Valid Parentheses

Problem Link

20. Valid Parentheses – LeetCode

Problem Statement

Given a string s containing just the characters '(', ')', '{', '}', '[', and ']', determine if the input string is valid.

A string is valid if:

1. Open brackets are closed by the same type of brackets.

2. Open brackets are closed in the correct order.

Examples

Example 1:

Input: s = "()"
Output: true

Example 2:

Input: s = "()[]{}"
Output: true

Example 3:

Input: s = "(]"
Output: false

Constraints

• 1 <= s.length <= 10⁴

• s consists of parentheses only: '()[]{}'

Intuition

The problem is based on balanced brackets. A valid parentheses string must maintain:

• Matching pairs for every opening bracket.

• Correct nesting and closing order.

A stack is ideal for this, because it follows Last-In-First-Out (LIFO) behavior:

• Push opening brackets.

• For closing brackets, pop and check for correct pairing.

Approach: Stack-Based Matching

1. Traverse the input string character by character.

2. If the character is an opening bracket, push it to the stack.

3. If it's a closing bracket:

   • Check if the stack is empty → if yes, return false.

   • Check the top of the stack:

     • If it's a matching opening bracket, pop it.

     • If not, return false.

4. After the loop, return true only if the stack is empty (all brackets matched).

Java Code

class Solution {
    public boolean isValid(String s) {
        if (s.charAt(0) == '}' || s.charAt(0) == ')' || s.charAt(0) == ']') return false;

        Stack<Character> stack = new Stack<>();
        for (int i = 0; i < s.length(); i++) {
            char ch = s.charAt(i);
            
            if (ch == '(' || ch == '{' || ch == '[') {
                stack.push(ch);
            } else {
                if (stack.isEmpty()) return false;
                
                char top = stack.peek();
                if ((ch == ')' && top == '(') ||
                    (ch == ']' && top == '[') ||
                    (ch == '}' && top == '{')) {
                    stack.pop();
                } else {
                    return false;
                }
            }
        }
        return stack.isEmpty();
    }
}

Time and Space Complexity

Dry Run

Input:

s = "{[()]}"

Stack State:

• { → push → [{]

• [ → push → [{ [ ]

• ( → push → [{ [ ( ]

• ) → match with ( → pop

• ] → match with [ → pop

• } → match with { → pop

Stack is empty → valid.

Output:

true

Follow-Up

To generalize or optimize:

• Use a HashMap for bracket pairs ({')': '(', ']': '[', '}': '{'}) to avoid repetitive if-else conditions.

• Use Deque (ArrayDeque) for faster stack operations in Java.

Conclusion

This problem is a foundational example of stack-based validation, especially useful in:

• Parsing expressions

• Validating XML/HTML

• Compiler design