solve: longestCommonSubsequence

Author: GaggleHelena

longest-common-subsequence/yolophg.py

@@ -0,0 +1,29 @@
+# Time Complexity: O(m * n) - each (i, j) pair is computed once and stored, reducing redundant calls.
+# Space Complexity: O(m * n) - memoization dictionary stores O(m * n) states.
+# - Recursion stack depth is O(m + n) in the worst case.
+
+class Solution:
+    def longestCommonSubsequence(self, t1: str, t2: str) -> int:
+        # to memoize results so we don't recompute the same subproblems
+        m = dict()
+
+        # recursive function to compute LCS
+        def s(i, j):
+            # base case: if we reach the end of either string, there's nothing left to compare
+            if i == len(t1) or j == len(t2):
+                return 0
+            
+            # if already computed this state, just return the cached value
+            if (i, j) in m:
+                return m[(i, j)]
+            
+            # if the characters match, we take this character and move diagonally
+            if t1[i] == t2[j]:
+                m[i, j] = 1 + s(i + 1, j + 1)
+            else:
+                # if they don't match, we either move forward in t1 or t2 and take the max
+                m[i, j] = max(s(i + 1, j), s(i, j + 1))
+            
+            return m[i, j]
+
+        return s(0, 0)