# Solving Leetcode Interviews in Seconds with AI: Valid Parentheses


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "20" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> Given a string s containing just the characters '(', ')', '{', '}', '[' and ']', determine if the input string is valid. An input string is valid if:  Open brackets must be closed by the same type of brackets. Open brackets must be closed in the correct order. Every close bracket has a corresponding open bracket of the same type.    Example 1:  Input: s = "()" Output: true  Example 2:  Input: s = "()[]{}" Output: true  Example 3:  Input: s = "(]" Output: false  Example 4:  Input: s = "([])" Output: true    Constraints:  1 <= s.length <= 104 s consists of parentheses only '()[]{}'.  

	# Explanation
	Here's the approach to solve this problem:

*   **Stack Data Structure:** Use a stack to keep track of the opening brackets encountered.
*   **Matching Pairs:** Use a dictionary to map closing brackets to their corresponding opening brackets.
*   **Iteration and Validation:** Iterate through the string. If an opening bracket is encountered, push it onto the stack. If a closing bracket is encountered, check if the stack is empty or if the top element of the stack does not match the corresponding opening bracket. If either of these conditions is true, the string is invalid.

*   **Runtime Complexity:** O(n), where n is the length of the string s.
*   **Storage Complexity:** O(n) in the worst-case scenario (e.g., "(((((((").

	
	# Code
	```python
	def isValid(s: str) -> bool:
    """
    Determines if the input string is valid based on bracket matching.
    """
    stack = []
    mapping = {')': '(', '}': '{', ']': '['}

    for char in s:
        if char in mapping:
            top_element = stack.pop() if stack else '#'  # Use '#' as a dummy value
            if mapping[char] != top_element:
                return False
        else:
            stack.append(char)

    return not stack  # If the stack is empty, it's valid
	```