Create 04 - Space Optimized | DP | Approach.cpp

Author: JawadSher

24 - Dynamic Programming Problems/38 - Edit Distance/04 - Space Optimized | DP | Approach.cpp

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+class Solution {
+public:
+    // Helper function to calculate the minimum edit distance using dynamic programming
+    int solve(string word1, string word2) {
+        int n = word1.length(); // Length of the first string
+        int m = word2.length(); // Length of the second string
+
+        // Create two 1D arrays to store the current and next row values of the dp table
+        vector<int> curr(m+1, 0); // Current row (for the current i-th character in word1)
+        vector<int> next(m+1, 0); // Next row (for the next i+1-th character in word1)
+
+        // Initialize the next row for the case when word1 is empty
+        for(int j = 0; j < m; j++) next[j] = m - j;
+
+        // Iterate over word1 from bottom to top (i = n-1 to i = 0)
+        for(int i = n-1; i >= 0; i--) {
+            // Iterate over word2 from right to left (j = m-1 to j = 0)
+            for(int j = m-1; j >= 0; j--) {
+                // Set the base case for when word2 is empty (fill in the current row's last column)
+                curr[m] = n - i;
+
+                int ans = 0;
+                // If characters match, take the result from the next row's next column
+                if(word1[i] == word2[j]) ans = next[j+1];
+                else {
+                    // Otherwise, calculate the costs of insertion, deletion, and replacement:
+                    int insertAns = 1 + curr[j+1];    // Insert a character from word2 into word1
+                    int deleteAns = 1 + next[j];      // Delete a character from word1
+                    int replaceAns = 1 + next[j+1];  // Replace a character in word1 with word2
+
+                    // Take the minimum of these three options
+                    ans = min({insertAns, deleteAns, replaceAns});
+                }
+
+                // Store the result for the current cell in the current row
+                curr[j] = ans;
+            }
+            // Update the next row to be the current row after finishing one iteration
+            next = curr;
+        }
+
+        // The final result is stored in curr[0], which represents the minimum edit distance
+        return curr[0];
+    }
+
+    // Main function to calculate the minimum edit distance between word1 and word2
+    int minDistance(string word1, string word2) {
+        // Edge case: if one of the strings is empty, the result is the length of the other string
+        if(word1.length() == 0) return word2.length();
+        if(word2.length() == 0) return word1.length();
+        
+        // Call the solve function to compute the result
+        return solve(word1, word2);
+    }
+};