find if 3rd string is formed by interleaving of first 2 strings

Author: Alagappan Maruthappan

interleave_str.py

@@ -0,0 +1,42 @@
+# https://leetcode.com/problems/interleaving-string/description/
+
+
+class Solution:
+
+    def isInterleaveUtil(self, s1, s2, s3, interleave_str, s1_pos, s2_pos, memo):
+        if s1_pos < len(s1) and s2_pos < len(s2):
+            if memo[s1_pos][s2_pos]:
+                return False
+            memo[s1_pos][s2_pos] = True
+
+        if s1_pos == len(s1) and s2_pos == len(s2) and interleave_str == s3:
+            return True
+
+        ans = False
+        if s1_pos < len(s1) and s3[s1_pos + s2_pos] == s1[s1_pos]:
+            ans |= self.isInterleaveUtil(
+                s1, s2, s3, interleave_str + s1[s1_pos], s1_pos + 1, s2_pos, memo)
+
+        if not ans and s2_pos < len(s2) and s3[s1_pos + s2_pos] == s2[s2_pos]:
+            ans |= self.isInterleaveUtil(
+                s1, s2, s3, interleave_str + s2[s2_pos], s1_pos, s2_pos + 1, memo)
+
+        return ans
+
+    def isInterleave(self, s1, s2, s3):
+        """
+        :type s1: str
+        :type s2: str
+        :type s3: str
+        :rtype: bool
+        """
+        memo = [[False] * (len(s2)) for i in range(0, len(s1))]
+        if len(s1) + len(s2) != len(s3):
+            return False
+        return self.isInterleaveUtil(s1, s2, s3, "", 0, 0, memo)
+
+
+s1 = "aabcc"
+s2 = "dbbca"
+s3 = "aadbbbaccc"
+assert not Solution().isInterleave(s1, s2, s3)