Anime Screenshot Series – SKET Dance

Hello Everyone!

It’s been a long time now, right? Yeah.. The reason is that I had sone serious health issues and lost my password while I was at it.

I only have a couple of screenshots right now because I was in between my two computers. Enjoy what I have!

Onihime

 

See you later ^_^

 

Anime Screenshot Series – Tsugumomo

Hello Everyone!

It’s been a while right?

The screenshots are from Tsugumomo. I couldn’t take HQ screenshots but I will update as soon as I get my hands on a higher quality video 🙂

This slideshow requires JavaScript.

 

And this one is from Renai Boukun

See you later ^_^

 

How to solve ‘removeDuplicateAdjacents’ in CodeFights

The Problem

Given a string s, recursively remove any adjacent duplicate characters that it contains.

Example

• For s = "cooodefightssforrrcodee", the output should be
  removeDuplicateAdjacent(s) = "cdefightfocod".
• For s = "acaaabbbacdddd", the output should be
  removeDuplicateAdjacent(s) = "acac".

Input/Output

• [time limit] 4000ms (py)

• [input] string s

  A string composed of lowercase English letters.

  Constraints:

  1 ≤ s.length ≤ 50.

• [output] string

  A string obtained by removing all adjacent duplicates from the input string.

The Solution

The Explanation

CodeFights Problems

Stay tuned for more solution/explanations.

If you have a question you want to see the solution of, comment below so that I can help you!

Also, I figured out how to put actual code snippets with indentation and all, so, the newer ones will have the code as a text instead of an image.

How to solve “sumOfTwo” in CodeFights

The Problem:

You have two integer arrays, a and b, and an integer target value v. Determine whether there is a pair of numbers, where one number is taken from a and the other from b, that can be added together to get a sum of v. Return true if such a pair exists, otherwise return false.ExampleFor a = [1, 2, 3], b = [10, 20, 30, 40], and v = 42, the output should be
sumOfTwo(a, b, v) = true.

I guess people like to read about the things that they want the most and experience the least.

Code Formatting

Hello everyone!

I just realized that my settings remove the indentation of the codes. I will fix that problem as soon as possible. Sorry about the confusion!

How to Solve ‘DifferentRightmostBit’ in CodeFights

The Problem

You’re given two integers, n and m. Find position of the rightmost bit in which they differ in their binary representations (it is guaranteed that such a bit exists), counting from right to left.

Return the value of 2position_of_the_found_bit (0-based).

Example

For n = 11 and m = 13, the output should be
differentRightmostBit(n, m) = 2.

1110 = 10112, 1310 = 11012, the rightmost bit in which they differ is the bit at position 1 (0-based) from the right in the binary representations.
So the answer is 21 = 2.

Input/Output

• [time limit] 4000ms (py)

• [input] integer nConstraints:
  0 ≤ n ≤ 230.

• [input] integer mConstraints:
  0 ≤ m ≤ 230,
  n ≠ m.

• [output] integer

Anime Screenshot Series – Resurrection Efforts

Hi pals!

It’s been real long time since I did this.

Here are two screenshots from my computer. Tho, I have some in my phone, I am too lazy to reach to my phone now

First one is from Bakuon!!

A little Python!

The programming language one, not the snake one!

Hello pals!
Some time ago, I came across a problem in which I had to flatten an array of objects. That is, the elements of the array can be primitives or arrays. The end results is to be an array of primitives.

I used python to implement this since it allows much more compact snippets for such jobs.

Here is the code 🙂

def flatten(lst):
 return sum( ([x] if not isinstance(x, list) else flatten(x)
 for x in lst), [] )
 
lst = [[1], 2, [[3,4], 5], [[[]]], [[[6]]], 7, 8, []]
print flatten(lst)