Related
This is the original link for the problem in hackerrank: https://www.hackerrank.com/challenges/the-birthday-bar/problem
I have been fighting with this problem in Ruby and I don't know why my counter always returns 1. This is the solution. I hope you can help me to understand what I'm making wrong.
s = [1, 2, 1, 3, 2]
d = 3
m = 2
def birthday(s, d, m)
array = []
cont = 0
sum = 0
m.times {array.push(s.shift)}
(m-1).times do
array.each {|i| sum = sum + i}
if sum == d
cont += 1
end
array.shift
array.push(s.shift)
end
return cont
end
birthday(s, d, m)
Though the following does not answer your question directly, it is a Ruby-like way of solving the problem, especially by making use of the methods Enumerable#each_cons and Enumerable#count.
def birthday(s, d, m)
s.each_cons(m).count { |a| a.sum == d }
end
s = [1, 2, 1, 3, 2]
d = 3
m = 2
birthday(s, d, m)
#=> 2 ([1, 2] and [2, ])
s = [2, 2, 1, 3, 2]
d = 4
m = 2
birthday(s, d, m)
#=> 2 ([2, 2] and [1, 3])
s = [2, 4, 3, 2, 1, 2, 6, 1]
d = 9
m = 3
birthday(s, d, m)
#=> 4 ([2, 4, 3], [4, 3, 2], [1, 2, 6] and [2, 6, 1])
Notice from the doc that when each_cons is used without a block it returns an enumerator:
s = [1, 2, 1, 3, 2]
d = 3
m = 2
enum = s.each_cons(m)
#=> #<Enumerator: [1, 2, 1, 3, 2]:each_cons(2)>
enum will generate elements and pass them to count until there are no more to generate, at which time it raises a StopIteration exception:
enum.next #=> [1, 2]
enum.next #=> [2, 1]
enum.next #=> [1, 3]
enum.next #=> [3, 2]
enum.next #=> StopIteration (iteration reached an end) <exception>
We can write1:
enum.count { |a| a.sum == d }
#=> 2
After enum generates the first value ([1, 2]) the block variable a is assigned its value:
a = enum.next
#=> [1, 2]
and the block calculation is performed. As
a.sum == d
#=> [1, 2].sum == 3 => true
the count is incremented (from zero) by one. enum then passes each of its remaining values to count and the process is repeated. When, for example, [1, 3].sum == 3 => false is executed, the count is not incremented.
1. Note that since I just stepped through all the elements of enum, enum.next would generate another StopIteration exception. To execute enum.count { |a| a.sum == d } I therefore must first redefine the enumerator (enum = s.each_cons(m)) or Enumerator#rewind it: enum.rewind.
I wanted to practise some algorithms... Why doesn't my solution work on leetcode website?!?!
PS: Would be grateful for other resources to learn algorithms and practise interview questions.
# #param {Integer[]} nums
# #param {Integer} target
# #return {Integer[]}
def two_sum(nums, target)
i,j = 0,nums.length-1
output = []
while i < nums.length-1
while j > i
if nums[i] + nums[j] == target
output << i << j
end
j-=1
end
i+=1
end
output
end
Result from the website:
Input:
[3,2,4]
6
Output: []
Expected:[1,2]
Now that your question has been answered, I would like to suggest a more Ruby-like method.
Code
def two_sum(nums, target)
(0...nums.size).to_a.combination(2).find { |i,j| nums[i]+nums[j] == target }
end
Example
nums = [1,5,2,3,4]
target = 8
two_sum(nums, target)
#=> [1,3]
Explanation
For the example above, the steps are as follows:
a = nums.size
#=> 5
b = a.times
#=> #<Enumerator: 5:times>
c = b.to_a
#=> [0, 1, 2, 3, 4]
d = c.combination(2)
#=> #<Enumerator: [0, 1, 2, 3, 4]:combination(2)>
We can see the elements that are generated by the enumerator d by converting it to an array.
d.to_a
#=> [[0, 1], [0, 2], [0, 3], [0, 4], [1, 2], [1, 3], [1, 4], [2, 3], [2, 4], [3, 4]]
d.find { |i,j| nums[i]+nums[j] == target }
#=> [1, 3]
Note that (0...nums.size).to_a could be replaced by any of the following.
[*0...nums.size]
nums.each_index.to_a
nums.size.times.to_a
0.upto(nums.size-1).to_a
Array.new(nums.size) { |i| i }
Array.new(nums.size, &:itself)
Your error is that you don't reinitialize j when it reaches i which means that your algorithm just tries (0,n-1), (0,n-2), …, (0, 2), (0, 1) and then stops.
Say I have an array [1,2,3] and I want every combination of these numbers that don't exceed 4. So I would have [1,2,3].someMethod(4) and it would give me:
[1,1,1,1]
[1,1,2]
[1,3]
[2,2]
So far I have:
(1..4).flat_map{|size| [1,2,3].repeated_combination(size).to_a }
but this gives me every possible combinations, including the ones that exceed my given limit. Is there an good way to either only get combinations that add up to my limit?
arr = [1,2,3]
(arr+[0]).repeated_combination(4).select { |a| a.reduce(:+) == 4 }.map { |a| a - [0] }
#=> [[1, 3], [2, 2], [1, 1, 2], [1, 1, 1, 1]]
Change == to <= if desired.
This answer, like the others, assumes arr contains natural numbers, including 1.
results = (1..4).each.with_object([]) do |size, results|
[1,2,3].repeated_combination(size) do |combo|
results << combo if combo.reduce(:+) == 4
end
end
p results
--output:--
[[1, 3], [2, 2], [1, 1, 2], [1, 1, 1, 1]]
Parameterizing the algorithm:
def do_stuff(values, target_total)
(1..target_total).each.with_object([]) do |size, results|
values.repeated_combination(size) do |combo|
results << combo if combo.reduce(:+) == 4
end
end
end
p do_stuff([1, 2, 3], 4)
You can filter out the arrays you don't want by using the select method. Just select all the arrays that have a sum == 4 (the sum is calculated by the inject method).
all_arrs = (1..4).flat_map do |size|
[1,2,3].repeated_combination(size).to_a
end
valid_arrs = all_arrs.select do |arr|
arr.inject { |a, b| a + b } == 4
end
print valid_arrs
# Output:
# [[1, 3], [2, 2], [1, 1, 2], [1, 1, 1, 1]]
A recursive approach.
def some_method(a, n)
return [[]] if n == 0
a.select { |e| e <= n }.\
flat_map { |e| some_method(a,n-e).map { |es| ([e] + es).sort } }.\
sort.\
uniq
end
p some_method([1,2,3], 4)
# => [[1, 1, 1, 1], [1, 1, 2], [1, 3], [2, 2]]
EDIT: Here is another recursive version without filtering duplicates but with opposite order. I added comments to make it clearer.
def some_method(a, n)
return [[]] if n == 0 # bottom (solution) found
return [] if a.empty? || n < 0 # no solution
max = a.max
# search all solutions with biggest value
l = some_method(a, n-max).map { |e| [max] + e }
# search all solutions without biggest value
r = some_method(a-[max],n)
l + r
end
p some_method([1,2,3], 4)
# => [[3, 1], [2, 2], [2, 1, 1], [1, 1, 1, 1]]
I want to get all items from an array, which satisfy a predicate. Once I see an element that doesn't satisfy, I should stop iterating. For example:
[1, 4, -9, 3, 6].select_only_first { |x| x > 0}
I'm expecting to get: [1, 4]
This is how you want :
arup#linux-wzza:~> pry
[1] pry(main)> [1, 4, -9, 3, 6].take_while { |x| x > 0}
=> [1, 4]
[2] pry(main)>
Here is the documentation :
arup#linux-wzza:~> ri Array#take_while
= Array#take_while
(from ruby site)
------------------------------------------------------------------------------
ary.take_while { |arr| block } -> new_ary
ary.take_while -> Enumerator
------------------------------------------------------------------------------
Passes elements to the block until the block returns nil or false, then stops
iterating and returns an array of all prior elements.
If no block is given, an Enumerator is returned instead.
See also Array#drop_while
a = [1, 2, 3, 4, 5, 0]
a.take_while { |i| i < 3 } #=> [1, 2]
lines 1-20/20 (END)
If you're exploring other solution, this works too:
[1, 4, -9, 3, 6].slice_before { |x| x <= 0}.to_a[0]
You have to change x > 0 to x <=0.
That was an excellent answer Arup. My method is slightly more complicated.
numbers = [1,4,-9,3,6]
i = 0
new_numbers = []
until numbers[i] < 0
new_numbers.push(numbers[i])
i+= 1
end
=> [1,4]
Let's say I am trying to remove elements from array a = [1,1,1,2,2,3]. If I perform the following:
b = a - [1,3]
Then I will get:
b = [2,2]
However, I want the result to be
b = [1,1,2,2]
i.e. I only remove one instance of each element in the subtracted vector not all cases. Is there a simple way in Ruby to do this?
You may do:
a= [1,1,1,2,2,3]
delete_list = [1,3]
delete_list.each do |del|
a.delete_at(a.index(del))
end
result : [1, 1, 2, 2]
[1,3].inject([1,1,1,2,2,3]) do |memo,element|
memo.tap do |memo|
i = memo.find_index(e)
memo.delete_at(i) if i
end
end
Not very simple but:
a = [1,1,1,2,2,3]
b = a.group_by {|n| n}.each {|k,v| v.pop [1,3].count(k)}.values.flatten
=> [1, 1, 2, 2]
Also handles the case for multiples in the 'subtrahend':
a = [1,1,1,2,2,3]
b = a.group_by {|n| n}.each {|k,v| v.pop [1,1,3].count(k)}.values.flatten
=> [1, 2, 2]
EDIT: this is more an enhancement combining Norm212 and my answer to make a "functional" solution.
b = [1,1,3].each.with_object( a ) { |del| a.delete_at( a.index( del ) ) }
Put it in a lambda if needed:
subtract = lambda do |minuend, subtrahend|
subtrahend.each.with_object( minuend ) { |del| minuend.delete_at( minuend.index( del ) ) }
end
then:
subtract.call a, [1,1,3]
A simple solution I frequently use:
arr = ['remove me',3,4,2,45]
arr[1..-1]
=> [3,4,2,45]
a = [1,1,1,2,2,3]
a.slice!(0) # remove first index
a.slice!(-1) # remove last index
# a = [1,1,2,2] as desired
For speed, I would do the following, which requires only one pass through each of the two arrays. This method preserves order. I will first present code that does not mutate the original array, then show how it can be easily modified to mutate.
arr = [1,1,1,2,2,3,1]
removals = [1,3,1]
h = removals.group_by(&:itself).transform_values(&:size)
#=> {1=>2, 3=>1}
arr.each_with_object([]) { |n,a|
h.key?(n) && h[n] > 0 ? (h[n] -= 1) : a << n }
#=> [1, 2, 2, 1]
arr
#=> [1, 1, 1, 2, 2, 3, 1]
To mutate arr write:
h = removals.group_by(&:itself).transform_values(&:count)
arr.replace(arr.each_with_object([]) { |n,a|
h.key?(n) && h[n] > 0 ? (h[n] -= 1) : a << n })
#=> [1, 2, 2, 1]
arr
#=> [1, 2, 2, 1]
This uses the 21st century method Hash#transform_values (new in MRI v2.4), but one could instead write:
h = Hash[removals.group_by(&:itself).map { |k,v| [k,v.size] }]
or
h = removals.each_with_object(Hash.new(0)) { | n,h| h[n] += 1 }