solves scramble strings (#87) in python

Author: sumanth-botlagunta

python/scramble_strings.py

@@ -0,0 +1,46 @@
+# https://leetcode.com/problems/scramble-string/description/
+# T: O(n^4) where n is the number of characters in the string
+# S: O(n^3) where n is the number of characters in the string
+
+from collections import defaultdict
+
+class Solution:
+    def isScramble(self, s1: str, s2: str) -> bool:
+        cache = dict()
+        return self.helper(s1, s2, cache)
+    
+    def helper(self, s1: str, s2: str, cache: dict) -> bool:
+        key = (s1,s2)
+        key_r = (s2,s1)
+        if key in cache:
+            return cache[key]
+        if key_r in cache:
+            return cache[key_r]
+        # If not cached
+        n = len(s1)
+        # Base case
+        if sorted(s1) != sorted(s2):
+            cache[key] = False
+            return False
+        if n <= 3:
+            cache[key] = True
+            return True
+        # split sting for comparision
+        count_s1 = defaultdict(int)
+        count_s2 = defaultdict(int)
+        count_s2_r = defaultdict(int)
+        
+        for i in range(1, n):
+            count_s1[s1[i-1]] += 1
+            count_s2[s2[i-1]] += 1
+            count_s2_r[s2[-i]] += 1
+
+            if count_s1 == count_s2:
+                cache[key] = self.helper(s1[0:i], s2[0:i], cache) and self.helper(s1[i:n], s2[i:n], cache)
+                if cache[key]:
+                    return True
+            if count_s1 == count_s2_r:
+                cache[key] = self.helper(s1[0:i], s2[n-i:n], cache) and self.helper(s1[i:n], s2[0:n-i], cache)
+                if cache[key]:
+                    return True
+        return False