This approach uses a stack to efficiently match opening and closing brackets. The stack stores opening brackets, and for each closing bracket encountered, it checks if it closes the last opened bracket (top of the stack).
Time Complexity: O(n), where n is the length of the string, since we process each character once.
Space Complexity: O(n) in the worst case if the string consists only of opening brackets.
1import java.util.Stack;
2
3public class Solution {
4 public boolean isValid(String s) {
5 Stack<Character> stack = new Stack<>();
6 for (char c : s.toCharArray()) {
7 if (c == '(' || c == '{' || c == '[') {
8 stack.push(c);
9 } else {
10 if (stack.isEmpty()) return false;
11 char top = stack.pop();
12 if ((c == ')' && top != '(') ||
13 (c == '}' && top != '{') ||
14 (c == ']' && top != '['))
15 return false;
16 }
17 }
18 return stack.isEmpty();
19 }
20}The Java solution employs a stack to process and match each bracket in the string. If there's any imbalance or mismatched pair, it returns false.
This approach uses two pointers with stack-like behavior internally, taking advantage of simple list operations for push and pop.
Time Complexity: O(n)
Space Complexity: O(n/2) = O(n)
1def isValid(s: str) -> bool:
2 if len(s) % 2 != 0:
3 return False
4
5 stack =[]
6 for char in s:
7 if char in ['(', '{', '[']:
8 stack.append(char)
9 else:
10 if len(stack) == 0:
11 return False
12 top = stack.pop()
13 if not ((char == ')' and top == '(') or
14 (char == '}' and top == '{') or
15 (char == ']' and top == '[')):
16 return False
17 return len(stack) == 0The Python code employs both length parity and stack growth inside looping. It performs element-wise checks against frequent boundary conditions, using a single output for validity.