Here is a task i got:
Cards are laid out on the table in a row, each card has a natural number written on it. In one move, it is allowed to take a card either from the left or from the right end of the row. In total, you can make k moves. The final score is equal to the sum of the numbers on the selected cards. Determine what is the maximum score you can get at the end of the game.
Here`s my code:
def card_counter(arr, k):
if len(arr) == k:
return sum(arr)
rang = len(arr) // 2
left = arr[:rang]
right = list(reversed(arr[rang:]))
c = 0
for _ in range(k):
min_arr = left if sum(left) >= sum(
right) and len(left) > 0 else right
c += min_arr.pop(0)
return c
if __name__ == '__main__':
assert card_counter([1, 2, 3, 4, 5], 5) == 15
assert card_counter([0, 0, 0], 1) == 0
assert card_counter([150], 1) == 150
This code works on all variants that I have come up with, including extreme cases. But the system does not accept this option, automatic tests do not pass. Where can there be a mistake?
I cannot comment on the algorithm you implemented as you did not state it in non-algorithm terms and I am not familiar with the language you are using (which I am guessing is Python). I will present a simple solution written in Ruby, hoping that the description I give will make it understood by readers who do not know Ruby.
Suppose
deck = [1, 2, 5, 7, 1, 4, 6, 3]
nbr_moves = 4
ds = deck.size
#=> 8
After removing nbr_moves from the ends,
m = ds - nbr_moves
#=> 4
consecutive cards will remain. In Ruby we could write
best = (0..ds-1).each_cons(m).max_by { |arr| deck.values_at(*arr).sum }
#=> [3, 4, 5, 6]
to obtain the indices of deck, for which the sum of the associated values is maximum:
deck.values_at(*best).sum
#=> 18
where
deck.values_at(*best)
#=> [7, 1, 4, 6]
In view of the value of best ([3, 4, 5, 6]), we need to remove left = best.first #=> 3 elements from the left and nbr_moves - left #=> 1 element from the right. The order of the removals is not relevant.
Note that
enum = (0..ds-1).each_cons(m)
#=> #<Enumerator: 0..7:each_cons(4)>
returns an enumerator. We can convert this enumerator to an array to see the values it will generate.
enum.to_a
#=> [[0, 1, 2, 3], [1, 2, 3, 4], [2, 3, 4, 5], [3, 4, 5, 6], [4, 5, 6, 7]]
When, for example,
arr = [2, 3, 4, 5]
then
a = deck.values_at(*arr)
#=> [5, 7, 1, 4]
a.sum
#=> 17
Note that, having computed, for example,
t = [deck[i], deck[i+1],..., deck[j]].sum
the sum of
[deck[i+1], deck[i+2],..., deck[j+1]]
is seen to equal
t - deck[i] + deck[j+1]
This suggests a more efficient way to perform the calculations when m is large.
Related
I have a hash whose keys are a range of integers (lets say [1..5]) and its corresponding 5 values are all nil. I have also an array of integers (lets say [1,2,3,4,5]. What I want to do is very specific: I want to take every single key and add it to every single of the array elements, giving me a hash that has the original keys, but has now for values the entire shifted array.
After spending a few hours I have concluded that this is impossible through a really laconic expression, because it is leading to .each shadowing statements.
I think that the only way to go through with this is to create 5 almost identical methods and call them separately.
def a1
array.each do |x|
x+1
end
end
def a2
array.each do |x|
x+2
end
end
and so on..
The end product I want to achieve is this:
{1=>[2,3,4,5,6],2=>[3,4,5,6,7],3=>[4,5,6,7,8],4=>[5,6,7,8,9],5=>[6,7,8,9,10]}
It feels like there should be a more DRY way to achieve this. Any ideas?
Assuming these initial conditions:
h = {1=>nil, 2=>nil, 3=>nil, 4=>nil, 5=>nil}
arr = [1,2,3,4,5]
...it's pretty straightforward:
h.keys.each do |key|
h[key] = arr.map {|i| i+key}
end
# h is now: {1=>[2, 3, 4, 5, 6], 2=>[3, 4, 5, 6, 7], 3=>[4, 5, 6, 7, 8], 4=>[5, 6, 7, 8, 9], 5=>[6, 7, 8, 9, 10]}
(However, it may be that your question is about achieving the initial conditions. If so, I didn't grasp that, and I didn't worry about it; I just started with what I took to be your initial conditions and ended up with your desired result.)
Why don't you do this
h = {}
rng.each{|i| h[i] = ary.map{|j| j + i}}
That should work where rng is the range and ary is the array.
For example
h = {}
(1..5).each{|i| h[i] = [1,2,3,4,5].map{|j| j+i}}
results in
h = {1=>[2, 3, 4, 5, 6], 2=>[3, 4, 5, 6, 7], 3=>[4, 5, 6, 7, 8], 4=>[5, 6, 7, 8, 9], 5=>[6, 7, 8, 9, 10]}
I am trying to output an array of 10-element arrays containing permutations of 1 and 2, e.g.:
[[1,1,1,1,1,1,1,1,1,1],
[1,1,1,1,1,1,1,1,1,2],
[1,2,1,2,1,2,1,2,1,2],
...etc...
[2,2,2,2,2,2,2,2,2,2]]
I have done this with smaller arrays, but with (10):
a = [1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2,2]
a = a.permutation(10).to_a
print a.uniq
...this apparently is too big a calculation- after an hour running, it hadn't completed and the ruby process was sitting on 12GB of memory. Is there another way to come at this?
First, check size of that permutation
a.permutation(10).size
=> 670442572800
It's huge. What you can do instead is to use Array#repeated_permutations on smaller array. Check it here:
b = [1, 2] # Only unique elements
b.repeated_permutation(10) # Returns enumerable
b.repeated_permutation(10).to_a # Creates array with all of permutations
Those permutations are already unique (which u can check by printing it size with and without Array#uniq )
The approach you've gone down does indeed generate a lot of data - 20!/10! or about 600 billion arrays. This is clearly very wasteful when the output you are after only has 1024 arrays
what you are after is closer to the product method on array
[1,2].product(*([[1,2]]*9))
The product method produces all the possible combinations by picking one element from each of the receiver and its arguments. The use of splats and the * method on array is just to avoid writing [1,2] 9 times.
Edit: Don't use this... it's slower by a factor (10 times slower) then running: b= [1,2]; b.repeated_permutation(10).to_a
What you're looking for is actually a binary permutation array...
So taking that approach, we know we have 2**10 (== 1024) permutations, which translated to the numbers between 0 and 1023.
try this:
1024.times.with_object([]) {|i, array| array << ( ("%10b" % (i-1) ).unpack("U*").map {|v| (v == 49) ? 2 : 1} ) }
Or this (slightly faster):
(0..1023).each.with_object([]) {|i, array| array << ( (0..9).each.with_object([]) {|j, p| p << (i[j] +1)} ) }
You take the number of options - 1024. For each option (i) you assign a number (i-1) and extract the binary code that comprises that number.
The binary code is extracted, in my example, by converting it to a string 10 digits long using "%10b" % (i-1) and then unpacking that string to an array. I map that array, replacing the values I get from the string (white space == 32 && zero == 48) with the number 1 or (the number 1 == 49) with the number 2.
Voila.
There should be a better way to extract the binary representation of the numbers, but I couldn't think of one, as I'm running ob very little sleep.
Here's another way that is similar to the approach taken by #Myst, except I've used Fixnum#to_s to convert an integer to a string representation of its binary equivalent.
There are 2**n integers with n digits (including leading zeros), when each digit equals 1 or 2 (or equals 0 or 1). We therefore can 1-1 map each integer i between 0 and 2**n-1 to one of the integers containing only digits 1 and 2 by converting 0-bits to 1 and 1-bits to 2.
In Ruby, one way to convert 123 to binary is:
123.to_s(2).to_i #=> 1111011
As
2**9 #=> 512
(2**9-1).to_s(2) #=> "111111111"
(2**9-1).to_s(2).to_i #=> 111111111
we see that when comparing numbers between 0 and 511 (2**9-1), the binary representation of 123 would have two leading zeros. As we need those leading zeros to make the conversion to 1's and 2's, it is convenient to leave the binary representation of each numbers as a string, and pad the string with zeros:
str = 123.to_s(2).rjust(9,'0') #=> "001111011"
That allows us to write:
str.each_char.map { |c| c.eql?("0") ? 1 : 2 } }
#=> [1, 1, 2, 2, 2, 2, 1, 2, 2]
We can wrap this in a method:
def combos(n)
(2**n).times.map { |i| i.to_s(2)
.rjust(n,'0')
.each_char
.map { |c| c.eql?("0") ? 1 : 2 } }
end
combos(1)
#=> [[1], [2]]
combos(2)
#=> [[1, 1], [1, 2], [2, 1], [2, 2]]
combos(3)
#=> [[1, 1, 1], [1, 1, 2], [1, 2, 1], [1, 2, 2],
# [2, 1, 1], [2, 1, 2], [2, 2, 1], [2, 2, 2]]
combos(4)
#=> [[1, 1, 1, 1], [1, 1, 1, 2], [1, 1, 2, 1], [1, 1, 2, 2],
# [1, 2, 1, 1], [1, 2, 1, 2], [1, 2, 2, 1], [1, 2, 2, 2],
# [2, 1, 1, 1], [2, 1, 1, 2], [2, 1, 2, 1], [2, 1, 2, 2],
# [2, 2, 1, 1], [2, 2, 1, 2], [2, 2, 2, 1], [2, 2, 2, 2]]
combos(10).size
#=> 1024
Simple example but I want to understand how it is done so I can apply it else where I have a main array with 6 elements. I want to take 3 of the elements from the main array and put it in a array and then take the other 3 from main array and put them in b array. I will use this to apply it to dealing cards to two players
main = [1, 2, 3, 4, 5, 6]
a = [ ]
b = [ ]
main = [1, 2, 3, 4, 5, 6]
#=> [1, 2, 3, 4, 5, 6]
main.first(3)
#=> [1, 2, 3]
main.last(3)
#=> [4, 5, 6]
a = [1, 2, 3, 4, 5, 6]
#=> [1, 2, 3, 4, 5, 6]
b = a.take(3)
#=> [1, 2, 3]
c = a.drop(3)
#=> [4, 5, 6]
All may have given the right answer, But as I understood from your question (I will use this to apply it to dealing cards to two players) When you dealing cards, as you deal cards to player main array should remove that element from self array to overcome Redundancy Problem (duplication). When you deal the all cards main array must be empty.
For this solution have a look at Array#shift
> main = [1,2,3,4,5,6] # I have 6 cards on my hand before dealing cards to players
=> [1, 2, 3, 4, 5, 6]
> a = main.shift(3) # given 3 cards to Player a
=> [1, 2, 3]
> b = main.shift(3) # given 3 cards to Player b
=> [4, 5, 6]
> main # after dealing all cards to two players I should not have any card on my hand
=> []
You have many ways to do the same thing in Ruby. Splitting arrays isn't an exception. Many answers (and comments) told you some of the ways to do that. If your program is dealing cards, you won't stop there. First, you'll probably have more than 6 cards. Second, you're probably going to have more than 2 players. Let's say the cards are C and the players are P. You need to write a method that, no matter how many Cs or Ps there are, the method is going to give each Player an equal number of Ccards (or return an error if it can't give it an equal number of cards). So for 6 cards and 2 players, it will give 3 cards each. For 12 cards and 3 players, 4 cards each. For 3 cards and 2 players, it's going to produce an error because the cards can't be evenly split:
def split_cards_evenly_between_players(cards, players)
if cards.size % players != 0
raise 'Cannot split evenly!'
else
groups_to_split_into = cards.size / players
cards.each_slice(groups_to_split_into).to_a
end
end
Let's go through the code. If the cards can't be evenly split between players, then the remainder by dividing them won't be 0 (6 cards / 3 players = remainder 0. 7 cards / 3 players = remainder 1). That's what line 2 checks. If the cards CAN be split, then we first find the groups to split into (which is dividing the number of cards by the number of players). Then we just split the array into that many groups with Enumerable#each_slice. Finally, since this doesn't produce an array, we need .to_a to convert it. The return value in Ruby is always the value of the last expression executed. The only expression in this method is the if/then expression which also returns the value of the last expression executed (which is the line where each_slice is). Let's try it out:
p split_cards_evenly_between_players([1,2,3,4,5,6,7,8,9,10,11,12],2) #=> [[1, 2, 3, 4, 5, 6], [7, 8, 9, 10, 11, 12]]
p split_cards_evenly_between_players([4,5,1,2,5,3], 3) #=> [[4, 5], [1, 2], [5, 3]]
p split_cards_evenly_between_players([1,2,3],2) #=> Error: Cannot split evenly!
The nice thing about Ruby is its simple syntax and the fact it tries to get out of your way while solving a problem so you can focus more on the actual problem than the code.
I know that ruby has myarray.sample(i) to sample i elements from an array. My problem is that the number of elements are not integers.
i.e I would like a method mysample such that if I call myarray.mysample(1.5) 10 times, the number of elements I get should be close to 15.
With sample, I will get either 10 or 20 depending on the int conversion. Similarly, if I call myarray.mysample(.25) I want it to return an element with 0.25 probability (that is, it should return one element one times out of four, and three times out of four, it should return an empty array/nil).
How do I do this?
My attempt so far:
def mysample(array,s)
ints = array.sample(s.floor)
if (Random.rand > s - s.floor)
ints << array.sample
end
return ints
end
Is there a better way?
Basing my answer off of this:
if I call myarray.mysample(1.5) 10 times, the number of elements I get should be close to 15.
Extending Array yields the following:
class Array
def mysample(num)
self.sample( ( num + rand() ).floor )
end
end
> [1, 2, 3, 4, 5].mysample(2.5)
=> [1, 3]
> [1, 2, 3, 4, 5].mysample(2.5)
=> [4, 2, 5]
> [1, 2, 3, 4, 5].mysample(0.5)
=> []
> [1, 2, 3, 4, 5].mysample(0.5)
=> [3]
etc.
To optimal argument is there to decide the spread of randomness for numbers above 1.
class Array
def my_sample(number, deviation=0.3)
if number < 1
return sample rand(100) < number * 100 ? 1 : 0
end
speard = (number*deviation).to_i
randomness = rand(-speard..speard)
sample(number+randomness)
end
end
p [1,2,3,4,5,6,7,8,9,10].my_sample(0.5) #=> []
p [1,2,3,4,5,6,7,8,9,10].my_sample(0.5) #=> [3]
p [1,2,3,4,5,6,7,8,9,10].my_sample(5) #=> [9, 2, 1, 4, 10, 7, 3]
p [1,2,3,4,5,6,7,8,9,10].my_sample(5) #=> [7, 2, 3, 8]
I have integers i.e. 9, 5, 4, 3, 1, 6, 7, 8. I want to return the index where a sequence of three descending or ascending integers exists. In the example above I would get indices 1 and 5. What is the ruby code for this?
def seq
array = [9,5,4,3,1,6,7,8]
array.each_with_index |val, index|
if (val < (array[index + 1]).val < (array[index + 1]).val)
puts "#{index}"
# Skip two indexes
end
end
I think the logic behind your solution is almost correct, but your syntax is pretty far off from valid Ruby.
Here are a pair of pretty verbose solutions that will (hopefully) be fairly obvious:
numbers = [9, 6, 5, 4, 3, 1, 6, 7, 8]
# Find non-overlapping sets
i = 0
until i > numbers.length - 2
a, b, c = numbers[i..i + 2]
if (a - b == b - c) && (a - b).abs == 1
puts "#{i} (#{a},#{b},#{c})"
# Skip next two indexes
i += 3
else
i += 1
end
end
# Find overlapping sets (same solution, but don't skip indexes)
(0...numbers.length - 2).each do |i|
a, b, c = numbers[i..i + 2]
if (a - b == b - c) && (a - b).abs == 1
puts "#{i} (#{a},#{b},#{c})"
end
end
Since the question is not clear enough. I will assume the question is about finding 3 ascending or descending continuous numbers. If the length of the satisfied sequence it longer than 3, e.g [2, 3, 4, 5], it returns 0 and 1.
Here is the algorithm, do list[index] - list[index - 1] for all elements, and repeat it for another time, the answer will be the index of 0 elements after the calculation.
Intuitively,
original 9, 5, 4, 3, 1, 6, 7, 8
first pass -4, -1, -1, -2, 5, 1, 1
2nd pass 3, 0, -1, 7, 4, 0 -> the answer will be the indexes of 0's, which is 1, 5
Algorithm:
lst = [9, 5, 4, 3, 1, 6, 7, 8]
lst1 = lst.each_cons(2).map{ |a, b| b-a }
lst2 = lst1.each_cons(2).map{ |a, b| b-a }
result = lst2.each_index.select{|i| lst2[i] == 0}
result = [1, 5]
Here’s a solution using each_cons(3).with_index:
[9,5,4,3,1,6,7,8].each_cons(3).with_index.select { |s, i| s[0] < s[1] && s[1] < s[2] }.map(&:last)