From 51a53ef6a292b25e2ab0fbeebe3a61d805108bf6 Mon Sep 17 00:00:00 2001
From: ImAlpha <arcticjcb@gmail.com>
Date: Sun, 23 Jul 2023 17:58:51 -0700
Subject: [PATCH] moving on from question 856

---
 Top150Interview/856_Score_of_Parenthesis.py | 60 ++++++++++++++++-----
 Top150Interview/Questions_to_revisit.md     |  1 +
 2 files changed, 47 insertions(+), 14 deletions(-)

diff --git a/Top150Interview/856_Score_of_Parenthesis.py b/Top150Interview/856_Score_of_Parenthesis.py
index 6cb322a..d0e36a8 100644
--- a/Top150Interview/856_Score_of_Parenthesis.py
+++ b/Top150Interview/856_Score_of_Parenthesis.py
@@ -3,19 +3,51 @@ def scoreOfParentheses(s):
     :type s: str
     :rtype: int
     """
-    score = left = right = 0
+    # Attempted Solution
+    # score = left = right = 0
 
-    for x in s:
-        if x == "(":
-            left += 1
-            score += 1
-        elif x == ")":
-            right += 1
-            score += 1
-    if left > right:
-        score = 2 * left
-    else:
-        score /= 2
-    return score
+    # for x in s:
+    #     if x == "(":
+    #         left += 1
+    #         score += 1
+    #     elif x == ")":
+    #         right += 1
+    #         score += 1
+    # if (left + right) % 2 == 0:
+    #     score /= 2
+    # return score
 
-print(scoreOfParentheses("()"))
+    # 96% faster solution
+    # Initialize an empty list 'a', the length of the input string 's' as 'n'
+    a, n = [], len(s)
+    
+    # Initialize 'c' and 't' to 0.
+    c = t = 0
+    
+    # Loop through the characters of the input string, starting from the second character (index 1).
+    for i in range(1, n):
+        print(f'c is {c}')
+        print(t)
+        # If the current character is an opening parenthesis '('
+        if s[i] == '(':
+            # Increment the temporary variable 't' to keep track of the depth of nesting.
+            t += 1
+        # If the current character is a closing parenthesis ')', and the previous character was an opening parenthesis '('
+        elif s[i - 1] == '(':
+            # Calculate the score for the current balanced pair of parentheses and add it to 'c'.
+            # The score is 2 raised to the power of 't' (2^t).
+            c += 1 << t
+            
+            # Decrease the temporary variable 't' as the current nested pair is closed.
+            t -= 1
+        else:
+            # If the current character is a closing parenthesis ')' and the previous character was also a closing parenthesis ')',
+            # simply decrease the temporary variable 't' to indicate the decrease in the depth of nesting.
+            t -= 1
+    
+    # Return the total score of the balanced parentheses expression, stored in 'c'.
+    return c
+
+# print(scoreOfParentheses("()")) # 1
+print(scoreOfParentheses("(()(()))")) # 6
+# print(scoreOfParentheses("()()()")) # 3
diff --git a/Top150Interview/Questions_to_revisit.md b/Top150Interview/Questions_to_revisit.md
index 5236fe4..3004e6e 100644
--- a/Top150Interview/Questions_to_revisit.md
+++ b/Top150Interview/Questions_to_revisit.md
@@ -1,5 +1,6 @@
 # Questions that required help to solve
 ## Strings
 ### [22_Generate_Parenthesis](https://leetcode.com/problems/generate-parentheses/description/)
+###[856_Score_of_Parenthesis](https://leetcode.com/problems/score-of-parentheses/description/)
 ## Two Pointers
 ### 
\ No newline at end of file