REVERSE WORDS IN A STRING | LEETCODE 151 | PYTHON SOLUTION

[@dnm9931]

Guys it looks like ‘i-1’ isn’t needed, ,y solution passes all test cases when I submit it without that line! Thanks so much,Cush cracking faang for the vid! You probably see me commenting on all your vids now!

[@arkachaudhuri7339]

Hey, Can someone explain the purpose for the i -= 1 after parsing the string builder? Shouldn't it anyway iterate forth to the next if it encounters a whitespace?

[@elias043011]

He's moving it back WITHIN the if statement because after exiting the if conditional block, he adds to "i" again.. you could get around this by checking ahead in the nested while statement while i + 1 < len(s) and s[i + 1] != " ":

[@kingfriizy5707]

its a medium because the follow up is solve in constant memory

[@barybary42]

which is not that straightforward

[@Ammar23217]

very good solution

[@crackfaang]

Thanks mate, glad you found it helpful

[@wentaogao1118]

Thanks., very nice solution

[@crackfaang]

Thanks for the kind words! Make sure to subscribe so you don’t miss future videos

[@pshi1989]

thank you 😊

[@MohitJain-dt4nb]

why can't we simply do this-: class Solution: def reverseWords(self, s: str) -> str: words = s.split() return ' '.join(reversed(words))

[@GiaBìnhVũ-w4s]

yoooo that is much better, ty !!

[@luuc]

That's definitely one way to solve it. Though, the idea behind the question is to practice a specific competitive programming pattern (two-pointer). Can you solve it without any memory usage?