samidarko

    componentWillUpdate() {
        this.scrollY = window.scrollY;
    }

    componentDidUpdate() {
        if (this.scrollY) {
            window.scroll(0, this.scrollY);
        }
    }

samidarko

def countOccurrences (s: String): String = s span (x => x == s.head) match {
  case ("", "") => ""
  case (fw, rest) => fw.length.toString + fw.head ++ countOccurrences(rest)
}

countOccurrences("")        // ""
countOccurrences("aaa")     // "3a"
countOccurrences("aaabbb")  // "3a3b"
countOccurrences("aaabbbccchhhhhhddddhhhheee")  // "3a3b3c6h4d4h3e"

samidarko

import qualified Data.Char as Char

countOccurrences :: String -> String
countOccurrences "" = "" 
countOccurrences s = let (fw, rest) = span (\x -> x == head s) s in (Char.intToDigit $ length $ fw) : (head s) : (countOccurrences rest)

countOccurrences "aaaaabbbbccccccaaaaaaa"
-- "5a4b6c7a"

samidarko

# python 3

def span(string, char):
    i = 0
    while i < len(string) and string[i] == char:
        i += 1
    return string[0:i], string[i:]


def count_occurrences(string):

samidarko

def solution(arr):
    """
    Take a list of days and return optimal solution for transport fees where
    Tickets are: day (2), weekly (7) and monthly (25) tickets
    And month is bounded to 30 days, first day is 1
    :param arr: list of days
    :type arr: list
    :return: transport month fees
    :type: int

samidarko

module Main exposing (..)

import Html exposing (Html)
import Html.App
import Widget


-- MODEL

samidarko

from math import log
from matplotlib import pyplot as plt
import numpy as np

def fn_a(n):
  return 64*n*log(n)

def fn_b(n):
  return 8*n*n

samidarko

def cities_by_distance(arr):

    result = [None for _ in arr]
    temp = [0 for _ in range(len(arr) - 1)]

    for i, _ in enumerate(arr):
        origin = i
        pos = i
        hop = 0

samidarko

/**
  * Created by vincentdupont on 26/8/16.
  */

import util.Random

object InsertionSort extends App {
  def sortImp(a: Array[Int]) : Array[Int] = {
    var key, i : Int = 0
    for (j <- 1 until a.length) {

samidarko

-- give all the ways of giving the change of an amount of money based on a list of coins
countChange :: Int -> [Int] -> Int
countChange money coins
    | money < 0     = 0
    | money == 0    = 1
    | coins == []   = 0
    | otherwise     = countChange (money - (head coins)) coins + countChange money (tail coins)
    
-- countChange 10 [1, 5] == 3