Remove Outermost Parentheses - Video Solutions

Problem Overview: You are given a valid parentheses string s. The string can be split into multiple primitive valid parentheses substrings. The task is to remove the outermost pair of parentheses from every primitive part and return the final combined string.

Approach 1: Using a Counter to Track Depth (O(n) time, O(1) space)

This approach tracks the current nesting depth while iterating through the string once. Increment the depth when encountering '(' and decrement it when encountering ')'. The key insight: characters should only be added to the result when they are not the outermost parentheses. For an opening bracket, append it only if the current depth is already greater than zero before incrementing. For a closing bracket, decrement first and append only if the depth remains greater than zero. This effectively strips the first and last parentheses of each primitive group without explicitly splitting the string. Since the algorithm performs a single pass and only maintains an integer counter, it runs in O(n) time with O(1) extra space. This method is the most common solution when working with string traversal problems involving balanced parentheses.

Approach 2: Stack-Based Iteration (O(n) time, O(n) space)

A more explicit method uses a stack to simulate the structure of nested parentheses. Push an index or character when encountering '(' and pop when encountering ')'. The stack size represents the current nesting level. When processing characters, skip the parentheses that correspond to the outermost layer of each primitive substring. In practice, append characters only when the stack size after the operation indicates you are inside a nested level. While this approach mirrors how many stack-based parsing algorithms work, it uses additional memory proportional to the nesting depth. The runtime remains O(n) because each character is pushed and popped at most once.

Recommended for interviews: The depth counter solution is the expected answer. It shows you recognize that the stack structure is unnecessary because the nesting level alone determines whether a parenthesis is outermost. The stack approach demonstrates understanding of stack-driven parsing, but the counter approach is cleaner, uses constant space, and signals strong problem-solving instincts.