**ObjecÂtive:** **– **Given “n”, generate all valid parenthesis strings of length “2n”.**Example**:

Given n=2 Output: (()) ()()

**Approach:**

- Recursion is the key here.
- Divide the N into N/2 and N/2 (Count for open and closed parentheses ).
- Select the
parentheses, add it to the result string and reduce its count and make a recursive call.*open* - Select the
parentheses, add it to the result string and reduce its count and make a recursive call.*close* - To print only valid parentheses, make sure at any given point of time, close parentheses count is not less that open parentheses count because it means close parentheses has been printed with its respective open parentheses.
- See picture for better explanation.

**Code:**

**Output:**

(()) ()()

Great analysis. If you want to see another insightful discussion with code, check https://goo.gl/ALZBh0.!

Time Complexity O(logN) ?

Space Complexity O(1) or should I consider the stackframe as space complexity O(logN)?

if(openP >closeP) // close parantheses is more than open ? how greater than sign is towards the openP