added edit distance problem in cpp

DSA SOLUTIONS (TILL TREE & BASIC GRAPH CONCEPTS)/DP/EditDistance.cpp

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+/**
+ * @file edit_distance.cpp
+ * @brief Calculates the minimum number of operations to convert one string to another
+ * 
+ * Problem: Given two strings word1 and word2, return the minimum number of 
+ * operations required to convert word1 to word2.
+ * 
+ * Allowed operations:
+ * - Insert a character
+ * - Delete a character
+ * - Replace a character
+ * 
+ * Time Complexity: O(m*n), where m and n are lengths of word1 and word2
+ * Space Complexity: O(m*n)
+ * 
+ * Dynamic Programming Approach:
+ * 1. Create a 2D DP table to store minimum edit distances
+ * 2. Initialize base cases for empty string transformations
+ * 3. Fill the table by comparing characters and choosing minimum operations
+ */
+
+#include <bits/stdc++.h>
+using namespace std;
+
+// Function to calculate minimum edit distance between two strings
+int minDistance(string word1, string word2) {
+    int a = word1.length(); 
+    int b = word2.length(); 
+
+    // DP table to store edit distances
+    vector<vector<int>> dp(a + 1, vector<int>(b + 1, 0));
+
+    // Base case: transforming empty string
+    for (int i = 0; i <= a; i++) {
+        dp[i][0] = i;  // Delete i characters
+    }
+
+    for (int j = 0; j <= b; j++) {
+        dp[0][j] = j;  // Insert j characters
+    }
+
+    // Fill DP table
+    for (int i = 1; i <= a; i++) {
+        for (int j = 1; j <= b; j++) {
+            // If characters match, no operation needed
+            if (word1[i-1] == word2[j-1]) {
+                dp[i][j] = dp[i-1][j-1];
+            } 
+            else {
+                // Choose minimum of three operations:
+                // 1. Insert (dp[i][j-1])
+                // 2. Delete (dp[i-1][j])
+                // 3. Replace (dp[i-1][j-1])
+                int insOp = dp[i][j-1];     // Insert
+                int delOp = dp[i-1][j];     // Delete
+                int repOp = dp[i-1][j-1];   // Replace
+
+                // Add 1 to the minimum operation cost
+                dp[i][j] = 1 + min({insOp, delOp, repOp});
+            }
+        }
+    }
+
+    // Return minimum edit distance
+    return dp[a][b];
+}
+
+// Example usage and test cases
+int main() {
+    // Test Case 1: Different strings
+    cout << "Edit distance between 'horse' and 'ros': " 
+         << minDistance("horse", "ros") << endl;  // Expected output: 3
+
+    // Test Case 2: Same strings
+    cout << "Edit distance between 'hello' and 'hello': " 
+         << minDistance("hello", "hello") << endl;  // Expected output: 0
+
+    // Test Case 3: Empty strings
+    cout << "Edit distance between '' and 'abc': " 
+         << minDistance("", "abc") << endl;  // Expected output: 3
+
+    return 0;
+}
+
+/* 
+ * Approach Explanation:
+ * - Uses Dynamic Programming to solve the problem efficiently
+ * - DP table stores minimum operations to transform prefixes
+ * - Each cell represents min operations to convert word1[0:i] to word2[0:j]
+ * 
+ * Example walkthrough for "horse" to "ros":
+ * 1. Replace 'h' with 'r' (1 operation)
+ * 2. Replace 'r' with 'o' (1 operation)
+ * 3. Delete 's' and 'e' (2 operations)
+ * Total: 3 operations
+ */