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

Author: JawadSher

24 - Dynamic Programming Problems/36 - Longest Common Subsequence/02 - Top-Down | DP | Approach.cpp

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+class Solution {
+public:
+    // Helper function to recursively find the length of the longest common subsequence with memoization
+    int solve(string &text1, string &text2, int i, int j, vector<vector<int>>& dp) {
+        // Base case: if either string is exhausted, no subsequence is possible
+        if (i >= text1.length() || j >= text2.length()) return 0;
+
+        // Check if the result for this state has already been computed
+        if (dp[i][j] != -1) return dp[i][j];
+
+        int ans = 0; // Initialize the answer for the current state
+
+        // Case 1: Characters match, include them in the subsequence
+        if (text1[i] == text2[j]) 
+            return dp[i][j] = 1 + solve(text1, text2, i + 1, j + 1, dp);
+        else {
+            // Case 2: Characters don't match, explore two possibilities:
+            // a) Skip the current character of text1
+            // b) Skip the current character of text2
+            // Take the maximum of these two possibilities
+            ans = max(solve(text1, text2, i + 1, j, dp), solve(text1, text2, i, j + 1, dp));
+        }
+
+        // Store the computed result in the memoization table
+        return dp[i][j] = ans;
+    }
+
+    // Main function to compute the longest common subsequence
+    int longestCommonSubsequence(string text1, string text2) {
+        // Create a 2D vector for memoization initialized with -1
+        // dp[i][j] stores the result of solve for indices i and j
+        vector<vector<int>> dp(text1.size() + 1, vector<int>(text2.size() + 1, -1));
+        
+        // Start the recursive computation from the beginning of both strings
+        return solve(text1, text2, 0, 0, dp);
+    }
+};