Time: 265 ms (5.11%), Space: 18.4 MB (59.09%) - LeetHub

Author: hogan-tech

0087-scramble-string/0087-scramble-string.py

@@ -0,0 +1,31 @@
+# time complexity: O(n^4)
+# space complexity: O(n^3)
+class Solution:
+    def isScramble(self, s1: str, s2: str) -> bool:
+        n = len(s1)
+        dp = [[[False for _ in range(n)] for _ in range(n)]
+              for _ in range(n + 1)]
+        for i in range(n):
+            for j in range(n):
+                dp[1][i][j] = s1[i] == s2[j]
+        for length in range(2, n + 1):
+            for i in range(n + 1 - length):
+                for j in range(n + 1 - length):
+                    for newLength in range(1, length):
+                        dp1 = dp[newLength][i]
+                        dp2 = dp[length - newLength][i + newLength]
+                        dp[length][i][j] |= dp1[j] and dp2[j + newLength]
+                        dp[length][i][j] |= (
+                            dp1[j + length - newLength] and dp2[j])
+        return dp[n][0][0]
+
+
+s1 = "great"
+s2 = "rgeat"
+print(Solution().isScramble(s1, s2))
+s1 = "abcde"
+s2 = "caebd"
+print(Solution().isScramble(s1, s2))
+s1 = "a"
+s2 = "a"
+print(Solution().isScramble(s1, s2))