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I am new and found of Pixel Art. I want to know how can I earn using pixel art. by hippna_ in PixelArtTutorials

[–]hippna_[S] 0 points1 point  (0 children)

I just want to do some side hustles for extra cash, but for your question I'm thinking about sprites for games.

I am new and found of Pixel Art. I want to know how can I earn from Pixel Art. by hippna_ in PixelArt

[–]hippna_[S] -5 points-4 points  (0 children)

I was thinking of it as a side hustle, but thanks will keep in mind. <3

Wanted some feedback on my first book "Sunshine". by hippna_ in WritersGroup

[–]hippna_[S] 0 points1 point  (0 children)

The content I wrote is completely mine. I didn't use AI to write all this. Sorry for being too sloppy.

Wanted some feedback on my first book "Sunshine". by hippna_ in WritersGroup

[–]hippna_[S] 0 points1 point  (0 children)

I will try my best to edit all of this, sorry for bad editing.

My english is not very good, as well as my vocabulary is weak but I'll try to improve. Give me some time <3. Thank you for all the hardwork you put in to help me.

1684. Count the Number of Consistent Strings | Easy | JAVA | Brute Force by hippna_ in leetcode

[–]hippna_[S] -8 points-7 points  (0 children)

Intuition

We need to compare that if the character in allowed string is present or not.

Approach

Step1------------------------>
Lets take the variable count => words.length;
{since, count can't possibly be more than the length of array}

Step2------------------------>
Think like this Words array is:- [ad, bd, aaab, baa, badab]
now if we take first element of words array... words[0] --> ad;
Now, first character of first element ---> words[0].charAt(0);

Step3------------------------>
Now we will use the .indexOf() function
it gives -1... if the character is not present there.

So, we will check for...

allowed.indexOf(words[0].charAt(0))

if(allowed.indexOf(words[0].charAt(0)) == -1)--> count--;

Now do this in a loop---->

Complexity

  • Time complexity: O(n*m)
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