avanishgiri’s gists

avanishgiri / merge_unique.js

Last active December 30, 2015 02:39

merge two sorted arrays, ignoring duplicates.

	var merge_unique = function(array1, array2){
	var len1 = array1.length;
	var len2 = array2.length;
	var buff = []
	var i = 0, j = 0

	while(i != len1 && j != len2){
	if(array1[i] < array2[j])
	buff.push(array1[i++]);
	else

avanishgiri / promise.js

Created November 21, 2013 04:59

	function Promise(){
	this.callbacks = [];
	this.value = undefined;
	}

	Promise.prototype.success = function(callback){
	this.callbacks.push(callback);
	if(this.value)
	this.execute_callbacks();
	}

avanishgiri / cipher.rb

Last active December 28, 2015 23:09

	def is_lower(c)
	c.ord >= 'a'.ord && c.ord <= 'z'.ord
	end

	def is_upper(c)
	c.ord >= 'A'.ord && c.ord <= 'Z'.ord
	end

	def rotx(x, string, encrypt = true)
	x = encrypt ? x : -x

avanishgiri / promise.js

Created November 21, 2013 04:18

	function Promise(){
	this.callbacks = [];
	this.value = undefined;
	}

	Promise.prototype.success = function(callback){
	this.callbacks.push(callback);
	if(this.value)
	this.execute_callbacks();
	}

avanishgiri / updated.rb

Last active December 28, 2015 22:59

	### ALGORITHM ###

	# for each index 0 .. last_index
	#for each index i .. 0
	# during each iteration of the inner loop
	# we store one member of a diagonal from the upper-left half
	# and the element that reflects it across the line y=x (the middle diagonal)
	# the member from the upper-left half is at index [i-j][i]
	# we push this member into the final array at index i

avanishgiri / all_diagonals.rb

Created November 21, 2013 00:40

	matrix = ('a'..'y').each_slice(5).to_a

	def all_diagonals(m)
	len = m[0].length
	final = Array.new(2*len - 1){[]}

	0.upto(len-1) do \|i\|
	i.downto(0) do \|j\|
	final[i] << m[i-j][j]
	final[2len-i-2] << m[len-1-j][j+len-i-1] if 2len-i-2 != len-1

avanishgiri / max_sum.rb

Created November 19, 2013 08:01

dynamic programming solution to robbers problem -- runs in linear time in the size of the array. tested against my previous brute force solution.

	a = [447, 462, 9, 272, 242, 0, 206, 998, 35, 414]

	def max(a, i, memo = {})
	return 0 if i == 0
	return a[i] if i == 1
	return memo[i] if memo[i]
	return memo[i] = [max(a,i-2) + a[i], max(a,i-1)].max
	end

	p max(a, 9)

avanishgiri / index.html

Last active December 28, 2015 07:19

Problem 4 solution. Uses jQuery for DOM manipulation.

	<!DOCTYPE HTML>
	<html>
	<head>
	<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.10.1/jquery.min.js"></script>
	<script type="text/javascript" src="timer.js"></script>
	</head>
	<input id="seconds" type="text" value="0">
	<div id="countdown">
	<span class="days">0 days</span>
	<span class="hours">0 hours</span>

avanishgiri / ruby_solutions.rb

Last active December 28, 2015 07:09

ruby solutions to problems 1 & 2.

	##################################### P1

	def reverse(string)
	string.length == 0 ? "" : string[-1] + reverse(string[0..-2])
	end

	##################################### P2

	def tokenize(url)
	url.split(/&/).map { \|i\| i.split(/=/) }

avanishgiri / hash_url.rb

Created November 14, 2013 04:47

ruby solution for converting url into dictionary

	def tokenize(url)
	url.split(/&/).map { \|i\| i.split(/=/) }
	end

	def dict(array_of_pairs)
	array_of_pairs.inject(Hash.new{[]}) { \|hash,pair\| hash[pair.first] += [pair.last] ; hash }
	end

	def scrub(hash)
	hash.map { \|k,v\| hash[k] = v.first if v.length == 1 }