Create 02 - Top-Down | DP | Approach.cpp

Author: JawadSher

24 - Dynamic Programming Problems/38 - Edit Distance/02 - Top-Down | DP | Approach.cpp

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+class Solution {
+public:
+    // Helper function to calculate the minimum distance (edit distance) using recursion and memoization
+    int solve(string word1, string word2, int i, int j, vector<vector<int>> &dp) {
+        int n = word1.length(); // Length of the first string
+        int m = word2.length(); // Length of the second string
+
+        // Base cases: If one string is fully traversed, the remaining length of the other string is the answer
+        if (i >= n) return m - j;  // If word1 is fully traversed, we need to insert remaining characters of word2
+        if (j >= m) return n - i;  // If word2 is fully traversed, we need to delete remaining characters of word1
+
+        // If the result for this subproblem is already computed (cached in dp table), return it
+        if (dp[i][j] != -1) return dp[i][j];
+        
+        int ans = 0;
+
+        // If characters at the current positions match, move to the next character in both strings
+        if (word1[i] == word2[j]) return solve(word1, word2, i + 1, j + 1, dp);
+        else {
+            // If characters do not match, consider the three possible operations:
+            // 1. Insert a character into word1 (move j ahead in word2)
+            int insertAns = 1 + solve(word1, word2, i, j + 1, dp);
+
+            // 2. Delete a character from word1 (move i ahead in word1)
+            int deleteAns = 1 + solve(word1, word2, i + 1, j, dp);
+
+            // 3. Replace a character in word1 (move both i and j ahead)
+            int replaceAns = 1 + solve(word1, word2, i + 1, j + 1, dp);
+
+            // The minimum of these three options gives the optimal edit distance at this step
+            ans = min({insertAns, deleteAns, replaceAns});
+        }
+
+        // Store the computed result in the dp table and return it
+        return dp[i][j] = ans;
+    }
+
+    // Main function that calculates the minimum distance between word1 and word2
+    int minDistance(string word1, string word2) {
+        // Create a dp table initialized to -1 (indicating that no subproblem has been solved yet)
+        // dp[i][j] represents the minimum edit distance between word1[0..i-1] and word2[0..j-1]
+        vector<vector<int>> dp(word1.length() + 1, vector<int>(word2.length() + 1, -1));
+
+        // Call the solve function starting from the beginning of both strings
+        return solve(word1, word2, 0, 0, dp);
+    }
+};