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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.
1public class Main {
2 public static int minAddToMakeValid(String s) {
3 int balance = 0, result = 0;
4 for (char c : s.toCharArray()) {
5 balance += (c == '(') ? 1 : -1;
6 if (balance == -1) {
7 result++;
8 balance++;
9 }
10 }
11 return result + balance;
12 }
13
14 public static void main(String[] args) {
15 String s = ")((";
16 System.out.println(minAddToMakeValid(s));
17 }
18}The Java version uses for-each over a character array representation of the string to perform similar logic.
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.
Here, we maintain a stack variable to track open parentheses and an invalid count for unmatched closing parentheses. This eliminates the need for an actual stack structure, only integers are enough.