I LOVE This EASY Coding Interview Question! | Length of Last Word - Leetcode 58

dynamic programming, leetcode, coding interview question, data structures, data structures and algorithms, faang

[GregHogg]

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[stanleykoh]

Wouldn't it be faster since you're already accessing the back of the string to just start counting until you hit a space then return the length of the word

[tomalator]

That's an awful solution.

Start at the end, go towards the beginning until you find the first letter, the increase the count for each letter you find, and return the count once you find a space

[JoshBlasy]

I feel like this channel deliberately gives suboptimal solutions to bait engagement in the comments

[Dissimulate]

Or just start at the index -1 to get the last character and count backwards until you hit a space.. This becomes more efficient the longer your string is.

[ElmoSeesYou]

You could also just say: return len(s.split()[-1]) i believe

[haraldschioberg]

Not a good answer. You can traverse right to left.

[as]

I was struggling with my homework, you fixed it!

[XxThatGuyMysteryXx]

let str = "Hello my name is Jeff. ";
str = str.trim(); // Remove leading and trailing whitespace
let words = str.split(' '); // Split the string into an array of words
let lastWord = words[words.length - 1]; // Access the last element of the array
console.log(lastWord.length); // Print the last word

[neupanes]

why not do the same thing from the back, return when the length is higher than 0 (avoiding space)? (Assuming, a single word is considered as the last word) 🤔

[anubis]

return len(input.strip().split(' ')[-1])

[bjni]

idk just split the string, check the last array index if its empty, check the next one until you hit something thats not empty, then check the length of that

[AvgBlue]

I fill like you give the best solution so people will go into the comment and try to fix it and then you will have more engagement on the video

[hunorfekete]

One line prob same complexity would be to say [st for st s.split(" ") if st][-1]
If u go from the end with the same strat as from front but if u see a space and count is not none return it won't go through the whole str wich helps a lot for huge texts

[ilm_yanfa]

simply invert the string and count characters until seeing the first space:

def lent(s:str):
count = 0
for i in s[::-1]:
if i == " ":
return count
count += 1
return -1

[ARKGAMING]

Why would you go first to last?
Assuming they don't want you to use built in functions (cause if they did it's just _words = sentence.trim().split(' '); return words[words.length - 1].length;_ )

You just start from the end of the string, start counting when you reach a non-space character, and then when you reach a space and the counter is more than 0, that means you just found the length of the last word...

[iamtheonlyjohn]

Couldn't you revert the string and count the first word that doesn't have the length 0?

[aaronrockwell]

split string by spaces, check for length greater than 0, get last index with length > 0

[MZGamessuper]

Why not just return len(arr.split()[-1])

[mufasum]

You can also just the strip and split method to do this in a few lines of code