Leetcode 856

Author: Kondareddy1209

Leetcode/Leetcode_856.py

@@ -0,0 +1,58 @@
+class Solution:
+    def scoreOfParentheses(self, s: str) -> int:
+        """
+        LeetCode 856 - Score of Parentheses
+
+        Approach (Stack):
+        - Use a stack to store the score at each level of nesting.
+        - Initialize the stack with 0, representing the score of the
+          outermost level.
+        - Traverse each character in the string:
+            1. If the character is '(':
+               - Start a new nested level by pushing 0 onto the stack.
+            2. If the character is ')':
+               - Pop the score of the current level.
+               - If the popped score is 0, it represents "()", whose score is 1.
+               - Otherwise, it represents "(A)", whose score is 2 * A.
+               - Add the calculated score to the previous level.
+
+        Why stack = [0]?
+        - The initial 0 acts as the base level.
+        - Every completed parenthesis contributes its score to its parent level.
+        - Without this base level, the first completed pair would have
+          nowhere to store its score.
+
+        Example:
+        Input: s = "(()(()))"
+
+        Stack Evolution:
+        Start      -> [0]
+        (          -> [0, 0]
+        (          -> [0, 0, 0]
+        )          -> [0, 1]
+        (          -> [0, 1, 0]
+        (          -> [0, 1, 0, 0]
+        )          -> [0, 1, 1]
+        )          -> [0, 3]
+        )          -> [6]
+
+        Output: 6
+
+        Time Complexity: O(n)
+            - Each character is processed exactly once.
+
+        Space Complexity: O(n)
+            - In the worst case, the stack stores one score for each level
+              of nested parentheses.
+        """
+
+        stack = [0]
+
+        for ch in s:
+            if ch == "(":
+                stack.append(0)
+            else:
+                current_score = stack.pop()
+                stack[-1] += max(2 * current_score, 1)
+
+        return stack[0]