keif

var charCodes= {
	"a": ". _",
	"b": "_ . . .",
	"c": "_ . _ .",
	"d": "_ . .",
	"e": ".",
	"f": ". . _ .",
	"g": "_ _ .",
	"h": ". . . .",
	"i": ". .",

keif

// ["2", "1", "+", "3", "*"] -> ((2 + 1) * 3) -> 9
// ["4", "13", "5", "/", "+"] -> (4 + (13 / 5)) -> 6
var arr1 = ["2", "1", "+", "3", "*"]; // 9
var arr2 = ["4", "13", "5", "/", "+"]; // 6

// lifted from jQuery
var inArray = function (elem, arr, i) {
    var len;

    if (arr) {

keif

// naive approach
// Naively, we can simply examine every substring and check if it is palindromic.
// The time complexity is O(n^3).// Therefore, this approach is just a start, we need a better algorithm.
var longestPalindrome1 = function (string) {
	var maxPalinLength = 0,
		longestPalindrome = null,
		length = string.length;

	// check all possible sub strings
	for (var i = 0; i < length; i++) {

keif

/**
 *  Computes the longest palindromic substring in linear time
 *  using Manacher's algorithm.
 *
 *  Credits: The code is lifted from the following excellent reference
 *  http://www.leetcode.com/2011/11/longest-palindromic-substring-part-ii.html
 **/

var Manacher = (function () {
    var palindromic = [],  // p[i] = length of longest palindromic substring of t, centered at i

keif

// Given a string s and a dictionary of words dict, determine if s can be segmented into a space-separated sequence of one or more dictionary words.
// For example, given
var string = "leetcode",
	dict = ["leet", "code"];
// Return true because "leetcode" can be segmented as "leet code".

// 1. Naive Approach
// This problem can be solve by using a naive approach, which is trivial. A discussion can always start from that though.
// Time is O(n^2) and exceeds the time limit.
var wordBreak = function (string, dict) {

keif

// Given a string s and a dictionary of words dict, add spaces in s to construct a sentence where each
// word is a valid dictionary word. Return all such possible sentences. For example, given:
// s = "catsanddog", dict = ["cat", "cats", "and", "sand", "dog"], the solution is ["cats and dog", "cat sand dog"].

// need to track the actual matched words
var wordBreak = function (string, dict) {
    //create an array of ArrayList<String>
    var dp = [string.length + 1];
    dp[0] = [];
 

keif

// old school double loop
// quadratic O(n^2) inefficiency
var unique = function(array) {
    var arr = [],
    	len = array.length;

    for (var idx = 0; idx < len; i++) {
		for(var jdx = i + 1; jdx < len; jdx++) {
			if (array[idx] === array[jdx]) {
				j = ++i;

keif

#!/bin/sh

# Alot of these configs have been taken from the various places
# on the web, most from here
# https://github.com/mathiasbynens/dotfiles/blob/master/.osx

# Set the colours you can use
black='\033[0;30m'
white='\033[0;37m'
red='\033[0;31m'

keif

// variables for panel splitter;
var panelSplitterWidth = {
  defaultW: null,
  currentW: null
};

// fires on mousedown on panel splitter draggable area
var panelSplitterClick = function(event) {
  var $dragPanel = $(event.Event.target).closest("td:not('.af_panelSplitter_horizontal-icon-container')");

keif

#!/bin/sh

# Install Homebrew
which -s brew
if [[ $? != 0 ]] ; then
    /usr/bin/ruby -e "$(curl -fsSL https://raw.githubusercontent.com/Homebrew/install/master/install)"
else
    brew update
fi