Valid Parentheses - LeetCode #20 - Python, JavaScript, Java, C++

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Valid Parentheses with Python, JavaScript, Java and C++, LeetCode #20!

Welcome to my channel! In this video, we're going to tackle the "Valid Parentheses" problem, a classic algorithmic challenge in computer science.

The problem statement goes like this: Given a string s containing parentheses, curly braces, and square brackets, we need to determine if the arrangement of these characters forms a valid sequence of parentheses.

We'll walk through a Python solution step by step, exploring how to use a stack data structure to efficiently check the validity of the parentheses arrangement. Along the way, we'll discuss the time and space complexity of our solution, ensuring it's both efficient and practical.

So, if you're ready to level up your algorithm skills and master the art of checking valid parentheses arrangements, then stay tuned! Let's dive into the world of "Valid Parentheses" together.

CodingNinja offers other videos on programming, algorithms, and data structures as well. Subscribe to the channel so you don't miss out on new content!

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■ Timeline
0:04 HashMap and Stack Solution
3:53 Coding
5:28 Time Complexity and Space Complexity
6:08 Stack Solution
9:35 Coding
11:46 Time Complexity and Space Complexity

■ Blind 75 LeetCode questions

■ Twitter

■ Instagram

■ Problem Link

■ Codes in the video
- Python, JavaScript, Java and C++ are available from here.

■ Other Playlists

#leetcode #python #interview

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I had the same idea but my code became convoluted and it started to cause Output Limit exceeded error in some test cases :

class Solution:
def isValid(self, s: str) -> bool:

stack = []

opening_brackets = ["(", "{", "["]
closing_brackets = [")", "}", "]"]

length = len(s)

stack.append(s[0])

for i in range(1, length) :

bracket = s[i]
# print(bracket)

if stack == [] :
if bracket in closing_brackets :
print("False 1")
return False
else :
stack.append(bracket)

else :
# print("Tag 1")
last_bracket = stack[-1]
print(last_bracket)

if (bracket == ")" and last_bracket == "(") or (bracket == "}" and last_bracket == "{") or (bracket == "]" and last_bracket == "["):
stack.pop()
print(stack)
elif bracket in closing_brackets and last_bracket in opening_brackets :
print("False 2")
return False
else :
print("Tag 2")
stack.append(bracket)

print(stack)

if stack != [] :
print("False 3")
return False

return True

curiously-cinnamon

Not really sure what is going on around the 2:00 mark. Really poorly explained.

JohnDoe-wxzm

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