## Problem Statement

Given a string `s`

containing only the characters '(', ')', '{', '}', '[', and ']', we need to determine if the input string is valid. An input string is considered valid if the following conditions are met:

- Open brackets must be closed by the same type of brackets.
- Open brackets must be closed in the correct order.
- Every closing bracket has a corresponding open bracket of the same type.

Let's walk through the steps to solve this problem using Python.

## Solution Approach

The given solution utilizes a stack data structure to efficiently check the validity of the input string. Here's a breakdown of the steps involved:

- Initialize an empty stack to keep track of opening brackets encountered in the input string.
- Create a mapping dictionary (
`map`

) to map closing brackets to their corresponding opening brackets. For example, ')' corresponds to '('. - Iterate through each character in the input string.
- For each character:

- If the stack is not empty, the current character is a closing bracket, and the top of the stack contains the corresponding opening bracket, pop the element from the stack as they form a valid pair.
- Otherwise, if the character is not a closing bracket or if it doesn't match the corresponding opening bracket at the top of the stack, push it onto the stack.

- After processing all characters in the string, check if the stack is empty. If it is, return
`True`

, indicating that the input string is valid; otherwise, return`False`

.

Now, let's dive into the code and provide detailed commentary for better understanding.

```
class Solution:
def isValid(self, s: str) -> bool:
# Initialize an empty stack to keep track of opening brackets
stack = []
# Create a mapping dictionary to map closing brackets to their corresponding opening brackets
mapping = {")": "(", "]": "[", "}": "{"}
# Iterate through each character in the input string
for char in s:
# If the stack is not empty and the current character is a closing bracket,
# and the top of the stack contains the corresponding opening bracket,
# pop the element from the stack as they form a valid pair
if stack and char in mapping and stack[-1] == mapping[char]:
stack.pop()
# Otherwise, if the character is not a closing bracket or if it doesn't match
# the corresponding opening bracket at the top of the stack, push it onto the stack
else:
stack.append(char)
# After processing all characters in the string, check if the stack is empty
# If it is, return True indicating that the input string is valid; otherwise, return False
return len(stack) == 0
```

This solution efficiently solves the problem with a time complexity of O(n), where n is the length of the input string `s`

. By using a stack, we can maintain the order of brackets and quickly check for validity.

I hope this article has provided you with a clear understanding of how to approach and solve the valid parentheses problem. Happy coding!