railsstudent

class VigenereCipher
  constructor: (@key, @abc) ->
    @dict = {}    
    i = 0
    for x in @abc
      shift = @abc[i..] + @abc[0...i]
      @dict[x] = shift
      i++;
      
  encode: (str) ->  

railsstudent

function VigenèreCipher(key, abc) {
  
  this.encode = function (str) {
    var encodeStr = ''
    for (var i in str) {
      var c = str[i];
      var k = key[i % key.length];
      if (abc.indexOf(c) >= 0) { 
        encodeStr += abc[(abc.indexOf(c) + abc.indexOf(k)) % abc.length];
      } else {

railsstudent

// http:// stackoverflow.com/questions/23202489/how-does-this-code-find-the-number-of-trailing-zeros-from-any-base-number-factor
// https://comeoncodeon.wordpress.com/2010/02/17/number-of-zeores-and-digits-in-n-factorial-in-base-b/
function zeroes (base, number) {
  
  var noz = Number.MAX_VALUE;
  // Now we can break the Base B as a product of primes :
  // B = a^p1 * b^p2 * c^p3 * …
  //Then the number of trailing zeroes in N factorial in Base B is given by the formulae
  // min{1/p1(n/a + n/(a*a) + ….), 1/p2(n/b + n/(b*b) + ..), ….}.
  var j = base;

railsstudent

var Sudoku = function(data) 
{
  //   Private methods
  // -------------------------
  var N = data.length;
  var rootN = Math.floor(Math.sqrt(N));
  
  var isValidSubSudoku = function(leftX, leftY, size) {
    var sudokuTracker = {};
    for (var j = leftX; j < leftX+size; j++) {

railsstudent

// https://codemyroad.wordpress.com/2014/05/01/solving-sudoku-by-backtracking/
/*
  On the other hand, the backtracking algorithm fills up a blank cell with a 
  valid number (i.e. no two same numbers in any row, column or big box),
  moves on to the next cell, and then does the same thing. If all the possible 
  numbers from 1 to 9 are invalid for any cell that the algorithm is currently “at”, 
  the algorithm moves back to the previous cell and changes that cell’s value to another valid number. Afterwards, it moves back to the next cell and the whole process repeats.
*/
// https://en.wikipedia.org/wiki/Sudoku_solving_algorithms
// use backtracking algorithm

railsstudent

function handleAcromyn(accumulator, path, isLast) {
  var excluded = ["the","of","in","from","by","with","and", "or", "for", "to", "at", "a"];
  var acromynList = path.split('-');
  var shortener = '';
  for (var k in acromynList) {
    var term = acromynList[k];        
    if (excluded.indexOf(term) < 0) {
      shortener += term[0];
    }
  }

railsstudent

// http://www.geeksforgeeks.org/find-last-digit-of-ab-for-large-numbers/
//http://qa.geeksforgeeks.org/1061/given-big-number-form-string-how-do-take-mod-of-that-big-number

var asciiZero = "0".charCodeAt(0);
var findModulo = function(base, exponent) {
  var mod = 0;
  for (var i in exponent) {
    //mod = (mod*10 + b[i] - '0')%a;
    mod = (mod * 10 + exponent.charCodeAt(i) - asciiZero) % base;
  }

railsstudent

// http://stackoverflow.com/questions/5131497/find-the-index-of-a-given-permutation-in-the-sorted-list-of-the-permutations-of
// http://stackoverflow.com/questions/18644470/anagram-index-calculation/18646156#18646156
var f = [];
function factorial (n) {
  if (n == 0 || n == 1)
    return 1;
  if (f[n] > 0)
    return f[n];
  return f[n] = factorial(n-1) * n;
} 

railsstudent

// complete the function so that it returns the fastest route 
function navigate(numberOfIntersections, roads, start, finish) {
  var arrDrivingTimes = [];
  var arrRoutes = []
  
  // initialize driving time and route in matrixes
  for (var i = 0; i < numberOfIntersections; i++) {
      arrDrivingTimes.push([]);
      arrRoutes.push([]);
      for (var j = 0; j < numberOfIntersections; j++) {

railsstudent

function undoRedo(object) {
  //var storage = object;
  var actions = [];  // keep track of action performed
  // action schema -> { name, params, has_undone }
  // for non-undo acton, example of params { key, value }
  // for undo action, example of params { action, key, value }

  var findLastAction = function(status) {
    var lastAction = null; 
    for (var i = actions.length - 1; i >= 0; i--) {