Create 03 - Bottom-Up | DP | Approach.cpp

Author: JawadSher

24 - Dynamic Programming Problems/36 - Longest Common Subsequence/03 - Bottom-Up | DP | Approach.cpp

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+class Solution {
+public:
+    // Helper function to compute the longest common subsequence using bottom-up dynamic programming
+    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 a 2D DP table initialized with 0
+        // dp[i][j] represents the length of the LCS of text1[i..n-1] and text2[j..m-1]
+        vector<vector<int>> dp(n + 1, vector<int>(m + 1, 0));
+
+        // Iterate over the strings in reverse order to fill the DP table
+        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 + dp[i + 1][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(dp[i + 1][j], dp[i][j + 1]);
+                
+                // Store the result in the DP table
+                dp[i][j] = ans;
+            }
+        }
+
+        // The final answer (LCS of the entire strings) is stored in dp[0][0]
+        return dp[0][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);
+    }
+};