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Hi I'm doing the palindrome partition problem using recursion. This problem is return all possible palindrome partitions of a given string input.
Input: "aab"Output: [["aa", "b"], ["a", "a", "b"]]
A palindrome partition definition: given a string S, a partition is a set of substrings, each containing one or more characters, such that every substring is a palindrome
My code is below. The issue I'm having is that the result array never gets correctly populated. From a high level I feel like my logic makes sense, but when I try to debug it I'm not really sure what is going on.
def partition(string)
result = []
output = []
dfs(string, 0, output, result)
result
end
def dfs(string, start, output, result)
if start == string.length
result << output
return
end
(start..string.length-1).to_a.each do |i|
if is_palindrome(string, start, i)
output << string[start..(i-start+1)]
dfs(string, i+1, output, result)
output.pop
end
end
end
def is_palindrome(string, start, end_value)
result = true
while start < end_value do
result = false if string[start] != string[end_value]
start += 1
end_value -= 1
end
result
end
puts partition("aab")
Yes, you do want to use recursion. I haven't analyzed your code carefully, but I see one problem is the following in the method dfs:
if start == string.length
result << output
return
end
If the if condition is satisfied, return without an argument will return nil. Perhaps you want return result.
Here is a relatively compact, Ruby-like way of writing it.
def pps(str)
return [[]] if str.empty?
(1..str.size).each_with_object([]) do |i,a|
s = str[0,i]
next unless is_pal?(s)
pps(str[i..-1]).each { |b| a << [s, *b] }
end
end
def is_pal?(str)
str == str.reverse
end
pps "aab"
#=> [["a", "a", "b"],
# ["aa", "b"]]
pps "aabbaa"
#=> [["a", "a", "b", "b", "a", "a"],
# ["a", "a", "b", "b", "aa"],
# ["a", "a", "bb", "a", "a"],
# ["a", "a", "bb", "aa"],
# ["a", "abba", "a"],
# ["aa", "b", "b", "a", "a"],
# ["aa", "b", "b", "aa"],
# ["aa", "bb", "a", "a"],
# ["aa", "bb", "aa"],
# ["aabbaa"]]
pps "aabbbxaa"
#=> [["a", "a", "b", "b", "b", "x", "a", "a"],
# ["a", "a", "b", "b", "b", "x", "aa"],
# ["a", "a", "b", "bb", "x", "a", "a"],
# ["a", "a", "b", "bb", "x", "aa"],
# ["a", "a", "bb", "b", "x", "a", "a"],
# ["a", "a", "bb", "b", "x", "aa"],
# ["a", "a", "bbb", "x", "a", "a"],
# ["a", "a", "bbb", "x", "aa"],
# ["aa", "b", "b", "b", "x", "a", "a"],
# ["aa", "b", "b", "b", "x", "aa"],
# ["aa", "b", "bb", "x", "a", "a"],
# ["aa", "b", "bb", "x", "aa"],
# ["aa", "bb", "b", "x", "a", "a"],
# ["aa", "bb", "b", "x", "aa"],
# ["aa", "bbb", "x", "a", "a"],
# ["aa", "bbb", "x", "aa"]]
pps "abcdefghijklmnopqrstuvwxyz"
#=> [["a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m",
# "n", "o", "p", "q", "r", "s", "t", "u", "v", "w", "x", "y", "z"]]
The best way of understanding how this recursion works is add some puts statements and re-run it.
def pps(str)
puts "\nstr=#{str}"
return [[]] if str.empty?
rv = (1..str.size).each_with_object([]) do |i,a|
s = str[0,i]
puts "i=#{i}, a=#{a}, s=#{s}, is_pal?(s)=#{is_pal?(s)}"
next unless is_pal?(s)
pps(str[i..-1]).each { |b| puts "b=#{b}, [s,*b]=#{[s,*b]}"; a << [s, *b] }
puts "a after calling pps=#{a}"
end
puts "rv=#{rv}"
rv
end
pps "aab"
str=aab
i=1, a=[], s=a, is_pal?(s)=true
str=ab
i=1, a=[], s=a, is_pal?(s)=true
str=b
i=1, a=[], s=b, is_pal?(s)=true
str=
b=[], [s,*b]=["b"]
a after calling pps=[["b"]]
rv=[["b"]]
b=["b"], [s,*b]=["a", "b"]
a after calling pps=[["a", "b"]]
i=2, a=[["a", "b"]], s=ab, is_pal?(s)=false
rv=[["a", "b"]]
b=["a", "b"], [s,*b]=["a", "a", "b"]
a after calling pps=[["a", "a", "b"]]
i=2, a=[["a", "a", "b"]], s=aa, is_pal?(s)=true
str=b
i=1, a=[], s=b, is_pal?(s)=true
str=
b=[], [s,*b]=["b"]
a after calling pps=[["b"]]
rv=[["b"]]
b=["b"], [s,*b]=["aa", "b"]
a after calling pps=[["a", "a", "b"], ["aa", "b"]]
i=3, a=[["a", "a", "b"], ["aa", "b"]], s=aab, is_pal?(s)=false
rv=[["a", "a", "b"], ["aa", "b"]]
#=> [["a", "a", "b"], ["aa", "b"]]
Given a string, I want to find a word bar, b-a-r, b--a--r etc. where - can be any letter. But interval between letters must be the same.
All letters are lower case and there is no gap betweens.
For example bar, beayr, qbowarprr, wbxxxxxayyyyyrzzz should match this.
I tried /b[a-z]*a[a-z]*r/ but this matches bxar which is wrong.
I am wondering if I achieve this with regexp?
Here's is one way to get all matches.
Code
def all_matches_with_spacers(word, str)
word_size = word.size
word_arr = word.chars
str_arr = str.chars
(0..(str.size - word_size)/(word_size-1)).each_with_object([]) do |n, arr|
regex = Regexp.new(word_arr.join(".{#{n}}"))
str_arr.each_cons(word_size + n * (word_size - 1))
.map(&:join)
.each { |substring| arr << substring if substring =~ regex }
end
end
This requires word.size > 1.
Example
all_matches_with_spacers('bar', 'bar') #=> ["bar"]
all_matches_with_spacers('bar', 'beayr') #=> ["beayr"]
all_matches_with_spacers('bar', 'qbowarprr') #=> ["bowarpr"]
all_matches_with_spacers('bar', 'wbxxxxxayyyyyrzzz') #=> ["bxxxxxayyyyyr"]
all_matches_with_spacers('bobo', 'bobobocbcbocbcobcodbddoddbddobddoddbddob')
#=> ["bobo", "bobo", "bddoddbddo", "bddoddbddo"]
Explanation
Suppose
word = 'bobo'
str = 'bobobocbcbocbcobcodbddoddbddobddoddbddob'
then
word_size = word.size #=> 4
word_arr = word.chars #=> ["b", "o", "b", "o"]
str_arr = str.chars
#=> ["b", "o", "b", "o", "b", "o", "c", "b", "c", "b", "o", "c", "b", "c",
# "o", "b", "c", "o", "d", "b", "d", "d", "o", "d", "d", "b", "d", "d",
# "o", "b", "d", "d", "o", "d", "d", "b", "d", "d", "o", "b"]
If n is the number of spacers between each letter of word, we require
word.size + n * (word.size - 1) <= str.size
Hence (since str.size => 40),
n <= (str.size - word_size)/(word_size-1) #=> (40-4)/(4-1) => 12
We therefore will iterate over zero to 12 spacers:
(0..12).each_with_object([]) do |n, arr| .. end
Enumerable#each_with_object creates an initially-empty array denoted by the block variable arr. The first value passed to block is zero (spacers), assigned to the block variable n.
We then have
regex = Regexp.new(word_arr.join(".{#{0}}")) #=> /b.{0}o.{0}b.{0}o/
which is the same as /bar/. word with n spacers has length
word_size + n * (word_size - 1) #=> 19
To extract all sub-arrays of str_arr with this length, we invoke:
str_arr.each_cons(word_size + n * (word_size - 1))
Here, with n = 0, this is:
enum = str_arr.each_cons(4)
#=> #<Enumerator: ["b", "o", "b", "o", "b", "o",...,"b"]:each_cons(4)>
This enumerator will pass the following into its block:
enum.to_a
#=> [["b", "o", "b", "o"], ["o", "b", "o", "b"], ["b", "o", "b", "o"],
# ["o", "b", "o", "c"], ["b", "o", "c", "b"], ["o", "c", "b", "c"],
# ["c", "b", "c", "b"], ["b", "c", "b", "o"], ["c", "b", "o", "c"],
# ["b", "o", "c", "b"], ["o", "c", "b", "c"], ["c", "b", "c", "o"],
# ["b", "c", "o", "b"], ["c", "o", "b", "c"], ["o", "b", "c", "o"]]
We next convert these to strings:
ar = enum.map(&:join)
#=> ["bobo", "obob", "bobo", "oboc", "bocb", "ocbc", "cbcb", "bcbo",
# "cboc", "bocb", "ocbc", "cbco", "bcob", "cobc", "obco"]
and add each (assigned to the block variable substring) to the array arr for which:
substring =~ regex
ar.each { |substring| arr << substring if substring =~ regex }
arr => ["bobo", "bobo"]
Next we increment the number of spacers to n = 1. This has the following effect:
regex = Regexp.new(word_arr.join(".{#{1}}")) #=> /b.{1}o.{1}b.{1}o/
str_arr.each_cons(4 + 1 * (4 - 1)) #=> str_arr.each_cons(7)
so we now examine the strings
ar = str_arr.each_cons(7).map(&:join)
#=> ["boboboc", "obobocb", "bobocbc", "obocbcb", "bocbcbo", "ocbcboc",
# "cbcbocb", "bcbocbc", "cbocbco", "bocbcob", "ocbcobc", "cbcobco",
# "bcobcod", "cobcodb", "obcodbd", "bcodbdd", "codbddo", "odbddod",
# "dbddodd", "bddoddb", "ddoddbd", "doddbdd", "oddbddo", "ddbddob",
# "dbddobd", "bddobdd", "ddobddo", "dobddod", "obddodd", "bddoddb",
# "ddoddbd", "doddbdd", "oddbddo", "ddbddob"]
ar.each { |substring| arr << substring if substring =~ regex }
There are no matches with one spacer, so arr remains unchanged:
arr #=> ["bobo", "bobo"]
For n = 2 spacers:
regex = Regexp.new(word_arr.join(".{#{2}}")) #=> /b.{2}o.{2}b.{2}o/
str_arr.each_cons(4 + 2 * (4 - 1)) #=> str_arr.each_cons(10)
ar = str_arr.each_cons(10).map(&:join)
#=> ["bobobocbcb", "obobocbcbo", "bobocbcboc", "obocbcbocb", "bocbcbocbc",
# "ocbcbocbco", "cbcbocbcob", "bcbocbcobc", "cbocbcobco", "bocbcobcod",
# ...
# "ddoddbddob"]
ar.each { |substring| arr << substring if substring =~ regex }
arr #=> ["bobo", "bobo", "bddoddbddo", "bddoddbddo"]
No matches are found for more than two spacers, so the method returns
["bobo", "bobo", "bddoddbddo", "bddoddbddo"]
For reference, there is a beautiful solution to the overall problem that is available in regex flavors that allow a capturing group to refer to itself:
^[^b]*bar|b(?:[^a](?=[^a]*a(\1?+.)))+a\1r
Sadly, Ruby doesn't allow this.
The interesting bit is on the right side of the alternation. After matching the initial b, we define a non-capturing group for the characters between b and a. This group will be repeated with the +. Between the a and r, we will inject capture group 1 with \1`. This group was captured one character at a time, overwriting itself with each pass, as each character between b and a was added.
See Quantifier Capture where the solution was demonstrated by #CasimiretHippolyte who refers to the idea behind the technique the "qtax trick".
What is a good way to get from this:
['a','b','c',['d1','d2']]
to this:
[['a','b','c','d1']['a','b','c','d2']]
another example, from this:
[['a1','a2'],'b',['c1','c2']]
to this:
[['a1','b','c1'],['a1','b','c2'],['a2','b','c1'],['a2','b','c2']]
edit 1:
Sorry for the confusion and thanks for response so far, the individual contents of the array items doesn't matter but the order must be preserved. The method needs to work for both example because the nested array can be in any position of the outer array, and the nested array can have more the 2 elements.
It's sort of like a regex with multiple or conditions
ab(c|d)
expand to match abc and abd
It is a bit hard to know exactly what you want, but this produces something quite similar:
# Create a list:
a = [['a1','a2'],'b',['c1','c2']]
# Split it into sub-arrays and single values:
list, other = a.partition{|x|x.is_a? Array}
# Split the array in order to get the product:
list_first, list_rest = list.pop, list
# Get the product and add the others_values:
p list_first.product(*list_rest).map{|list| list+other}
#=> [["c1", "a1", "b"], ["c1", "a2", "b"], ["c2", "a1", "b"], ["c2", "a2", "b"]]
1st:
arr1 = ['a','b','c',['d1','d2']]
*a, b = arr1
# ["a", "b", "c", ["d1", "d2"]]
a
# ["a", "b", "c"]
b
# ["d1", "d2"]
b.map{|x| a+[x]}
# [["a", "b", "c", "d1"], ["a", "b", "c", "d2"]]
and 2nd:
a, b, c = [["a1", "a2"], "b", ["c1", "c2"] ]
a.product c
#=> [["a1", "c1"], ["a1", "c2"], ["a2", "c1"], ["a2", "c2"]]
a.product(c).map{|x| x<<b}
#=> [["a1", "c1", "b"], ["a1", "c2", "b"], ["a2", "c1", "b"], ["a2", "c2", "b"]]
#or little less readable:
a.product(c).map{|x| [ x[0], b, x[1] ]}
# [["a1", "b", "c1"], ["a1", "b", "c2"], ["a2", "b", "c1"], ["a2", "b", "c2"]]
hirolau got me really close, here is what I ended with so the order is preserved:
# given a sample list
sample = [['a','b'],'c','d',['e','f'],'g',['h','i']]
# partition out the arrays
arrs, non_arrays = sample.partition {|sample| sample.is_a? Array}
# work out all possible products
first_elem, *the_rest = arrs
products = first_elem.product(*the_rest)
# finally swap it back in to get all valid combinations with order preserved
combinations = []
products.each do |p|
combinations << sample.map {|elem| elem.is_a?(Array) ? p.shift : elem}
end
# combinations
=> [["a", "c", "d", "e", "g", "h"],
["a", "c", "d", "e", "g", "i"],
["a", "c", "d", "f", "g", "h"],
["a", "c", "d", "f", "g", "i"],
["b", "c", "d", "e", "g", "h"],
["b", "c", "d", "e", "g", "i"],
["b", "c", "d", "f", "g", "h"],
["b", "c", "d", "f", "g", "i"]]
I'm trying to build a multidimensional array dynamically. What I want is basically this (written out for simplicity):
b = 0
test = [[]]
test[b] << ["a", "b", "c"]
b += 1
test[b] << ["d", "e", "f"]
b += 1
test[b] << ["g", "h", "i"]
This gives me the error: NoMethodError: undefined method `<<' for nil:NilClass. I can make it work by setting up the array like
test = [[], [], []]
and it works fine, but in my actual usage, I won't know how many arrays will be needed beforehand. Is there a better way to do this? Thanks
No need for an index variable like you're using. Just append each array to your test array:
irb> test = []
=> []
irb> test << ["a", "b", "c"]
=> [["a", "b", "c"]]
irb> test << ["d", "e", "f"]
=> [["a", "b", "c"], ["d", "e", "f"]]
irb> test << ["g", "h", "i"]
=> [["a", "b", "c"], ["d", "e", "f"], ["g", "h", "i"]]
irb> test
=> [["a", "b", "c"], ["d", "e", "f"], ["g", "h", "i"]]
Don't use the << method, use = instead:
test[b] = ["a", "b", "c"]
b += 1
test[b] = ["d", "e", "f"]
b += 1
test[b] = ["g", "h", "i"]
Or better still:
test << ["a", "b", "c"]
test << ["d", "e", "f"]
test << ["g", "h", "i"]
Here is a simple example if you know the size of array you are creating.
#OneDArray=[1,2,3,4,5,6,7,8,9]
p_size=#OneDArray.size
c_loop=p_size/3
puts "c_loop is #{c_loop}"
left=p_size-c_loop*3
#TwoDArray=[[],[],[]]
k=0
for j in 0..c_loop-1
puts "k is #{k} "
for i in 0..2
#TwoDArray[j][i]=#OneDArray[k]
k+=1
end
end
result will be #TwoDArray= [[1,2,3].[3,4,5],[6,7,8]]
Given an array of arrays
[["B", "C", "E", "F"], ["A", "B", "C", "D"], ["G"]]
What is the simplest way to merge the array items that contain members that are shared by any two or more arrays items. For example the above should be
[["A", "B", "C", "D","E", "F"], ["G"]] since "B" and "C" are shared by the first and second array items.
Here are some more test cases.
[["B", "C", "E", "F"], ["A", "B", "C", "D"], ["F", "G"]]
=> [["A", "B", "C", "D", "E", "F", "G"]]
[["B", "C", "E", "F"], ["A", "B", "C", "D"], ["G"], ["G", "H"]]
=> [["A", "B", "C", "D", "E", "F"], ["G", "H,"]]
Here is my quick version which can be optimized I am sure :)
# array = [["B", "C", "E", "F"], ["A", "B", "C", "D"], ["G"]]
# array = [["B", "C", "E", "F"], ["A", "B", "C", "D"], ["F", "G"]]
array = [["B", "C", "E", "F"], ["A", "B", "C", "D"], ["G"], ["G", "H"]]
array.collect! do |e|
t = e
e.each do |f|
array.each do |a|
if a.index(f)
t = t | a
end
end
end
e = t.sort
end
p array.uniq
Edit: Martin DeMello code was fixed.
When running Martin DeMello code (the accepted answer) I get:
[["B", "C", "E", "F"], ["A", "B", "C", "D"], ["F", "G"]] =>
[["B", "C", "E", "F", "A", "D", "G"], ["A", "B", "C", "D"], ["F", "G"]]
and
[["B", "C", "E", "F"], ["A", "B", "C", "D"], ["G"], ["G", "H"]] =>
[["B", "C", "E", "F", "A", "D"], ["A", "B", "C", "D"], ["G", "H"], ["G", "H"]]
which does not seem to meet your spec.
Here is my approach using a few of his ideas:
a = [["B", "C", "E", "F"], ["A", "B", "C", "D"], ["F", "G"]]
b = [["B", "C", "E", "F"], ["A", "B", "C", "D"], ["G"], ["G", "H"]]
def reduce(array)
h = Hash.new {|h,k| h[k] = []}
array.each_with_index do |x, i|
x.each do |j|
h[j] << i
if h[j].size > 1
# merge the two sub arrays
array[h[j][0]].replace((array[h[j][0]] | array[h[j][1]]).sort)
array.delete_at(h[j][1])
return reduce(array)
# recurse until nothing needs to be merged
end
end
end
array
end
puts reduce(a).to_s #[["A", "B", "C", "D", "E", "F", "G"]]
puts reduce(b).to_s #[["A", "B", "C", "D", "E", "F"], ["G", "H"]]
Different algorithm, with a merge-as-you-go approach rather than taking two passes over the array (vaguely influenced by the union-find algorithm). Thanks for a fun problem :)
A = [["A", "G"],["B", "C", "E", "F"], ["A", "B", "C", "D"], ["B"], ["H", "I"]]
H = {}
B = (0...(A.length)).to_a
def merge(i,j)
A[j].each do |e|
if H[e] and H[e] != j
merge(i, H[e])
else
H[e] = i
end
end
A[i] |= A[j]
B[j] = i
end
A.each_with_index do |x, i|
min = A.length
x.each do |j|
if H[j]
merge(H[j], i)
else
H[j] = i
end
end
end
out = B.sort.uniq.map {|i| A[i]}
p out
Not the simplest ,may be the longest :)
l = [["B", "C", "E", "F"], ["A", "B", "C", "D"], ["G"]]
puts l.flatten.inject([[],[]]) {|r,e| if l.inject(0) {|c,a| if a.include?(e) then c+1 else c end} >= 2 then r[0] << e ; r[0].uniq! else r[1] << e end ; r}.inspect
#[["B", "C"], ["E", "F", "A", "D", "G"]]
l = [["B", "C", "E", "F"], ["A", "B","C", "D"], ["G"]]
p l.inject([]){|r,e|
r.select{|i|i&e!=[]}==[]&&(r+=[e])||(r=r.map{|i|(i&e)!=nil&&(i|e).sort||i})
}
im not sure about your cond.
The simplest way to do it would be to take the powerset of an array (a set containing every possible combination of elements of the array), throw out any of the resulting sets if they don't have a common element, flatten the remaining sets and discard subsets and duplicates.
Or at least it would be if Ruby had proper Set support. Actually doing this in Ruby is horribly inefficient and an awful kludge:
power_set = array.inject([[]]){|c,y|r=[];c.each{|i|r<<i;r<<i+[y]};r}.reject{|x| x.empty?}
collected_powerset = power_set.collect{|subset| subset.flatten.uniq.sort unless
subset.inject(subset.last){|acc,a| acc & a}.empty?}.uniq.compact
collected_powerset.reject{|x| collected_powerset.any?{|c| (c & x) == x && x.length < c.length}}
Power set operation comes from here.
Straightforward rather than clever. It's destructive of the original array. The basic idea is:
go down the list of arrays, noting which array each element appears in
for every entry in this index list that shows an element in more than one array, merge all those arrays into the lowest-indexed array
when merging two arrays, replace the lower-indexed array with the merged result, and the higher-indexed array with a pointer to the lower-indexed array.
It's "algorithmically cheaper" than intersecting every pair of arrays, though the actual running speed will depend on what ruby hands over to the C layer.
a = [["B", "C", "E", "F"], ["A", "B", "C", "D"], ["G"], ["G", "H"]]
h = Hash.new {|h,k| h[k] = []}
a.each_with_index {|x, i| x.each {|j| h[j] << i}}
b = (0...(a.length)).to_a
h.each_value do |x|
x = x.sort_by {|i| b[i]}
if x.length > 1
x[1..-1].each do |i|
b[i] = [b[i], b[x[0]]].min
a[b[i]] |= a[i]
end
end
end
a = b.sort.uniq.map {|i| a[i]}
def merge_intersecting(input, result=[])
head = input.first
tail = input[1..-1]
return result if tail.empty?
intersection = tail.select { |arr| !(head & arr).empty? }
unless intersection.empty?
merged = head | intersection.flatten
result << merged.sort
end
merge_intersecting(tail, result)
end
require 'minitest/spec'
require 'minitest/autorun'
describe "" do
it "merges input array" do
input = [["B", "C", "E", "F"], ["A", "B", "C", "D"], ["F", "G"]]
output = [["A", "B", "C", "D", "E", "F", "G"]]
merge_intersecting(input).must_equal output
end
it "merges input array" do
input = [["B", "C", "E", "F"], ["A", "B", "C", "D"], ["G"], ["G", "H"]]
output = [["A", "B", "C", "D", "E", "F"], ["G", "H"]]
merge_intersecting(input).must_equal output
end
end