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This approach involves iterating through the string and using a balance counter to track the validity of the sequence.
We increment the balance for every opening parenthesis '(' and decrement it for a closing parenthesis ')'. If, at any point, the balance becomes negative, it means we have an extra closing parenthesis, and we need to increase the count of steps required to balance the string. By the end of the iteration, the balance will also tell us how many opening parentheses need to be closed to make the string valid.
Time Complexity: O(n) - We traverse the string once.
Space Complexity: O(1) - Only a few integer variables are used.
1#include <iostream>
2#include <string>
3
4int minAddToMakeValid(std::string s) {
5 int balance = 0, result = 0;
6 for (char ch : s) {
7 balance += (ch == '(') ? 1 : -1;
8 if (balance == -1) {
9 result++;
10 balance++;
11 }
12 }
13 return result + balance;
14}
15
16int main() {
17 std::string s = ")((";
18 std::cout << minAddToMakeValid(s) << std::endl;
19 return 0;
20}The C++ solution adopts a similar approach to C, using a range-based for loop to iterate over each character in the string.
In this approach, we leverage a stack structure to simulate parentheses placement, assisting in counting the balance.
Iterate over the string, pushing opening parentheses onto the stack and popping for each encountered closing parenthesis when possible. When a closing parenthesis finds no partner to pop, increase the invalid count, simulating the need for one more opening parenthesis. In the end, the stack length represents unmatched opening parentheses.
Time Complexity: O(n) - Iterate over the string once.
Space Complexity: O(1) - Uses a few integer counters.
The Python version uses simple integer counters to track stack size and invalid closes.