Longest common prefix and suffix of arrays

问题

What is the best way to get the longest common prefix (sub-array that starts from index 0 of the original) and suffix (sub-array that ends with index -1 of the original) of two arrays? For example, given two arrays:

[:foo, 1, :foo, 0, nil, :bar, "baz", false]
[:foo, 1, :foo, 0, true, :bar, false]

the longest common prefix of these arrays is:

[:foo, 1, :foo, 0]

and the longest common suffix of these arrays is:

[false]

When the elements at index 0/-1 differ in the original arrays, then the common prefix/suffix should be an empty array.

回答1:

One possible solution:

a1 = [:foo, 1, 0, nil, :bar, "baz", false]
a2 = [:foo, 1, 0, true, :bar, false]

a1.zip(a2).take_while { |x, y| x == y }.map(&:first)
#=> [:foo, 1, 0]

Reversing the input arrays and the output array finds a common suffix:

a1.reverse.zip(a2.reverse).take_while { |x, y| x == y }.map(&:first).reverse
#=> [false]

Edge case: zip fills "argument" arrays with nil values:

a1 = [true, nil, nil]
a2 = [true]

a1.zip(a2).take_while { |x, y| x == y }.map(&:first)
#=> [true, nil, nil]

This can be avoided by truncating the initial array to the second array's length:

a1[0...a2.size].zip(a2).take_while { |x, y| x == y }.map(&:first)

回答2:

Another solution without edge case:

a1 = [:foo, 1, 0, nil, :bar, "baz", false]
a2 = [:foo, 1, 0, true, :bar, false]

a1.take_while.with_index {|v,i| a2.size > i && v == a2[i] }

EDIT: Improved performance

class Array
  def common_prefix_with(other)
    other_size = other.size
    take_while.with_index {|v,i| other_size > i && v == other[i] }
  end
end

a1.common_prefix_with a2

回答3:

Three solutions: brutish (#3, initial answer), better (#2, first edit) and best (#1, variant of @Stefan's answer, second edit).

a = [:foo, 1, :foo, 0, nil, :bar, "baz", false]
b = [:foo, 1, :foo, 0, true, "baz", false]

c = [:foo,1,:goo]
d = [:goo,1,:new]

Note b is slightly changed from the OP's example.

Unless otherwise noted below, the common suffix would be calculated by reversing the arrays, applying common_prefix and then reversing the result.

Variant of Stefan's answer that rids it of zip and map (and retains his technique of truncating one array to at most the length of the other):

def common_prefix(a,b)
  a[0,b.size].take_while.with_index { |e,i| e == b[i] }
end

common_prefix(a,b)
  #=> [:foo, 1, :foo, 0]
common_prefix(c,d)
  #=> []

def common_prefix(a,b)
  any, arr = a.zip(b).chunk { |e,f| e==f }.first
  any ? arr.map(&:first) : []
end

def common_suffix(a,b)
  any, arr = a[a.size-b.size..-1].zip(b).chunk { |e,f| e==f }.to_a.last
  any ? arr.map(&:first) : []
end

common_prefix(a,b)
  #=> [:foo, 1, :foo, 0]
  # Nore: any, arr = a.zip(b).chunk { |e,f| e==f }.to_a
  #  => [[true, [[:foo, :foo], [1, 1], [:foo, :foo], [0, 0]]],
  #      [false, [[nil, true], [:bar, :baz], ["baz", false], [false, nil]]]]

common_suffix(a,b)
  #=> ["baz", false]
  # Note: any, arr = a[a.size-b.size..-1].zip(b).chunk { |e,f| e==f }.to_a
  #  => [[false, [[1, :foo], [:foo, 1], [0, :foo], [nil, 0], [:bar, true]]],
  #      [true, [["baz", "baz"], [false, false]]]]

When :first is sent to the enumerator Enumerable#chunk the first element of the enumerator is returned. It therefore should be comparable in efficiency to using Enumerable#take_while.

common_prefix(c,d)
  #=> []
common_suffix(c,d)
  #=> []

def common_prefix(a,b)
  a[0,(0..[a.size, b.size].min).max_by { |n| (a[0,n]==b[0,n]) ? n : -1 }]
end

common_prefix(a,b)
  #=> [:foo, 1, :foo, 0]

common_prefix(c,d)
  #=> []

回答4:

Gentlemen, start your engines!

Methods tested

I tested the second method given by @stefan, both methods proposed by @BroiSatse and the three methods I offered.

class Array #for broiSatse2
  def common_prefix_with(other)
    other_size = other.size
    take_while.with_index {|v,i| other_size > i && v == other[i] }
  end
end

class Cars
  def self.stefan(a1,a2)
    a1[0...a2.size].zip(a2).take_while { |x, y| x == y }.map(&:first)
  end

  def self.broiSatse1(a1,a2)
    a1.take_while.with_index {|v,i| a2.size > i && v == a2[i] }
  end

  def self.broiSatse2(a1,a2)
    a1.common_prefix_with(a2)
  end

  def self.cary1(a,b)
    a[0,b.size].take_while.with_index { |e,i| e == b[i] }
  end

  def self.cary2(a,b)
    any, arr = a.zip(b).chunk { |e,f| e==f }.first
    any ? arr.map(&:first) : []
  end

  def self.cary3(a,b)
    a[0,(0..[a.size, b.size].min).max_by { |n| (a[0,n]==b[0,n]) ? n : -1 }]
  end
end

I did not include the solution given by @6ftDan (not to be confused with 5' Dan or 7' Dan) because it did not pass all the tests.

Construction of test arrays

random_arrays(n) constructs one pair of arrays. n is the size of the smaller of the two. The larger is of size n+1.

def random_arrays(n)
  m = rand(n)
  # make arrays the same for the first m elements
  a = Array.new(m,0)
  b = Array.new(m,0)
  if m < n
    # make the m+1 elements different
    a << 0
    b << 1
    # randomly assign 0s and 1a to the remaining elements
    (n-m-1).times { a << rand(2); b << rand(2) }  if m < n - 1
  end
  # make arrays unequal in length
  (rand(2) == 0) ? a << 0 : b << 0
  [a,b]
end

Confirm methods tested give identical results

N = 10000 #size of smaller of two input arrays
methods = Cars.methods(false)

# ensure are methods produce the same results and that
# all deal with edge cases properly
20.times do |i|
  test = case i
         when 0 then [[0,1],[1,1]]
         when 1 then [[0],[]]
         when 1 then [[0,0,0],[0,0]]
         else
         random_arrays(N)
         end  
  soln = Cars.send(methods.first, *test)
  methods[1..-1].each  do |m|
    unless soln == Cars.send(m, *test)
      puts "#{m} has incorrect solution for #{test}"
      exit
    end
  end
end

puts "All methods yield the same answers\n"

Benchmarking

require 'benchmark'

I = 1000 #number of array pairs to test

@arr = I.times.with_object([]) { |_,a| a << random_arrays(N) } #test arrays

#test method m 
def testit(m)
  I.times { |i| Cars.send(m, *@arr[i]) }
end    

Benchmark.bm(12) { |bm| methods.each { |m| bm.report(m) { testit(m) } } end

Results

All methods yield the same answers

                   user     system      total        real
stefan        11.260000   0.050000  11.310000 ( 11.351626)
broiSatse1     0.860000   0.000000   0.860000 (  0.872256)
broiSatse2     0.720000   0.010000   0.730000 (  0.717797)
cary1          0.680000   0.000000   0.680000 (  0.684847)
cary2         13.130000   0.040000  13.170000 ( 13.215581)
cary3         51.840000   0.120000  51.960000 ( 52.188477)

回答5:

This works on unique arrays.

Prefix

x = [:foo, 1, 0, nil, :bar, "baz", false]
y = [:foo, 1, 0, true, :bar, false]
(x & y).select.with_index {|item,index| x.index(item) == index}

OUTPUTS: => [:foo, 1, 0]

Run reverse on Arrays for Affix

(x.reverse & y.reverse).select.with_index {|item,index| x.reverse.index(item) == index}

OUTPUTS: => [false]

来源：https://stackoverflow.com/questions/25406102/longest-common-prefix-and-suffix-of-arrays

标签

ruby

arrays

longest-prefix