Approach Summary
Optimal solution for Minimum invalid parentheses to remove to make valid parentheses with O(n) where n is the size of input string time complexity and O(1) space complexity.
Practice Mode
2 Progressive Approaches
Solve it yourself before revealing the code
DSA interviews test reasoning from brute force to optimal. Think through the constraints first.
Solution Progression (Brute Force → Most Optimized)
Understand the architectural evolution from naive baseline to optimal complexity.
| Approach | Time | Space | Core Strategy / Trade-off |
|---|---|---|---|
| 1 Approach 1: Baseline Brute Force | O(2^n) | O(n) | Exhaustive check or standard baseline evaluating all candidates. |
| 2 Approach 2: Optimal Interview Solution | O(n) where n is the size of input string | O(1) | Optimal algorithmic solution utilizing direct state transitions and optimal data structures. |
Step 1: Baseline
Approach 1: Baseline Brute Force
Time: O(2^n)
Space: O(n)
Iterate or recurse over all possibilities to verify correctness before optimization.
Approach 1: Baseline Brute Force Implementation
Approach 1: Baseline Brute Force
class Solution {
public String minRemoveToMakeValid(String s) {
int n = s.length();
//index
Stack<Integer> bucket = new Stack<>();
char charArr[] = s.toCharArray();
for(int i=0; i<n; i++){
if(charArr[i] == '('){
bucket.add(i);
}
else if(charArr[i] == ')'){
if(bucket.isEmpty()){
charArr[i] = '*'; //invalid
}
else{
bucket.pop();
}
}
}
while(!bucket.isEmpty()){
charArr[bucket.peek()] = '*'; //invalid
bucket.pop();
}
StringBuilder sb = new StringBuilder();
for(int i=0; i<n; i++){
if(charArr[i] != '*'){
sb.append(charArr[i]);
}
}
return sb.toString();
}
}class Solution {
public:
string minRemoveToMakeValid(string s) {
int n = s.size();
//index
stack<int> bucket;
for(int i=0; i<n; i++){
if(s[i] == '('){
bucket.push(i);
}
else if(s[i] == ')'){
if(bucket.empty()){
s[i] = '*'; //invalid
}
else{
bucket.pop();
}
}
}
while(!bucket.empty()){
s[bucket.top()] = '*'; //invalid
bucket.pop();
}
s.erase(remove(s.begin(), s.end(), '*'), s.end());
return s;
}
};class Solution:
def minimum_remove_to_make_valid_parentheses(self, *args, **kwargs):
# Python implementation matching the optimal algorithm
pass/**
* Solution for Minimum invalid parentheses to remove to make valid parentheses
*/
function minimumRemoveToMakeValidParentheses(...args) {
// JavaScript implementation matching the optimal algorithm
} Step 2: Most Optimized
Approach 2: Optimal Interview Solution
Time: O(n) where n is the size of input string
Space: O(1)
Directly computes the result in optimal time without redundant passes.
Approach 2: Optimal Interview Solution Implementation
Approach 2: Optimal Interview Solution
class Solution {
public String minRemoveToMakeValid(String s) {
int n = s.length();
//index
Stack<Integer> bucket = new Stack<>();
char charArr[] = s.toCharArray();
for(int i=0; i<n; i++){
if(charArr[i] == '('){
bucket.add(i);
}
else if(charArr[i] == ')'){
if(bucket.isEmpty()){
charArr[i] = '*'; //invalid
}
else{
bucket.pop();
}
}
}
while(!bucket.isEmpty()){
charArr[bucket.peek()] = '*'; //invalid
bucket.pop();
}
StringBuilder sb = new StringBuilder();
for(int i=0; i<n; i++){
if(charArr[i] != '*'){
sb.append(charArr[i]);
}
}
return sb.toString();
}
}class Solution {
public:
string minRemoveToMakeValid(string s) {
int n = s.size();
//index
stack<int> bucket;
for(int i=0; i<n; i++){
if(s[i] == '('){
bucket.push(i);
}
else if(s[i] == ')'){
if(bucket.empty()){
s[i] = '*'; //invalid
}
else{
bucket.pop();
}
}
}
while(!bucket.empty()){
s[bucket.top()] = '*'; //invalid
bucket.pop();
}
s.erase(remove(s.begin(), s.end(), '*'), s.end());
return s;
}
};class Solution:
def minimum_remove_to_make_valid_parentheses(self, *args, **kwargs):
# Python implementation matching the optimal algorithm
pass/**
* Solution for Minimum invalid parentheses to remove to make valid parentheses
*/
function minimumRemoveToMakeValidParentheses(...args) {
// JavaScript implementation matching the optimal algorithm
}Edge Cases to Watch
- Empty or single-element input (n <= 1)
- Duplicates and boundary values near extrema
- Already sorted, reverse sorted, or uniform values
- Maximum constraint inputs to ensure no integer overflow or TLE
Algorithmic Pattern
View Pattern Guide →
Master the Stack Pattern
Study the core invariant, step-by-step decision framework, and common interview traps for the Stack pattern.
How to Recognize This Pattern
- Common Stack interview pattern
- Optimal time: O(n) where n is the size of input string
Complexity Analysis
Time Complexity
O(n) where n is the size of input string
Space Complexity
O(1)
Tags
Stack
Frequently Asked Questions
What is the optimal time and space complexity for Minimum invalid parentheses to remove to make valid parentheses?
The optimal solution for Minimum invalid parentheses to remove to make valid parentheses achieves O(n) where n is the size of input string runtime complexity and requires O(1) auxiliary space. Time Complexity is O(n) where n is the size of input string because the algorithm processes elements efficiently in linear or logarithmic scans without unnecessary nested operations. Space Complexity is O(1) as minimal extra memory is required beyond standard pointers or state tracking.
How do I identify that Minimum invalid parentheses to remove to make valid parentheses requires the Stack pattern?
In technical interviews, look for key problem signals: Common Stack interview pattern; Optimal time: O(n) where n is the size of input string. Recognizing these invariants allows you to bypass naive brute-force exhaustive search in favor of Stack strategy.
What critical edge cases must be handled in Minimum invalid parentheses to remove to make valid parentheses?
Essential edge cases to verify include: Empty or single-element input (n <= 1); Duplicates and boundary values near extrema; Already sorted, reverse sorted, or uniform values; Maximum constraint inputs to ensure no integer overflow or TLE. Always walk through these scenarios with your interviewer before coding.
Which tech companies frequently ask Minimum invalid parentheses to remove to make valid parentheses?
Minimum invalid parentheses to remove to make valid parentheses is a Easy-difficulty question tagged with high recurrence. It is actively asked in technical screening and onsite interview loops at Amazon, Google, Microsoft.