Here's the prompt I've been given:
Catsylvanian money is a strange thing: they have a coin for every
denomination (including zero!). A wonky change machine in
Catsylvania takes any coin of value N and returns 3 new coins,
valued at N/2, N/3 and N/4 (rounding down).
Write a method wonky_coins(n) that returns the number of coins you
are left with if you take all non-zero coins and keep feeding them
back into the machine until you are left with only zero-value coins.
Difficulty: 3/5
describe "#wonky_coins" do
it "handles a simple case" do
wonky_coins(1).should == 3
end
it "handles a larger case" do
wonky_coins(5).should == 11
# 11
# => [2, 1, 1]
# => [[1, 0, 0], [0, 0, 0], [0, 0, 0]]
# => [[[0, 0, 0], 0, 0], [0, 0, 0], [0, 0, 0]]
end
it "handles being given the zero coin" do
wonky_coins(0).should == 1
end
end
Maybe it's because of the tests given that involve arrays, but I couldn't get my mind off of them! So my solution so far is as follows:
def wonky_coins(n)
arr = []
arr << n/2 << n/3 << n/4
#base case?
if arr.all?{|coin| coin == 0}
return arr.flatten.length
else
arr.map{|x| wonky_coins(x)}
end
end
p wonky_coins(5)
Except I get [[3,3,3],3,3] as an output if I map it. It's not actually recurring, but even before that, it's giving a strange output that I can't for the life of me understand why the output is this way!
I know it's because I'm using the map method, is it because I'm mutating it while iterating it through wonky_coins again that I'm getting this strange output I can't explain?
I've since looked at the solution and realized that arrays made it needlessly complicated, but I'm still wondering what's going on here??
Here's what Seeing is Believing shows as the code runs:
def wonky_coins(n)
arr = [] # => [], [], [], [], [], [], []
arr << n/2 << n/3 << n/4 # => [2, 1, 1], [1, 0, 0], [0, 0, 0], [0, 0, 0], [0, 0, 0], [0, 0, 0], [0, 0, 0]
#base case?
if arr.all?{|coin| coin == 0} # => false, false, true, true, true, true, true
return arr.flatten.length # => 3, 3, 3, 3, 3
else
arr.map{|x| wonky_coins(x)} # => [3, 3, 3], [[3, 3, 3], 3, 3]
end
end
p wonky_coins(5) # => [[3, 3, 3], 3, 3]
# >> [[3, 3, 3], 3, 3]
Seeing is Believing is a great tool and can help dig out weirdness in code.
Related
I'm working on the following algorithm from Leetcode:
Numbers can be regarded as product of its factors. For example,
8 = 2 x 2 x 2;
= 2 x 4.
Write a function that takes an integer n and return all possible combinations of its factors.
Note:
You may assume that n is always positive.
Factors should be greater than 1 and less than n.
Examples:
input: 1
output:
[]
input: 37
output:
[]
input: 12
output:
[
[2, 6],
[2, 2, 3],
[3, 4]
]
input: 32
output:
[
[2, 16],
[2, 2, 8],
[2, 2, 2, 4],
[2, 2, 2, 2, 2],
[2, 4, 4],
[4, 8]
]
Here's the code that I have thus far:
def get_factors(n)
factors = []
(2...n).each do |candidate|
if n % candidate == 0
factors << [candidate, (n/candidate)]
get_factors(n/candidate).each do |factor_set|
factors << [candidate] + factor_set
end
end
end
factors
end
This code works really well, but doesn't handle duplicates (e.g [3, 2, 2] will be inserted along with [2, 2, 3]). I tried using a Set with the following code,
def get_factors(n)
seen = Set.new
factors = []
(2...n).each do |candidate|
if n % candidate == 0 && !seen.include?(candidate)
factors << [candidate, (n/candidate)]
get_factors(n/candidate).each do |factor_set|
factors << [candidate] + factor_set
end
end
seen << (n/candidate)
end
factors
end
but that only works to solve some test cases and not others. I'm not sure how to go about ensuring no duplicates in an efficient way? The really inefficient way is to generate some sort of hash value for each array depending on it's elements (and not dependent on order), and while this would work, there definitely should be a better way. Any ideas?
I think always going forward is a good policy (i.e when checking, say, with 5, do not check with 2, 3, 4 etc). That way, searching for duplicates can be eliminated.
Since the algorithm already uses a lot of time, I don't see a problem sorting each answer and removing duplicates. This requires no proof to ensure it works, which the answer provided by mac does.
Code
require 'prime'
def get_factors(n)
primes, nbr = Prime.prime_division(n).transpose
powers = nbr.map { |m| (0..m).to_a }
powers.shift.
product(*powers).
map { |pows| primes.zip(pows).reduce(1) { |t,(pr,po)| t * (pr**po) } }.
sort
end
The array returned includes 1 and n (which are factors of n). If those values should be excluded, replace .sort with .sort - [1, n].
Examples
get_factors(24)
#=> [1, 2, 3, 4, 6, 8, 12, 24]
get_factors(64)
#=> [1, 2, 4, 8, 16, 32, 64]
get_factors(90)
#=> [1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90]
Explanation
Consider
n = 60
The steps are as follows.
a = Prime.prime_division(30)
#=> [[2, 2], [3, 1], [5, 1]]
Ergo, the primes of 30 are 2, 3 and 5, and
60 = 2**2 * 3**1 * 5**1
See Prime::prime_division. Continuing,
primes, nbr = a.transpose
#=> [[2, 3, 5], [2, 1, 1]]
primes
#=> [2, 3, 5]
nbr
#=> [2, 1, 1]
powers = nbr.map { |m| (0..m).to_a }
#=> [[0, 1, 2], [0, 1], [0, 1]]
This means that each factor will be the product of 0, 1 or 2 2's, 0 or 1 3's and 0 or 1 5's.
b = powers.shift
#=> [0, 1, 2]
powers
#=> [[0, 1], [0, 1]]
c = b.product(*powers)
#=> [[0, 0, 0], [0, 0, 1], [0, 1, 0], [0, 1, 1], [1, 0, 0], [1, 0, 1],
# [1, 1, 0], [1, 1, 1], [2, 0, 0], [2, 0, 1], [2, 1, 0], [2, 1, 1]]
d = c.map { |pows| primes.zip(pows).reduce(1) { |t,(pr,po)| t * (pr**po) } }
#=> [1, 5, 3, 15, 2, 10, 6, 30, 4, 20, 12, 60]
d.sort
#=> [1, 5, 3, 15, 2, 10, 6, 30, 4, 20, 12, 60]
Consider now the calculation of d. The 10th element of c that is passed to the block is [2, 0, 1]. The block calculation for that element is as follows.
pows = [2, 0, 1]
e = primes.zip(pows)
#=> [[2, 2], [3, 0], [5, 1]]
e.reduce(1) { |t,(pr,po)| t * (pr**po) }
#=> 20
The reduce calculation is equivalent to
2**2 * 3**0 * 5**1
#=> 4 * 1 * 5 => 20
The calculations for the other values of c passed to the block are similar.
A simple way is to replace the last line of your method with
factors.map(&:sort).uniq
which sorts all the subarrays and then eliminates duplicates.
def reverse_append(arr, n)
return arr if n < 0
reverse_append(arr, n-1)
arr << n
arr
end
reverse_append([],4) #=> [0, 1, 2, 3, 4]
I can't seem to understand this recursive method. It produces an array from 0 up to n.
Can someone explain this to me?
The method reverse_append([],4) is called
Since 4 >= 0, the return statement does not get called.
The method reverse_append([],3) is called.
Since 3 >= 0, the return statement does not get called.
The method reverse_append([],2) is called.
Since 2 >= 0, the return statement does not get called.
The method reverse_append([],1) is called.
Since 1 >= 0, the return statement does not get called.
The method reverse_append([],0) is called.
Since 0 >= 0, the return statement does not get called.
The method reverse_append([],-1) is called.
Since -1 < 0, the array ([]) is returned.
We pop up one level in our call stack, to where n = 0 and arr = [].
arr << n and arr is returned, so now arr = [0].
We pop up one level in our call stack, to where n = 1 and arr = [0].
arr << n and arr is returned, so now arr = [0, 1].
We pop up one level in our call stack, to where n = 2 and arr = [0, 1].
arr << n and arr is returned, so now arr = [0, 1, 2].
We pop up one level in our call stack, to where n = 3 and arr = [0, 1, 2].
arr << n and arr is returned, so now arr = [0, 1, 2, 3].
We pop up one level in our call stack, to where n = 4 and arr = [0, 1, 2, 3].
arr << n and arr is returned, so now arr = [0, 1, 2, 3, 4].
Finally, the "top-level" method returns, and we have our final result.
Well step through the code with the supplied parameters. The first step is to check if n < 0 which its not. If it isn't 0 reverse append with [], 3 and appends the that array the number and then returns the array.
So it takes the array, adds 4 to it after it has gone through the step of dealing with [], 3, [], 2, [],1 and [], 0. So the first call that will succeed is just returning the array when it gets below 0, next is 0 gets appended, then one, then 2, then 3 and lastly the original call with 4 gets added arr << n.
There's a nice tool you can add to many editors called "Seeing Is Believing", which lets you see what is happening as code runs:
def reverse_append(arr, n)
return arr if n < 0 # => false, false, false, false, true
reverse_append(arr, n-1) # => [], [0], [0, 1], [0, 1, 2]
arr << n # => [0], [0, 1], [0, 1, 2], [0, 1, 2, 3]
arr # => [0], [0, 1], [0, 1, 2], [0, 1, 2, 3]
end
reverse_append([], 3) # => [0, 1, 2, 3]
However, with a name like "reverse_append" it seems like you should see a result that is descending in values:
def reverse_append(arr, n)
return arr if n < 0 # => false, false, false, false, true
reverse_append(arr, n-1) # => [], [0], [1, 0], [2, 1, 0]
arr.unshift n # => [0], [1, 0], [2, 1, 0], [3, 2, 1, 0]
arr # => [0], [1, 0], [2, 1, 0], [3, 2, 1, 0]
end
reverse_append([], 3) # => [3, 2, 1, 0]
In either case, there are a lot of easier ways to generate such an array without relying on recursion:
[*0..3] # => [0, 1, 2, 3]
(0..3).to_a # => [0, 1, 2, 3]
[*0..3].reverse # => [3, 2, 1, 0]
(0..3).to_a.reverse # => [3, 2, 1, 0]
I have the following code I wrote which takes a 2d array with arrays of varying length and appends the value 0 to every index until all arrays are of the same length.
Can this code be written shorter or more efficient and modular?
a = [[1,7],[2,3],[5,1,2],[3],[1],[]]
l = a.map(&:length).max
a2 = a.each{|e| e.push(Array.new(l - e.length, 0) )}
a2.each{|e| e.flatten!}
#=> [[1, 7, 0], [2, 3, 0], [5, 1, 2], [3, 0, 0], [1, 0, 0], [0, 0, 0]]
UPDATE
Because I feel unable to award an accepted answer to the two answers given thus far based on the simplicity of the written code itself, I have decided to award it based on the efficiency of the running speed of the code.
100 test cases running code block 10000 times.
#My Code:
#average run time of test case -> 0.24267sec
#standard deviation -> 0.00735sec
#Stefan’s Code:
#average run time of test case -> 0.06389sec
#standard deviation -> 0.00756sec
#steenslag’s Code:
#average run time of test case -> 0.0577sec
#standard deviation -> 0.00413sec
tests were conducted under the same conditions using a custom written ruby timer class I made and are only relative to each other and my crappy 2010-macbookpro on which I ran them.
Another option using Array#fill:
l = a.map(&:length).max
a.each { |e| e.fill(0, e.length...l) }
a #=> [[1, 7, 0], [2, 3, 0], [5, 1, 2], [3, 0, 0], [1, 0, 0], [0, 0, 0]]
The flatten could be avoided by splatting the array:
l = a.max_by(&:size).size
a2 = a.each{|e| e.push(*Array.new(l - e.length, 0) )}
One more way, four days late:
len = a.map(&:size).max
a.map { |e| Array.new(len) { |i| e[i].to_i } }
#=> [[1,7,0],[2,3,0],[5,1,2],[3,0,0],[1,0,0],[0,0,0]]
I need to get all the possible number combinations from denom_arr which equal the amt.
denom_arr = [4,3,1]
amt = 10
This case would produce:
[4, 4, 1, 1]
[3, 3, 3, 1]
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1]
[4, 3, 1, 1, 1]
[4, 3, 3]
. . . (other cases...)
Problem is the code I wrote is breaking after 1-3 and I'm not sure how to make it loop over the same index to get case 4-6+
set, sets = [], []
i = 0
loop do
i = 0 if denom_arr[i].nil?
loop do
set << denom_arr[i]
break if set.inject(:+) > amt
end
set.pop if set.inject(:+) > amt
if set.inject(:+) == amt
sets << set
set = []
denom_arr.shift
end
i += 1
sets
break if denom_arr.empty?
end
UPDATE
I know this can be done with recursion with memoization/dynamic programming techniques, but I am trying to do this strictly in a loop for the sake of testing a theory.
I would do this recursively
def possible_sums(arr, amt)
return [[]] if amt == 0
return [] if amt < 0
arr.reduce([]) do |sums, e|
sums.concat(
possible_sums(arr, amt-e)
.map { |sum| sum.unshift(e).sort }
)
end.uniq
end
p possible_sums([4,3,1], 10)
# => [
# [1, 1, 4, 4], [3, 3, 4], [1, 1, 1, 3, 4], [1, 1, 1, 1, 1, 1, 4],
# [1, 3, 3, 3], [1, 1, 1, 1, 3, 3], [1, 1, 1, 1, 1, 1, 1, 3],
# [1, 1, 1, 1, 1, 1, 1, 1, 1, 1]
# ]
Although this is potentially inefficient in that it repeats work, this can be alleviated by using dynamic programming (essentially, memoizing the results of the recursive function).
UPDATE Here is an iterative solution:
def possible_sums_it(arr, amt)
sums = Array.new(amt+1) { [] }
sums[0] << []
(1..amt).each do |i|
arr.each do |e|
if i-e >= 0
sums[i].concat(
sums[i-e].map { |s| [e, *s].sort }
)
end
end
sums[i].uniq!
end
sums[amt]
end
This is in fact the dynamic programming algorithm for the problem.
So if you squint at it just right, you'll see that essentially what it is doing, is calculating all the possible sums for 0 up to amt into the sums array, using what is basically the recursive algorithm, but instead of the recursive call, we lookup a value in sums that we have calculated beforehand.
This works because we know that we won't need sums[i] before sums[j] for j < i.
I want to program a counter which is represented by an array of numbers, starting with:
[0, 0, 0]
The constraint here is, that each position has a different cap, so it's not necessarily 9 or something else, but it is given. For instance:
[4, 2, 1]
Which would lead to the following incrementation sequence:
[0, 0, 0]
[0, 0, 1]
[0, 1, 0]
[0, 1, 1]
[0, 2, 0]
[0, 2, 1]
[1, 0, 0]
.
.
.
Of course I can think of a solution using modulo and adding each carryover onto the next position. But has someone an idea how to implement this efficiently, respectively with nice Ruby syntax without cluttering it too much?
That is my naive implementation:
max = [10, 1, 1, 1, 10]
counter = [0, 0, 0, 0, 0]
i = counter.length-1
while counter != max do
counter[i] = counter[i] + 1
while counter[i] > max[i]
counter[i] = 0
i = i - 1
counter[i] = counter[i] + 1
end
i = counter.length-1
end
I'm not sure about efficiency but here's my shot at it:
start = [0, 0, 0]
cap = [4, 2, 1]
start.zip(cap).map{ |i, c| (i..c).to_a }.reduce(&:product).map &:flatten
Produces something like:
[[0, 0, 0],
[0, 0, 1],
[0, 1, 0],
[0, 1, 1],
[0, 2, 0],
[0, 2, 1],
[1, 0, 0],
[1, 0, 1],
[1, 1, 0],
[1, 1, 1],
[1, 2, 0],
[1, 2, 1],
[2, 0, 0],
[2, 0, 1]...]
Edit: I was writing this before you made your edit. It seemed like you wanted a counter object, not just to output a list.
1) I would recommend specifying not the limits but (limit+1) of each of the digits. For example, for a [second, minute, hour, day, year] counter it makes more sense (to me) to write [60, 60, 24, 365] instead of [59,59,23,364].
2) You'll have to figure out what to do if your counter overflows the last limit of your array. I added an extra position that counts to infinity.
3) I would also recommend reversing the order of the array, at least in the internal representation to avoid inverting subscripts. If you don't want it like that, you can .reverse the bases in initialize and #digits in to_s
class MyCounter
def initialize bases
#bases = bases
#bases << 1.0/0 # Infinity
#digits = Array.new(bases.size, 0)
prod = 1
#digit_values = [1] + #bases[0..-2].map { |b| prod *= b }
end
attr_reader :digit_values
def to_s
#digits
end
def increment(digit=0)
v = #digits[digit] + 1
if v < #bases[digit]
#digits[digit] = v
else
#digits[digit] = 0
increment(digit+1)
end
self
end
def +(integer)
(#digits.size - 1).step(0,-1).each do |i|
#digits[i] += integer / #digit_values[i]
integer = integer % #digit_values[i]
end
self
end
end
c1 = MyCounter.new [2,3,5]
20.times { c1.increment; p c1 }
c2 = MyCounter.new [2,3,5]
c2 += 20
p c2
Create an array for each cap, with values from 0 upto cap. Take the first array and calculate the Cartesian product with the rest of the arrays.
caps = [4, 2, 1]
arrs = caps.map{|cap| (0..cap).to_a} #=>[[0, 1, 2, 3, 4], [0, 1, 2], [0, 1]]
p arrs.shift.product(*arrs)
# =>[[0, 0, 0], [0, 0, 1], [0, 1, 0], [0, 1, 1], [0, 2, 0], [0, 2, 1], ...
If you don't want a memory-consuming array with the results, then provide a block. product will yield each element to it, one by one.
arrs = caps.map{|cap| (0..cap).to_a}
arrs.shift.product(*arrs){|el| puts el.join} #no resulting array
#000
#001
#010
#011
#...