D. J.:

- Added the leetcode problem and solution for 394

Author: nobodyPerfecZ

README.md

@@ -226,6 +226,7 @@
 - [380 Insert Delete GetRandom O(1)](https://leetcode.com/problems/insert-delete-getrandom-o1/description/)
 - [383 Ransom Note](https://leetcode.com/problems/ransom-note/description/)
 - [392 Is Subsequence](https://leetcode.com/problems/is-subsequence/description/)
+- [394 Decode String](https://leetcode.com/problems/decode-string/description/)
 - [399 Evaluate Division](https://leetcode.com/problems/evaluate-division/description/)
 - [412 Fizz Buzz](https://leetcode.com/problems/fizz-buzz/description/)
 - [415 Add Strings](https://leetcode.com/problems/add-strings/description/)

awesome_python_leetcode/_394_decode_string.py

@@ -0,0 +1,46 @@
+class Solution:
+    """Base class for all LeetCode Problems."""
+
+    def decodeString(self, s: str) -> str:
+        """
+        Given an encoded string, return its decoded string.
+
+        The encoding rule is: k[encoded_string], where the encoded_string inside the
+        square brackets is being repeated exactly k times. Note that k is guaranteed to
+        be a positive integer.
+
+        You may assume that the input string is always valid; there are no extra white
+        spaces, square brackets are well-formed, etc. Furthermore, you may assume that
+        the original data does not contain any digits and that digits are only for those
+        repeat numbers, k. For example, there will not be input like 3a or 2[4].
+
+        The test cases are generated so that the length of the output will never exceed
+        10^5.
+        """
+        decode = ""
+        i = 0
+        while i < len(s):
+            # Check if character is a string
+            if s[i] >= "a" and s[i] <= "z":
+                decode += s[i]
+                i += 1
+                continue
+            # Get the occurence for the open bracket [
+            j = i
+            while j < len(s) and s[j] != "[":
+                j += 1
+            number = int(str(s[i:j]))
+
+            # Get the last occurence for the closed bracket ]
+            openBrackets = 1
+            k = j + 1
+            while k < len(s) and openBrackets > 0:
+                if s[k] == "[":
+                    openBrackets += 1
+                if s[k] == "]":
+                    openBrackets -= 1
+                k += 1
+            substr = self.decodeString(s[j + 1 : k - 1])
+            decode += number * substr
+            i = k
+        return decode

tests/test_394_decode_string.py

@@ -0,0 +1,17 @@
+import pytest
+
+from awesome_python_leetcode._394_decode_string import Solution
+
+
+@pytest.mark.parametrize(
+    argnames=["s", "expected"],
+    argvalues=[
+        ("3[a]2[bc]", "aaabcbc"),
+        ("3[a2[c]]", "accaccacc"),
+        ("2[abc]3[cd]ef", "abcabccdcdcdef"),
+    ],
+)
+def test_func(s: str, expected: str):
+    """Tests the solution of a LeetCode problem."""
+    decoded_s = Solution().decodeString(s)
+    assert decoded_s == expected