I'm working on an algorithm to count the number of ways to build 100 cents given an infinite amount of pennies, dimes, nickels, and quarters.
I ended up coming with the above (which AFAIK works):
def count_ways(amount)
num_ways(amount, 0)
end
def num_ways(amount, index)
return 1 if amount == 0
return 0 if index >= COINS.length || amount < 0
num_ways(amount - COINS[index], index) +
num_ways(amount, index + 1)
end
Now, I'd like to memoize this algorithm. An effective way I've found to think about memoizing is to consider what inputs do we pass into this function repeatedly. In this case, I'd like to memoize the combination of amount & index parameters.
Typically when I have two parameters, I build a two D array as a way to memoize but here that makes a lot less sense. Consequently, how can you memoize off of these two parameters? Does it make sense to do something like this?
def count_ways(amount)
memo = Hash.new { |h, k| h[k] = [] }
num_ways(amount, 0, memo)
end
def num_ways(amount, index, memo)
return 1 if amount == 0
return 0 if index >= COINS.length || amount < 0
memo[amount - COINS[index], index] ||= num_ways(amount - COINS[index], index)
memo[amount, index + 1] ||= num_ways(amount, index + 1)
memo[amount - COINS[index], index] +
memo[amount, index + 1]
end
I believe there is no common way to implement memoization when you solve algo task. Because of speed. Way you have to choose depends on you algo, input data and so on.
Few imperative rules:
avoid creation of many data structure instances: h = {}; h[ [1,2] ]
= 3 will produce COINS.size * amount Arrays just for keys
use Array for contiguous data and Hash for gapped one
use Hash instead of Array when you can't predict data size
create Array with needed values when you can predict your
contiguous data size
Using that rules, memoization (just in your case, where COINS.size << amount and both datas are contiguous) might look like:
COINS = [25, 10, 5, 1]
def count_ways(amount)
# memo is just COINS.size + 1 Array instances
# a[0] = 1 instead of return 1 if amount == 0
memo = Array.new(COINS.size){ Array.new(amount).tap{ |a| a[0] = 1 } }
num_ways(amount, 0, memo)
end
def num_ways(amount, index, memo)
return 0 if index >= COINS.size || amount < 0
memo[index][amount] ||=
num_ways(amount - COINS[index], index, memo) +
num_ways(amount, index + 1, memo)
end
P.S. dynamic-programming tag is unnecessary :)
Since identical arrays have different object ids, you couldn't just put the values as an array in a hash key. However you can convert to string to make a simple lookup structure, for example converting to YAML/JSON
require 'json'
memo = {}
def memoize(*args, &blk)
memo[args.to_json] ||= blk.call
end
def num_ways(amount, index)
memoize(amount, index) do
# original method body
end
end
edit
nevermind, I guess I am wrong about the need to use to_json. You could just use args inside memoize and not args.to_json.
Related
I am solving the Permutational Primes Kata using Ruby. I manage to find out Brute-force solution but it exceeds the time limits. I need to optimize my code but I don't have any idea how to do this. The Kata.
require 'prime'
def permutational_primes(n_max, k_perms)
result_h = {}
result_keys = []
Prime.each(n_max) do |prime|
perms = prime.digits.permutation.to_a.map(&:join).map(&:to_i).uniq
prime_no_length = prime.to_s.length
perms = perms.delete_if { |el| el.to_s.length < prime_no_length }
# elimianate number greater than n_max
perms = perms.delete_if { |el| el > n_max }
next if (perms & result_keys).any?
perms = perms.delete_if { |el| !Prime.prime?(el) }
# minus one because we include
if perms.count - 1 == k_perms
result_h[prime] = perms
result_keys.append(prime)
end
end
return [0, 0, 0] if result_keys.empty?
[result_h.count, result_keys[0], result_keys[result_keys.count-1]]
end
Some low-hanging fruit:
Benchmarking is one way to understand what parts are more computationally expensive than others (likely Prime.prime?)
Find work that you're doing repeatedly or checks for information that you already know every Prime.each loop and cache it, trading memory for computation, e.g. Prime.prime?
All the common Enumerable methods create new Arrays. Instead, reuse existing arrays, e.g. map!
I have not yet thoroughly tested the following, but this general approach should speed things up considerably. There are two main elements that improved efficiency: use of the method Prime::each and maintaining a set of prime permutations already tested, to avoid unnecessary testing of duplicates.
require 'prime'
require 'set'
def permutational_primes(nMax, permutes)
skips = Set.new
enum = Prime.each
arr = []
while (prime = enum.next) < nMax
next if skips.include?(prime)
a = prime.digits.permutation.with_object([]) do |perm,a|
next if perm[-1].zero?
n = perm.reverse.join.to_i
if n < nMax && !skips.include?(n) && Prime.prime?(n)
skips << n
a << n
end
end
a.each { |n| skips << n }
next if a.size != permutes + 1
arr << a.min
end
[arr.size, arr.size.zero? ? 0 : arr.min, arr.size.zero? ? 0 : arr.max]
end
permutational_primes(1000, 3)
#=> [3, 149, 379]
This passes all tests but times out. I'm working on another optimization.
Explanation is under construction...
Working on the following algorithm:
Given an array of non-negative integers, you are initially positioned
at the first index of the array.
Each element in the array represents your maximum jump length at that
position.
Determine if you are able to reach the last index.
For example:
A = [2,3,1,1,4], return true.
A = [3,2,1,0,4], return false.
Below is my solution. It tries every single potential step, and then memoizes accordingly. So if the first element is three, the code takes three steps, two steps, and one step, and launches three separate functions from there. I then memoized with a hash. My issue is that the code works perfectly fine, but it's timing out for very large inputs. Memoizing helped, but only a little bit. Am I memoizing correctly or is backtracking the wrong approach here?
def can_jump(nums)
#memo = {}
avail?(nums, 0)
end
def avail?(nums, index)
return true if nums.nil? || nums.empty? || nums.length == 1 || index >= nums.length - 1
current = nums[index]
true_count = 0
until current == 0 #try every jump from the val to 1
#memo[index + current] ||= avail?(nums, index + current)
true_count +=1 if #memo[index + current] == true
current -= 1
end
true_count > 0
end
Here's a 𝑂(𝑛) algorithm:
Initialize 𝑚𝑎𝑥 to 0.
For each number 𝑛𝑖 in 𝑁:
If 𝑖 is greater than 𝑚𝑎𝑥, neither 𝑛𝑖 nor any subsequent number can be reached, so
return false.
If 𝑛𝑖+𝑖 is greater than 𝑚𝑎𝑥, set 𝑚𝑎𝑥 to 𝑛𝑖+𝑖.
If 𝑚𝑎𝑥 is greater than or equal to the last index in 𝑁
return true.
Otherwise return false.
Here's a Ruby implementation:
def can_jump(nums)
max_reach = 0
nums.each_with_index do |num, idx|
return false if idx > max_reach
max_reach = [idx+num, max_reach].max
end
max_reach >= nums.size - 1
end
p can_jump([2,3,1,1,4]) # => true
p can_jump([3,2,1,0,4]) # => false
See it on repl.it: https://repl.it/FvlV/1
Your code is O(n^2), but you can produce the result in O(n) time and O(1) space. The idea is to work backwards through the array keeping the minimum index found so far from which you can reach index n-1.
Something like this:
def can_jump(nums)
min_index = nums.length - 1
for i in (nums.length - 2).downto(0)
if nums[i] + i >= min_index
min_index = i
end
end
min_index == 0
end
print can_jump([2, 3, 1, 1, 4]), "\n"
print can_jump([3, 2, 1, 0, 4]), "\n"
This question was asked somewhere else, but I just wanted to check if what I did was applicable given the rspec circumstances:
Write a method that takes two sorted arrays and produces the sorted array that combines both.
Restrictions:
Do not call sort anywhere.
Do not in any way modify the two arrays given to you.
Do not circumvent (2) by cloning or duplicating the two arrays, only to modify the copies.
Hint: you will probably need indices into the two arrays.
combine_arrays([1, 3, 5], [2, 4, 6]) == [1, 2, 3, 4, 5, 6]
Can you just combine the two arrays into a single array and then run a typical bubble sort?
def combine_arrays(arr1,arr2)
final = arr1 + arr2
sorted = true
while sorted do
sorted = false
(0..final.length - 2).each do |x|
if final[x] > final[x+1]
final[x], final[x+1] = final[x+1], final[x]
sorted = true
end
end
end
final
end
p combine_arrays([1,3,5],[2,4,6]) => [1, 2, 3, 4, 5, 6]
Here is a variant which relies solely on Ruby's enumerators. The result is short and sweet.
# merge two sorted arrays into a sorted combined array
def merge(a1, a2)
[].tap do |combined|
e1, e2 = a1.each, a2.each
# The following three loops terminate appropriately because
# StopIteration acts as a break for Kernel#loop.
# First, keep appending smaller element until one of
# the enumerators run out of data
loop { combined << (e1.peek <= e2.peek ? e1 : e2).next }
# At this point, one of these enumerators is "empty" and will
# break immediately. The other appends all remaining data.
loop { combined << e1.next }
loop { combined << e2.next }
end
end
The first loop keeps grabbing the minimum of the two enumerator values until one of the enumerators runs out of values. The second loop then appends all remaining (which may be none) values from the first array's enumerator, the third loop does the same for the second array's enumerator, and tap hands back the resulting array.
Sure, you can do that but you are overlooking a real gimmee - the two arrays you are given will already be sorted.
def combine_arrays(A1, A2)
retVal = Array.CreateInstance(System::Int32, A1.Length + A2.Length - 1)
i = 0
j = 0
while i < A1.Length | j < A2.Length
if i < A1.Length and self.A1(i) < self.A2(j) then
self.retVal(i + j) = self.A1(i)
i += 1
else
self.retVal(i + j) = self.A2(j)
j += 1
end
end
return retVal
end
This is based on the same logic as Dale M's post, but in proper ruby:
def combine_arrays(arr1,arr2)
[].tap do |out|
i1 = i2 = 0
while i1 < arr1.size || i2 < arr2.size
v1 = arr1[i1]
v2 = arr2[i2]
if v1 && (!v2 || v1 < v2)
out << v1
i1 += 1
else
out << v2
i2 += 1
end
end
end
end
combine_arrays([1,3,5], [2,4,6])
Take a look at this one:
def merge(arr1, arr2)
arr2.each { |n| arr1 = insert_into_place(arr1, n) }
arr1.empty? ? arr2 : arr1
end
def insert_into_place(array, number)
return [] if array.empty?
group = array.group_by { |n| n >= number }
bigger = group[true]
smaller = group[false]
if bigger.nil?
number > smaller.last ? smaller << number : smaller.unshift(number)
else
(smaller << number) + bigger
end
end
So this code will count the total number of pairs of numbers whose difference is K. it is naive method and I need to optimize it. suggestions?
test = $stdin.readlines
input = test[0].split(" ")
numbers = test[1].split(" ")
N = input[0]
K = input[1]
count = 0
for i in numbers
current = i.to_i
numbers.shift
for j in numbers
difference = (j.to_i - current).abs
if (difference == K)
count += 1
end
end
end
puts count
Would have been nice for you to give some examples of input and output, but I think this is correct.
require 'set'
def count_diff(numbers, difference)
set = Set.new numbers
set.inject 0 do |count, num|
set.include?(num+difference) ? count+1 : count
end
end
difference = gets.split[1].to_i
numbers = gets.split.map { |num| num.to_i }
puts count_diff(numbers, difference)
Untested, hopefully actual Ruby code
Documentation for Set: http://www.ruby-doc.org/stdlib/libdoc/set/rdoc/classes/Set.html
require 'set'
numbers_set = Set.new
npairs = 0
numbers.each do |number|
if numbers_set.include?(number + K)
npairs += 1
end
if numbers_set.include?(number - K)
npairs += 1
end
numbers_set.add(number)
end
Someone deleted his post, or his post was deleted... He had the best solution, here it is :
test = $stdin.readlines
input = test[0].split(" ")
numbers = test[1].split(" ")
K = input[1]
count = 0
numbers.combination(2){|couple| couple.inject(:-).abs == K ? count++}
puts count
You don't even need N.
I do not know Ruby so I'll just give you the big idea:
Get the list
Keep a boolean array (call it arr), marking off numbers as true if the number exists in the list
Loop through the list and see if arr[num-K] and/or arr[num+K] is true where num is a number in your list
This uses up quite a bit of memory though so another method is to do the following:
Keep a hash map from an integer n to an integer count
Go through your list, adding num+K and num-K to the hash map, incrementing count accordingly
Go through your list and see if num is in the hash map. If it is, increment your counter by count
I am learning ruby and practicing it by solving problems from Project Euler.
This is my solution for problem 12.
# Project Euler problem: 12
# What is the value of the first triangle number to have over five hundred divisors?
require 'prime'
triangle_number = ->(num){ (num *(num + 1)) / 2 }
factor_count = ->(num) do
prime_fac = Prime.prime_division(num)
exponents = prime_fac.collect { |item| item.last + 1 }
fac_count = exponents.inject(:*)
end
n = 2
loop do
tn = triangle_number.(n)
if factor_count.(tn) >= 500
puts tn
break
end
n += 1
end
Any improvements that can be made to this piece of code?
As others have stated, Rubyists will use methods or blocks way more than lambdas.
Ruby's Enumerable is a very powerful mixin, so I feel it pays here to build an enumerable in a similar way as Prime. So:
require 'prime'
class Triangular
class << self
include Enumerable
def each
sum = 0
1.upto(Float::INFINITY) do |i|
yield sum += i
end
end
end
end
This is very versatile. Just checking it works:
Triangular.first(4) # => [1, 3, 7, 10]
Good. Now you can use it to solve your problem:
def factor_count(num)
prime_fac = Prime.prime_division(num)
exponents = prime_fac.collect { |item| item.last + 1 }
exponents.inject(1, :*)
end
Triangular.find{|t| factor_count(t) >= 500} # => 76576500
Notes:
Float::INFINITY is new to 1.9.2. Either use 1.0/0, require 'backports' or do a loop if using an earlier version.
The each could be improved by first checking that a block is passed; you'll often see things like:
def each
return to_enum __method__ unless block_given?
# ...
Rather than solve the problem in one go, looking at the individual parts of the problem might help you understand ruby a bit better.
The first part is finding out what the triangle number would be. Since this uses sequence of natural numbers, you can represent this using a range in ruby. Here's an example:
(1..10).to_a => [1,2,3,4,5,6,7,8,9,10]
An array in ruby is considered an enumerable, and ruby provides lots of ways to enumerate over data. Using this notion you can iterate over this array using the each method and pass a block that sums the numbers.
sum = 0
(1..10).each do |x|
sum += x
end
sum => 55
This can also be done using another enumerable method known as inject that will pass what is returned from the previous element to the current element. Using this, you can get the sum in one line. In this example I use 1.upto(10), which will functionally work the same as (1..10).
1.upto(10).inject(0) {|sum, x| sum + x} => 55
Stepping through this, the first time this is called, sum = 0, x = 1, so (sum + x) = 1. Then it passes this to the next element and so sum = 1, x = 2, (sum + x) = 3. Next sum = 3, x = 3, (sum + x) = 6. sum = 6, x = 4, (sum + x) = 10. Etc etc.
That's just the first step of this problem. If you want to learn the language in this way, you should approach each part of the problem and learn what is appropriate to learn for that part, rather than tackling the entire problem.
REFACTORED SOLUTION (though not efficient at all)
def factors(n)
(1..n).select{|x| n % x == 0}
end
def triangle(n)
(n * (n + 1)) / 2
end
n = 2
until factors(triangle(n)).size >= 500
puts n
n += 1
end
puts triangle(n)
It looks like you are coming from writing Ocaml, or another functional language. In Ruby, you would want to use more def to define your methods. Ruby is about staying clean. But that might also be a personal preference.
And rather than a loop do you could while (faction_count(traingle_number(n)) < 500) do but for some that might be too much for one line.