Determine if one array contains all elements of an

Question

I need to tell if an array contains all of the elements of another array with duplicates.

[1,2,3].contains_all? [1,2]   #=> true
[1,2,3].contains_all? [1,2,2] #=> false (this is where (a1-a2).empty? fails)
[2,1,2,3].contains_all? [1,2,2] #=> true

So the first array must contain as many or equal of the number of each unique element in the second array.

This question answers it for those using an array as a set, but I need to control for duplicates.

Update: Benchmarks

On Ruby 1.9.3p194

def bench
  puts Benchmark.measure {
    10000.times do
      [1,2,3].contains_all? [1,2]
      [1,2,3].contains_all? [1,2,2]
      [2,1,2,3].contains_all? [1,2,2]
    end
  }
end

Results in:

Rohit   0.100000   0.000000   0.100000 (  0.104486)
Chris   0.040000   0.000000   0.040000 (  0.040178)
Sergio  0.160000   0.020000   0.180000 (  0.173940)
sawa    0.030000   0.000000   0.030000 (  0.032393)

Update 2: Larger Arrays

@a1 = (1..10000).to_a
@a2 = (1..1000).to_a
@a3 = (1..2000).to_a

def bench
  puts Benchmark.measure {
    1000.times do
      @a1.contains_all? @a2
      @a1.contains_all? @a3
      @a3.contains_all? @a2
    end
  }
end

Results in:

Rohit    9.750000   0.410000  10.160000 ( 10.158182)
Chris   10.250000   0.180000  10.430000 ( 10.433797)
Sergio  14.570000   0.070000  14.640000 ( 14.637870)
sawa     3.460000   0.020000   3.480000 (  3.475513)

Answer 1

If you can't find a method, you can build one using ruby's include? method.

Official documentation: http://www.ruby-doc.org/core-1.9.3/Array.html#method-i-include-3F

Usage:

array = [1, 2, 3, 4]
array.include? 3       #=> true

Then, you can do a loop:

def array_includes_all?( array, comparision_array )
  contains = true
  for i in comparision_array do
    unless array.include? i
      contains = false
    end
  end
  return contains
end

array_includes_all?( [1,2,3,2], [1,2,2] )    #=> true

Answer 2

It seems you need a multiset. Check out this gem, I think it does what you need.

You can use is and do something like (if the intersection is equal to the second multiset then the first one includes all of its elements):

@ms1 & @ms2 == @ms2

Answer 3

Here's a naive and straightforward implementation (not the most efficient one, likely). Just count the elements and compare both elements and their occurrence counts.

class Array
  def contains_all? ary
    # group the arrays, so that 
    #   [2, 1, 1, 3] becomes {1 => 2, 2 => 1, 3 => 1}
    my_groups = group_and_count self
    their_groups = group_and_count ary

    their_groups.each do |el, cnt|
      if !my_groups[el] || my_groups[el] < cnt
        return false
      end
    end

    true
  end

  private
  def group_and_count ary
    ary.reduce({}) do |memo, el|
      memo[el] ||= 0
      memo[el] += 1
      memo
    end
  end

end

[1, 2, 3].contains_all? [1, 2]   # => true
[1, 2, 3].contains_all? [1, 2, 2] # => false
[2, 1, 2, 3].contains_all? [1, 2, 2] # => true
[1, 2, 3].contains_all? [] # => true
[].contains_all? [1, 2] # => false

Answer 4

Counting the number of occurrences and comparing them seems to be the obvious way to go.

class Array
   def contains_all? arr
       h = self.inject(Hash.new(0)) {|h, i| h[i] += 1; h}
       arr.each do |i|
           return false unless h.has_key?(i)
           return false if h[i] == 0
           h[i] -= 1
       end
       true
   end
end

Answer 5

Answering with my own implementation, but definitely want to see if someone can come up with a more efficient way. (I won't accept my own answer)

class Array
  def contains_all?(a2)
    a2.inject(self.dup) do |copy, el|
      if copy.include? el
        index = copy.index el
        copy.delete_at index
      else
        return false
      end
      copy
    end
    true
  end
end

And the tests:

1.9.3p194 :016 > [1,2,3].contains_all? [1,2]   #=> true
 => true 
1.9.3p194 :017 > [1,2,3].contains_all? [1,2,2] #=> false (this is where (a1-a2).empty? fails)
 => false 
1.9.3p194 :018 > [2,1,2,3].contains_all? [1,2,2] #=> true
 => true 

Answer 6

This solution will only iterate through both lists once, and hence run in linear time. It might however be too much overhead if the lists are expected to be very small.

  class Array
    def contains_all?(other)
      return false if other.size > size
      elem_counts = other.each_with_object(Hash.new(0)) { |elem,hash| hash[elem] += 1 }
      each do |elem|
        elem_counts.delete(elem) if (elem_counts[elem] -= 1) <= 0
        return true if elem_counts.empty?
      end
      false
    end
  end