added Reverse Substrings Between Each Pair of Parentheses (medium)

Author: minhyeong-joe

Medium/ReverseSubstringsBetweenParentheses/README.md

@@ -0,0 +1,38 @@
+
+# Reverse Substrings Between Each Pair of Parentheses
+[Leetcode Link](https://leetcode.com/problems/reverse-substrings-between-each-pair-of-parentheses/)
+
+## Problem:
+
+You are given a string `s` that consists of lower case English letters and brackets. 
+
+Reverse the strings in each pair of matching parentheses, starting from the innermost one.
+
+Your result should **not** contain any brackets.
+
+## Example:
+
+```
+Input: s = "(abcd)"
+Output: "dcba"
+```
+```
+Input: s = "(u(love)i)"
+Output: "iloveu"
+Explanation: The substring "love" is reversed first, then the whole string is reversed.
+```
+```
+Input: s = "(ed(et(oc))el)"
+Output: "leetcode"
+Explanation: First, we reverse the substring "oc", then "etco", and finally, the whole string.
+```
+```
+Input: s = "a(bcdefghijkl(mno)p)q"
+Output: "apmnolkjihgfedcbq"
+```
+
+## Note:
+
+- `0 <= s.length <= 2000`
+- `s` only contains lower case English characters and parentheses.
+- It's guaranteed that all parentheses are balanced.

Medium/ReverseSubstringsBetweenParentheses/solution.py

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+class Solution:
+ def reverseParentheses(self, s: str) -> str:
+ stack = list()
+ queue = list()
+ for c in s:
+ # reverse string on ')'
+ if c == ')':
+ top = stack.pop()
+ while top != '(':
+ queue.append(top)
+ top = stack.pop()
+ # insert reversed substr back into stack
+ while queue:
+ stack.append(queue.pop(0))
+ else:
+ stack.append(c)
+ return "".join(stack)
+
+
+# test driver
+sol = Solution()
+# s = "(ed(et(oc))el)"
+s = "a(bcdefghijkl(mno)p)q"
+print("Input:", s)
+print("Output:", sol.reverseParentheses(s))

README.md

@@ -59,6 +59,7 @@ Languages used: Java and Python
  - [Swap Nodes in Pairs](Medium/SwapNodesInPairs)
  - [All Paths From Source to Target](Medium/AllPathsFromSrcToTarget)
  - [Queens That Can Attack The King](Medium/QueensThatCanAttackKing)
+ - [Reverse Substrings Between Each Pair of Parentheses](Medium/ReverseSubstringsBetweenParentheses)
 - Hard
 
 ---