Adding Efficient Solution for Problem - 1021 - Remove Parantheses

Author: Vanditg

Remove_Parantheses/Efficient_Solution.py

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+##==================================
+## Leetcode
+## Student: Vandit Jyotindra Gajjar
+## Year: 2020
+## Problem: 1021
+## Problem Name: Remove Outermost Parantheses 
+##===================================
+#
+#A valid parentheses string is either empty (""), "(" + A + ")", or A + B, where A and B are valid parentheses strings, and + represents string concatenation.  For example, "", "()", "(())()", and "(()(()))" are all valid parentheses strings.
+#
+#A valid parentheses string S is primitive if it is nonempty, and there does not exist a way to split it into S = A+B, with A and B nonempty valid parentheses strings.
+#
+#Given a valid parentheses string S, consider its primitive decomposition: S = P_1 + P_2 + ... + P_k, where P_i are primitive valid parentheses strings.
+#
+#Return S after removing the outermost parentheses of every primitive string in the primitive decomposition of S.
+#
+#Example 1:
+#
+#Input: "(()())(())"
+#Output: "()()()"
+#Explanation: 
+#The input string is "(()())(())", with primitive decomposition "(()())" + "(())".
+#After removing outer parentheses of each part, this is "()()" + "()" = "()()()".
+#
+#Example 2:
+#
+#Input: "(()())(())(()(()))"
+#Output: "()()()()(())"
+#Explanation: 
+#The input string is "(()())(())(()(()))", with primitive decomposition "(()())" + "(())" + "(()(()))".
+#After removing outer parentheses of each part, this is "()()" + "()" + "()(())" = "()()()()(())".
+#
+#Example 3:
+#
+#Input: "()()"
+#Output: ""
+#Explanation: 
+#The input string is "()()", with primitive decomposition "()" + "()".
+#After removing outer parentheses of each part, this is "" + "" = "".
+class Solution:
+    def removeOuterParantheses(self, S):
+        tmp = ""	#Initialize tmp empty string 
+        final_S = ""	#Initialize final_S empty string which we'll return at the end.
+        count = 0	#Initialize count
+        for i in range(len(S)):	#Loop through our string 
+            tmp += S[i]	#tmp string will be tmp + S[i]. i.e: "" + "(" = "("
+            if S[i] == '(':	#Condition-check: If S[i] == '('
+                count += 1	#We'll increase our count by 1
+            else:	#Condition-check: else then
+                count -= 1	#We'll decrease count by 1
+            if count == 0:	#Condition-check: Now, if we reach count to zero. 
+                final_S += tmp[1:-1]	#final_S will be final_S + tmp[1:-1]. i.e. = "" + "(()())" = "" + "()()" = "()()"
+                tmp = ""	#Then we'll change previous tmp string to empty string
+        return final_S	#Finally we'll return final_S.