LeetCode Python Solutions: 20. Valid Parentheses #python #coding #leetcode смотреть онлайн

ZeroStress LeetCode Python Solutions: 20. Valid Parentheses #python #leetcode

1. Initialize a Stack and Define a Mapping:

Create an empty list called 'result,' which will serve as a stack to keep track of open brackets.
Define a dictionary called 'base' that maps closing brackets to their corresponding opening brackets. This dictionary is crucial for quickly checking if a closing bracket matches the most recent open bracket on the stack.

2. Iterate Through the Input String:

Loop through each character (bracket) in the input string 's.'

3. Handle Open Brackets:

If the current character is an open bracket ('(', '{', '['), push it onto the stack 'result.' This indicates that we have encountered an open bracket that needs to be matched with a closing bracket later.

4. Handle Closing Brackets:

If the current character is a closing bracket (')', '}', ']'), do the following:
Check if the 'result' stack is empty. If it's empty, return False immediately because there is no corresponding open bracket for this closing bracket.
If the stack is not empty, pop the most recent open bracket from the stack.
Compare the popped open bracket with the current closing bracket using the 'base' dictionary. If they don't match (e.g., '(' does not match with ']'), return False because the brackets are not valid.
If they match, continue processing the next character in the string.

5. Final Check:

After processing the entire string, check if the 'result' stack is empty. If it's empty, this means all open brackets have been correctly matched and closed, and the string is valid. Return True.
If there are unmatched open brackets left on the stack, return False because the string is not valid.

Approach Recap:

The key to this approach is using a stack ('result') to keep track of the open brackets as we encounter them. When we encounter a closing bracket, we check if it matches the most recent open bracket on the stack using the 'base' dictionary. If it matches, we pop the open bracket from the stack, indicating that it has been successfully matched and closed.

By following this approach, we ensure that open brackets are closed in the correct order, that every closing bracket has a corresponding open bracket of the same type, and that the string's overall bracket structure is valid. This solution elegantly handles the problem's requirements and efficiently determines the validity of the input string.

Time Complexity: O(n)

The time complexity of this solution is linear, where 'n' is the length of the input string 's.' This is because we iterate through each character in the string exactly once.

Space Complexity: O(n)

The space complexity is also linear and depends on the length of the input string 's.' In the worst case, if all characters in 's' are open brackets ('(', '{', '['), the 'result' stack could potentially contain all 'n' characters. Hence, the space complexity is O(n).

It's important to note that both the time and space complexities are linear, making this solution highly efficient for checking the validity of parentheses in a string. The use of the stack allows us to keep track of open brackets in the correct order and efficiently match them with closing brackets while ensuring that the algorithm runs in linear time.

Intuition:
At its core, this solution leverages the concept of a stack to efficiently check whether a given string of parentheses is valid. Here's how it works step by step:

We start by initializing an empty list called 'result,' which will serve as our stack. A stack is a data structure that follows the Last-In-First-Out (LIFO) principle, meaning the last element pushed onto the stack will be the first one popped off.

We define a dictionary called 'base' that maps closing brackets to their corresponding opening brackets. This dictionary is essential because it allows us to quickly check if a closing bracket matches the most recent open bracket on the stack.

We iterate through each character (bracket) in the input string 's.'

If we encounter an open bracket ('(', '{', '['), we push it onto the stack 'result.' This indicates that we have an open bracket that needs to be matched with a closing bracket later.

If we encounter a closing bracket (')', '}', ']'), we check whether the 'result' stack is empty or not. If it's empty, there's no corresponding open bracket, so we return False immediately. Otherwise, we compare the current closing bracket 'idx' with the expected open bracket using our 'base' dictionary. If they don't match, we also return False.

If the closing bracket matches the expected open bracket, we pop the most recent open bracket from the stack 'result' because it has been successfully matched and closed.

After processing the entire string, we check if the 'result' stack is empty. If it's empty, this means all open brackets have been correctly matched and closed, and the string is valid. We return True.