[m][n];
}

medium

A. Loops should start from 0, not 1

B. The cost calculation is incorrect

C. dp array size should be (m+1) x (n+1), not m x n

D. Return statement should be dp[m-1][n-1]

Solution

1. Step 1: Check dp array declaration

   dp is declared as dp[m][n], but indices go up to m and n inclusive, so size is insufficient.

2. Step 2: Understand indexing needs

   We need dp of size (m+1) x (n+1) to store base cases and full computation.

3. Final Answer:

   dp array size should be (m+1) x (n+1), not m x n -> Option C

4. Quick Check:

   dp size = (m+1)*(n+1) [OK]

Hint: DP table size must cover all indices including zero [OK]

Common Mistakes:

• Declaring dp too small causing out-of-bounds errors
• Misunderstanding loop ranges
• Incorrect return index

5. Given two strings "intention" and "execution", what is the minimum edit distance using Levenshtein algorithm? Consider insertions, deletions, and substitutions only.

hard

A. 7

B. 8

C. 5

D. 6

Solution

1. Step 1: Analyze string lengths and differences

   "intention" and "execution" differ in several characters and positions.

2. Step 2: Calculate minimum edits stepwise

   Edits include substitutions (i->e, n->x, t->e, e->c, n->u, etc.) totaling 6 changes.

3. Final Answer:

   6 -> Option D

4. Quick Check:

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

Hint: Count substitutions and insertions carefully [OK]

Common Mistakes:

• Underestimating substitutions needed
• Ignoring insertions or deletions
• Confusing similar characters as matches