lopespm

#!/bin/bash

build_folder="build"
source_image="source-image.png"

mkdir $build_folder

convert $source_image -scale 16 $build_folder/16.png
convert $source_image -scale 32 $build_folder/32.png
convert $source_image -scale 48 $build_folder/48.png

lopespm

const assert = require('assert');
const async = require('async');
const fs = require('fs');
let request = require('request');
const FileCookieStore = require('tough-cookie-filestore');

if (!fs.existsSync('cookies.json')) { fs.writeFileSync('cookies.json', '{}');}
let j = request.jar(new FileCookieStore('cookies.json'));
request = request.defaults({ jar : j });

lopespm

lopespm

Convert MOV to GIF using FFmpeg and ImageMagick

I tried a few different techniques to make a GIF via command-line and the following gives me the best control of quality and size. Once you're all setup, you'll be pumping out GIFs in no time!

Preparation

Install FFmpeg

• $ brew install ffmpeg [all your options]
  • Example: $ brew install ffmpeg --with-fdk-aac --with-ffplay --with-freetype --with-frei0r --with-libass --with-libvo-aacenc --with-libvorbis --with-libvpx --with-opencore-amr --with-openjpeg --with-opus --with-rtmpdump --with-schroedinger --with-speex --with-theora --with-tools

Install ImageMagick

lopespm

# Returns the size of the largest binary subtree that is complete, in Python - variant for exercise 9.1 on EPI (Elements of Programming Interviews)
# The gist of the solution is to keep track of the subtree's current number of nodes, current height and maximum height until that point.
# With the current number of nodes and height, one can calculate the root's own number of nodes and height via its direct childs respective information,
#   taking into to consideration the relation between the child heights and if they are perfect subtrees or not.
# Solution is O(n) time complexity and O(h) space complexity (function call stack corresponds from the root through the unique path to the current node)

from collections import namedtuple


class BTN():

lopespm

# (Variant for exercise 5.1 on EPI (Elements of Programming Interviews))
# The rationale behind it is to squeeze the forth color (middle right color) in between the middle left and right sub-arrays. It defines the colors as the algorithm progresses.
# It has a O(n) time complexity and O(1) space complexity

from typing import List

def dutch_variant_four_colors(array: List[int]) -> List[int]:
    left = array[0]
    mid_left = None
    right = None

lopespm

# (Variant #4 for exercise 11.1 on EPI (Elements of Programming Interviews))
# The rationale behind it to perform a binary search which only considers a given character index of the strings.
# If these are present, continue to the next character index. If any of the prefix characters cannot be found in any string, return immediatly
# Considering n strings and k prefix length:
#    Time complexity: O(k * log(n))
#    Space complexity: O(1)

from typing import List

def first(items: List[str], prefix: str, i: int, c: str, left: int, right: int):

lopespm

Keybase proof

I hereby claim:

• I am lopespm on github.
• I am lopespm (https://keybase.io/lopespm) on keybase.
• I have a public key ASBVOGTQ7FFo3e2z4-HGjCwpsxKjqroop4en_wFUDWevBwo

To claim this, I am signing this object:

lopespm

# Print tree by levels - using BFS
# The idea behind it is to use BFS and keep a level marker integer which marks the end the last node of the level. This is similar to Naresh's sentinel approach, but without the need of inserting a sentinel inside the queue, since this is accomplished by the level marker.
# Time complexity of O(n)
# Space complexity: O(2^tree_height)

from collections import deque

class Node:
    def __init__(self, data, left=None, right=None):
        self.data = data

lopespm

# Alternative python solution for 5.17 The Sudoku Check Problem on EPI (Elements of Programming Interviews)) (September 2018 edition)
# For an nxn Sudoku grid:
# Time complexity: O(n^2)
# Space complexity: O(n)

from typing import List
import math

from typing import List, Set
import math