Problem Solving Skills for LeetCode

When faced with the challenge of validating a string of parentheses, it is essential to approach the problem methodically. Let's break down the steps to solve this problem.

Understanding the Problem

To begin, whenever you don’t know how to solve a problem, always start by thinking of examples. In the case of validating parentheses, we deal with three types: () (round), [] (square), and () (curly). Each type must match its corresponding pair to be valid.

Here are a few straightforward examples:

Simple Matching:
```
() 
[]
()
```
All these examples are valid as each opened parenthesis is properly closed.
Nested Parentheses:
```
(([]))
```
Here, valid pairs are nested correctly. The most recently opened parenthesis is always closed first.
Wrong Order:
```
)( 
```
This is invalid as closing parenthesis ) comes before any opening parenthesis.
Mismatched Pairs:
```
(]
```
This is also invalid, as the closing ] does not match the nearest opened (.

Key Observations

From these examples, we gather key observations, particularly that a closing parenthesis must match the nearest open parenthesis. This critical insight allows us to employ a stack data structure to resolve the problem effectively.

Developing the Solution

Here’s a step-by-step approach using a stack:

Mapping Pairs: Map each closing parenthesis to its corresponding opening counterpart.
Iteration: Iterate over the input string:
- If you encounter an open parenthesis, push it onto the stack.
- If you encounter a closing parenthesis:
  - Check if the stack is empty. If it is, return false immediately.
  - Verify the top of the stack matches the corresponding opening parenthesis. If it doesn't, return false.
  - If it matches, pop the stack, indicating a successful match.
Final Check: After processing the entire string, if the stack is empty, it means all parentheses have been matched correctly, thus the string is valid.

Implementation

Here's a sample implementation in Python for better understanding:

def isValid(s: str) -> bool:
    # Mapping of closing to opening parentheses
    mapping = (')': '(', ')': '(', ']': '[')
    stack = []

    for char in s:
        if char in mapping:
            top_element = stack.pop() if stack else '#'
            if mapping[char] != top_element:
                return false
        else:
            stack.append(char)
    
    return not stack

Keywords

Parentheses Validation
Stack
Matching Pairs
Data Structure
Algorithm

FAQ

Q1: Why use a stack for this problem? A1: A stack helps efficiently manage the nested structure, ensuring that the most recent opening parenthesis is always matched with the current closing parenthesis.

Q2: What happens if there are unmatched opening parentheses? A2: If there are unmatched opening parentheses, they will remain in the stack. By checking if the stack is empty at the end, we ensure all parentheses are correctly matched.

Q3: Can this method be applied to other types of bracket matching problems? A3: Yes, this method is generic and can be applied to validate any types of nested structures, such as HTML tags or other bracket systems.

Q4: Is the provided code snippet efficient? A4: Yes, the provided code is efficient with a time complexity of O(n), where n is the length of the input string, as it involves a single pass through the string and uses space proportional to the number of unmatched opening parentheses.