🔒 Valid Paranthesis - Leet Code

[spark timber]

@late mica

Python help channel opened

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[violet wolf]

iterate over the string keeping count of the values for unique opening and unique closing brackets, if it any point the closing bracket count exceeds the opening bracket count of the same type, it is not valid

your code does not work because it will not consider ([]) to be valid because the ( is not immediately closed.

[late mica]

Got it
Thanks

[tropic python]

using this approach, ([)] will be considered valid

[violet wolf]

yeah i did realize that a minute ago. im pretty sure it can be fixed by checking that all open/close bracket counts are equal whenever a closing bracket is seen. which, if theres 3 types, unfortunately triples the operations, even if its still linear. wondering if im missing some clever "1-pass-1-check" solution

[tropic python]

there is a lot simpler solution: ||use stack of opened brackets||

[shell aspen]

!or

[spark timber]

The or-gotcha

When checking if something is equal to one thing or another, you might think that this is possible:

# Incorrect...
if favorite_fruit == 'grapefruit' or 'lemon':
    print("That's a weird favorite fruit to have.")

While this makes sense in English, it may not behave the way you would expect. In Python, you should have complete instructions on both sides of the logical operator.

So, if you want to check if something is equal to one thing or another, there are two common ways:

# Like this...
if favorite_fruit == 'grapefruit' or favorite_fruit == 'lemon':
    print("That's a weird favorite fruit to have.")

# ...or like this.
if favorite_fruit in ('grapefruit', 'lemon'):
    print("That's a weird favorite fruit to have.")

[violet wolf]

that was the one yeah

[spark timber]

Python help channel closed

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