diff --git a/DynamicProgramming/longest_common_subsequence.py b/DynamicProgramming/longest_common_subsequence.py
index 85d627e..12d298a 100644
--- a/DynamicProgramming/longest_common_subsequence.py
+++ b/DynamicProgramming/longest_common_subsequence.py
@@ -16,6 +16,35 @@ def longestCommonSubsequence(self, text1: str, text2: str) -> int:
         return 3
 
         Approach: Bottom up, Tabular
+        
+        T=O(mn), beats 93.88 %
+        S=O(n), beats 91.28 %
+        """
+        if not text1 or not text2:
+            return 0
+
+        l1 = len(text1)+1 if text1 else 0
+        l2 = len(text2)+1 if text2 else 0 
+        
+        curr = [0]*l2
+        prv = [0]*l2
+
+        for r in range(1, l1):
+            for c in range(1, l2):
+                if text1[r-1] == text2[c-1]:
+                    curr[c] = 1 + prv[c-1]
+                else:
+                    curr[c] = max(
+                        prv[c],
+                        curr[c-1]
+                        )
+            prv = curr.copy()
+
+        return prv[-1]
+        
+    def longestCommonSubsequence_full(self, text1: str, text2: str) -> int:
+        """
+        Approach: Bottom up, Full Tabular
         Make a table using Text1 as a row, Text2 as a column
         The first row is without any char from Text 1
         The first col is without any char from Text 2
@@ -29,21 +58,27 @@ def longestCommonSubsequence(self, text1: str, text2: str) -> int:
         c 0 1 2 2
         d 0 1 2 2
         e 0 1 2 3
+
+        T=O(mn) beats 79.45%, 
+        S=O(mn) beats 52.78%
         """
         if not text1 or not text2:
             return 0
 
         l1 = len(text1)+1 if text1 else 0
-        l2 = len(text2)+1 if text2 else 0
+        l2 = len(text2)+1 if text2 else 0 
         
-        longest = [[0]*l2 for _ in range(l1)]
+        max_length = [[0]*l2 for _ in range(l1)]
 
         for r in range(1, l1):
             for c in range(1, l2):
                 if text1[r-1] == text2[c-1]:
-                    longest[r][c] = 1 + longest[r-1][c-1]
+                    max_length[r][c] = 1 + max_length[r-1][c-1]
                 else:
-                    longest[r][c] = max(longest[r-1][c], longest[r][c-1])
+                    max_length[r][c] = max(
+                        max_length[r-1][c], 
+                        max_length[r][c-1]  
+                        )
 
-        return longest[r][c]
+        return max_length[-1][-1]