Create solution.md

Day-26/q2: Score of Parentheses/solution.md

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+## 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;
+
+    }
+};
+```