Edit Distance Problem Levenshtein in DSA C - Build from Scratch

Edit Distance Problem Levenshtein

📖 Scenario: You are building a simple spell checker that measures how different two words are by counting the minimum number of edits needed to change one word into the other. Edits can be inserting, deleting, or replacing a letter.This difference is called the Levenshtein distance.

🎯 Goal: Calculate the Levenshtein distance between two given words using dynamic programming in C.

📋 What You'll Learn

Create two strings word1 and word2 with exact values

Create an integer variable len1 for the length of word1 and len2 for the length of word2

Create a 2D integer array dp of size (len1+1) x (len2+1) to store intermediate results

Fill the dp array using the Levenshtein distance logic

Print the final edit distance stored in dp[len1][len2]

💡 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 dynamic programming and string algorithms like Levenshtein distance is important for software engineers working on text processing, search engines, and bioinformatics.

1

DATA SETUP: Create the input words

Create two strings called word1 with value "kitten" and word2 with value "sitting".

DSA C

# Your code here

Hint

Use char word1[] = "kitten"; and char word2[] = "sitting"; to create the words.

2

CONFIGURATION: Calculate lengths and declare dp array

Add two integer variables len1 and len2 to store the lengths of word1 and word2 using strlen. Then declare a 2D integer array dp of size len1 + 1 by len2 + 1.

DSA C

#include 
#include 

int main() {
    char word1[] = "kitten";
    char word2[] = "sitting";
    // Your code here
    return 0;
}

Hint

Use strlen(word1) and strlen(word2) to get lengths. Declare dp as int dp[len1 + 1][len2 + 1];.

3

CORE LOGIC: Fill dp array with Levenshtein distance values

Use two nested for loops with variables i from 0 to len1 and j from 0 to len2. Inside, fill dp[i][j] as follows: if i == 0, set dp[i][j] = j; else if j == 0, set dp[i][j] = i; else if word1[i-1] == word2[j-1], set dp[i][j] = dp[i-1][j-1]; otherwise, set dp[i][j] to 1 plus the minimum of dp[i-1][j], dp[i][j-1], and dp[i-1][j-1].

DSA C

#include 
#include 

int main() {
    char word1[] = "kitten";
    char word2[] = "sitting";
    int len1 = strlen(word1);
    int len2 = strlen(word2);
    int dp[len1 + 1][len2 + 1];
    // Your code here
    return 0;
}

Hint

Use nested loops and conditions to fill dp. Use a helper min function to find the smallest of three values.

4

OUTPUT: Print the final Levenshtein distance

Print the integer value stored in dp[len1][len2] using printf with the format "%d\n".

DSA C

#include 
#include 

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

int main() {
    char word1[] = "kitten";
    char word2[] = "sitting";
    int len1 = strlen(word1);
    int len2 = strlen(word2);
    int dp[len1 + 1][len2 + 1];

    for (int i = 0; i <= len1; i++) {
        for (int j = 0; j <= len2; j++) {
            if (i == 0) {
                dp[i][j] = j;
            } else if (j == 0) {
                dp[i][j] = i;
            } else if (word1[i - 1] == word2[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]);
            }
        }
    }

    // Your code here
    return 0;
}

Hint

Use printf("%d\n", dp[len1][len2]); to print the final edit distance.