1249. Minimum Remove to Make Valid Parentheses

Updated: 2024-04-15
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LeetCode problem 1249


The problem can be solved in two passes using a stack for the first pass and a simple iteration for the second:

  1. First Pass (Identify Invalid Parentheses):
  • Iterate through the string, and for every character:
    • If it’s ‘(’, push its index onto the stack.
    • If it’s ‘)’ and the stack is not empty (there is a matching ‘(’), pop from the stack. Otherwise, mark this ‘)’ as invalid.
  • After the iteration, any indices remaining in the stack represent unmatched ‘(’ that should be removed.
  1. Second Pass (Build the Result String):
  • Iterate through the string again, building the result string by including characters that are not marked as invalid.
class Solution:
    def minRemoveToMakeValid(self, s: str) -> str:
        stack = []                      # Stack to keep track of the indices of '('
        remove_indices = set()          # Set to keep track of indices to remove

        for i, char in enumerate(s):    # First pass to identify invalid parentheses
            if char == '(':
            elif char == ')':
                if stack:
        # Add indices of remaining '(' to remove
        remove_indices = remove_indices.union(set(stack))
        # Second pass to build the result string
        result = [s[i] for i in range(len(s)) if i not in remove_indices]
        return ''.join(result)