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+# Levenshtein Distance in C
+
+## Problem Statement
+The Levenshtein distance between two strings is the minimum number of single-character edits (insertions, deletions, or substitutions) required to transform one string into the other. This algorithm is commonly used in spell checkers, DNA sequencing, and natural language processing.
+
+### Examples
+
+#### Example 1
+**Input:**
+s1 = "kitten" s2 = "sitting"
+**Output:**
+3
+
+**Explanation:** 
+The minimum edits to transform "kitten" to "sitting" are:
+1. Substitute "k" with "s"
+2. Substitute "e" with "i"
+3. Append "g"
+
+#### Example 2
+**Input:**
+s1 = "flaw" s2 = "lawn"
+**Output:**
+2
+
+**Explanation:** 
+The minimum edits to transform "flaw" to "lawn" are:
+1. Substitute "f" with "l"
+2. Substitute "w" with "n"
+
+---
+
+## Approach
+This solution uses Dynamic Programming (DP) to calculate the Levenshtein distance between two strings efficiently.
+
+### DP Recurrence Relation
+Define `dp[i][j]` as the Levenshtein distance between the first `i` characters of `s1` and the first `j` characters of `s2`.
+
+1. **If characters match** (`s1[i-1] == s2[j-1]`):
+   - `dp[i][j] = dp[i-1][j-1]`
+2. **If characters do not match**:
+   - `dp[i][j] = 1 + min(dp[i-1][j], dp[i][j-1], dp[i-1][j-1])`
+   
+   where:
+   - `dp[i-1][j] + 1` represents a deletion.
+   - `dp[i][j-1] + 1` represents an insertion.
+   - `dp[i-1][j-1] + 1` represents a substitution.
+
+### Complexity
+- **Time Complexity:** \(O(n \times m)\), where `n` is the length of `s1` and `m` is the length of `s2`.
+- **Space Complexity:** \(O(n \times m)\), as we use a 2D array to store the distances.
+
+---
+
+## Code
+The full code is available in `program.c`.
+
+## Running the Code
+
+### Prerequisites
+Ensure you have a C compiler installed, such as GCC.
+
+### Instructions
+1. Clone this repository.
+2. Compile the C file:
+
+   ```bash
+   gcc program.c -o program
+
+3. Run the compiled program:
+   ./program
diff --git a/Dynamic Programming/Levenshtein Distance/program.c b/Dynamic Programming/Levenshtein Distance/program.c
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+#include <stdio.h>
+#include <string.h>
+#include <stdlib.h>
+
+// Function to calculate the minimum of three values
+int min(int a, int b, int c) {
+    int min = a;
+    if (b < min) min = b;
+    if (c < min) min = c;
+    return min;
+}
+
+// Function to calculate Levenshtein distance
+int levenshteinDistance(const char *s1, const char *s2) {
+    int len1 = strlen(s1);
+    int len2 = strlen(s2);
+
+    // Create a 2D array to store distances
+    int **dp = (int **)malloc((len1 + 1) * sizeof(int *));
+    for (int i = 0; i <= len1; i++) {
+        dp[i] = (int *)malloc((len2 + 1) * sizeof(int));
+    }
+
+    // Initialize base cases
+    for (int i = 0; i <= len1; i++) {
+        dp[i][0] = i; // Distance of any first string to an empty second string
+    }
+    for (int j = 0; j <= len2; j++) {
+        dp[0][j] = j; // Distance of any second string to an empty first string
+    }
+
+    // Fill the dp array
+    for (int i = 1; i <= len1; i++) {
+        for (int j = 1; j <= len2; j++) {
+            if (s1[i - 1] == s2[j - 1]) {
+                dp[i][j] = dp[i - 1][j - 1]; // Characters match, no cost
+            } else {
+                dp[i][j] = min(
+                    dp[i - 1][j] + 1,     // Deletion
+                    dp[i][j - 1] + 1,     // Insertion
+                    dp[i - 1][j - 1] + 1  // Substitution
+                );
+            }
+        }
+    }
+
+    int distance = dp[len1][len2];
+
+    // Free allocated memory
+    for (int i = 0; i <= len1; i++) {
+        free(dp[i]);
+    }
+    free(dp);
+
+    return distance;
+}
+
+int main() {
+    const char *s1 = "kitten";
+    const char *s2 = "sitting";
+
+    printf("Levenshtein distance between '%s' and '%s' is %d\n", s1, s2, levenshteinDistance(s1, s2));
+
+    return 0;
+}