Create 04 - Space Optimized | DP | Approach.cpp

Author: JawadSher

24 - Dynamic Programming Problems/36 - Longest Common Subsequence/04 - Space Optimized | DP | Approach.cpp

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+class Solution {
+public:
+    // Helper function to compute the longest common subsequence using optimized space
+    int solve(string &text1, string &text2) {
+        int n = text1.length(); // Length of the first string
+        int m = text2.length(); // Length of the second string
+
+        // Create two 1D arrays for the current and next rows in the DP table
+        // This reduces the space complexity to O(m) instead of O(n*m)
+        vector<int> curr(m + 1, 0); // Current row being processed
+        vector<int> next(m + 1, 0); // Next row (from the previous iteration)
+
+        // Iterate over the strings in reverse order
+        for (int i = n - 1; i >= 0; i--) {
+            for (int j = m - 1; j >= 0; j--) {
+                int ans = 0; // Variable to store the result for the current state
+                
+                // Case 1: Characters match, include them in the LCS
+                if (text1[i] == text2[j]) 
+                    ans = 1 + next[j + 1];
+                else
+                    // Case 2: Characters don't match, take the maximum:
+                    // a) Skip the current character of text1
+                    // b) Skip the current character of text2
+                    ans = max(next[j], curr[j + 1]);
+                
+                // Store the result in the current row
+                curr[j] = ans;
+            }
+
+            // Update the next row to the current row for the next iteration
+            next = curr;
+        }
+
+        // The final answer (LCS of the entire strings) is stored in curr[0]
+        return curr[0];
+    }
+
+    // Main function to compute the longest common subsequence
+    int longestCommonSubsequence(string text1, string text2) {
+        // Call the helper function and return the result
+        return solve(text1, text2);
+    }
+};