From 5c46ca9ae692fe5849b2aebf323351eb96c1c3d3 Mon Sep 17 00:00:00 2001
From: Sumiran Verma <112773734+sumiranverma@users.noreply.github.com>
Date: Sat, 27 Jan 2024 22:27:31 +0530
Subject: [PATCH] Create solution.md

---
 Day-26/q2: Score of Parentheses/solution.md | 48 +++++++++++++++++++++
 1 file changed, 48 insertions(+)
 create mode 100644 Day-26/q2: Score of Parentheses/solution.md

diff --git a/Day-26/q2: Score of Parentheses/solution.md b/Day-26/q2: Score of Parentheses/solution.md
new file mode 100644
index 0000000..808d0aa
--- /dev/null
+++ b/Day-26/q2: Score of Parentheses/solution.md	
@@ -0,0 +1,48 @@
+## Approach - Stacks
+1. Take a Stack, Iterate over the characters of string.
+2. For every `ith` character check if the character is `(` or not. If found to be true, then insert the character score into the stack.
+3. Initialize a stack to store the current traversed character score of inner balanced parenthesis.
+4. For every `i` check for the conditions: 
+  - if the current character is `(` push the current score into stack, enter the next inner layer and reset score to `0`.
+  - if the current character is `)` then `ind` score will be doubled and will be at least one.
+  - we exit the current level and set `ind = stack.top() + max(ind * 2, 1)`.
+  - pop the score from the stack.
+
+## Complexity Analysis
+
+`Time Complexity`: O(N)
+
+`Space Complexity`: O(N)
+
+## C++ Code
+
+```
+class Solution {
+public:
+    int scoreOfParentheses(string s) {
+        
+        stack<int> st;
+        int ind = 0;
+        
+        for(auto i : s)
+        {
+			// if we find open parenthesis
+			// push the current score into the stack
+            if(i == '(')
+            {
+                st.push(ind);
+				// reset the score to 0
+                ind = 0;
+            }
+            else // if we find close parenthesis
+            {
+                ind = st.top() + max(ind*2 ,1);
+                st.pop();
+            }
+                
+        }
+        return ind;
+        
+    }
+};
+```