Leetcode #921: Minimum Add to Make Parentheses Valid
In this guide, we solve Leetcode #921 Minimum Add to Make Parentheses Valid 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
A parentheses string is valid if and only if: It is the empty string, It can be written as AB (A concatenated with B), where A and B are valid strings, or It can be written as (A), where A is a valid string. You are given a parentheses string s.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Stack, Greedy, String
Intuition
A locally optimal choice leads to a globally optimal result for this structure.
That means we can commit to decisions as we scan without backtracking.
Approach
Sort or preprocess if needed, then repeatedly take the best available local choice.
Maintain the minimal state necessary to validate the greedy decision.
Steps:
- Sort or preprocess as needed.
- Iterate and pick the best local option.
- Track the current solution.
Example
Input: s = "())"
Output: 1
Python Solution
class Solution:
def minAddToMakeValid(self, s: str) -> int:
stk = []
for c in s:
if c == ')' and stk and stk[-1] == '(':
stk.pop()
else:
stk.append(c)
return len(stk)
Complexity
The time complexity is , and the space complexity is , where is the length of the string . The space complexity is , where is the length of the string .
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.