Leetcode #1249: Minimum Remove to Make Valid Parentheses
In this guide, we solve Leetcode #1249 Minimum Remove to Make Valid Parentheses in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
Given a string s of '(' , ')' and lowercase English characters. Your task is to remove the minimum number of parentheses ( '(' or ')', in any positions ) so that the resulting parentheses string is valid and return any valid string.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Stack, String
Intuition
The problem has a natural nested or last-in-first-out structure.
A stack lets us resolve matches in the correct order as we scan.
Approach
Push items as they appear and pop when you can finalize a decision.
The stack captures the unresolved part of the input.
Steps:
- Push elements as you scan.
- Pop when a rule or match is satisfied.
- Use the stack to compute results.
Example
Input: s = "lee(t(c)o)de)"
Output: "lee(t(c)o)de"
Explanation: "lee(t(co)de)" , "lee(t(c)ode)" would also be accepted.
Python Solution
class Solution:
def minRemoveToMakeValid(self, s: str) -> str:
stk = []
x = 0
for c in s:
if c == ')' and x == 0:
continue
if c == '(':
x += 1
elif c == ')':
x -= 1
stk.append(c)
x = 0
ans = []
for c in stk[::-1]:
if c == '(' and x == 0:
continue
if c == ')':
x += 1
elif c == '(':
x -= 1
ans.append(c)
return ''.join(ans[::-1])
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.