To solve this problem, we need to understand what it means to remove the outermost parentheses from a primitive group in a valid parentheses string.
A primitive parentheses string is one that is valid (i.e., balanced) and cannot be divided further into smaller valid parts. For example, "(())" and "()" are primitives, but "(()())(())" is not a primitive — it's made of multiple primitives.
Our goal is to go through the input and for every such primitive group, remove its first '(' and last ')'. For example, in "(()())", the primitive is the whole string, and removing its outermost parentheses gives "()()".
How We Approach This
We walk through the string character by character. As we go, we use a counter open
to track how deep we are in the parentheses nesting:
- When we see
'('
, we increase the nesting count.
- When we see
')'
, we decrease the nesting count.
But here's the trick: we don't include the first '(' when open
goes from 0 to 1 (this is the outermost), and we don't include the ')' that causes open
to go back to 0 (this is the closing of the outermost).
Example Cases
Let’s take "(()())(())"
as an example:
- The first primitive is "(()())" → we remove the outermost '(', ')' → becomes "()()"
- The second primitive is "(())" → becomes "()"
- Final result: "()()()"
Edge Cases
- If the input is an empty string
""
, there’s nothing to process, so we return ""
.
- If the string consists only of "()()", each pair is its own primitive, and both outer parentheses are removed, leaving an empty string.
- If the input is deeply nested like "((()))", then removing the outermost parentheses gives "(())".
This method works efficiently in one pass through the string and uses only a simple counter — making it an optimal solution.