I am trying to conceptualize the iteration of two loops
numbers_array = [1,2,3,4,5,6,7,8,9,10]
add_to_array = [1,2,3,4]
While the numbers_array iterates, add_to_array iterates simultaneously adding both elements together at the same time. The caveat is once add_to_array reaches the end, it starts over adding its element to the next index in numbers_array. So at numbers_array[4] we would be adding add_to_array[0] then adding numbers_array[5] to add_to_array[1] and so on. This process would repeat until we reach the end of the numbers_array.
Any input would be greatly appreciated!
You are looking for Enumerable#zip and Enumerable#cycle:
numbers_array = [1,2,3,4,5,6,7,8,9,10]
#⇒ [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
add_to_array = [1,2,3,4]
#⇒ [1, 2, 3, 4]
numbers_array.zip(add_to_array.cycle)
#⇒ [[1, 1], [2, 2], [3, 3], [4, 4], [5, 1],
# [6, 2], [7, 3], [8, 4], [9, 1], [10, 2]]
Now do whatever you want with the array returned. E.g. to reduce the zipped result summing elements, map ro Enumerable#sum:
numbers_array.zip(add_to_array.cycle).map(&:sum)
#⇒ [2, 4, 6, 8, 6, 8, 10, 12, 10, 12]
It works by using the % operator to cycle through the indexes.
numbers_array = [1,2,3,4,5,6,7,8,9,10]
add_to_array = [1,2,3,4]
numbers_array.map.with_index do |n, i|
n + add_to_array[i % add_to_array.length]
end
A cool method that's similar, if you didn't want to start over at the next array, would be .zip
https://apidock.com/ruby/Array/zip
add_to_array.zip(*numbers_array.each_slice(add_to_array.size)).
map { |a| a.sum { |e| e.to_i } }
#=> [16, 20, 13, 16]
e.to_i is needed to convert nil values to zeros. See NilClass#to_i.
Another option:
numbers_array.map { |e| e + add_to_array.rotate!.last }
# => [2, 4, 6, 8, 6, 8, 10, 12, 10, 12]
Drawback: add_to_array is mutated by rotate!
Related
So I am trying to get a solution to my two sum problem and I am stuck, I need to print the indices for the elements which add up to the target and my solution will return an element twice if it is one half of the target
def two_sum(nums, target)
num_hash = Hash.new(0)
nums.each_with_index do |num,idx|
num_hash[num] = idx
if num_hash.key?(target - num) && target % num != 0
return [num_hash[num], idx]
end
end
end
So I don't think the problem is related to the number being 1/2 of the target, it just seems to be "if a solution is found, it returns the same index twice". For instance, using the sample set [2, 7, 11, 15]
two_sum([2, 7, 11, 15], 14) # => [2, 7, 11, 15]
So, 7 is half of 14, which is the target, and instead of returning the index 1 twice, as you suggest it would, it returns the original input array (the result of nums.each_with_index. However, if we try passing a target of 9, it behaves as you describe:
two_sum([2, 7, 11, 15], 9) # => [1, 1]
The reason for this, is because of the line:
return [num_hash[num], idx]
you have already set num into the num_hash (num_hash[num] = idx) and then you are returning both the idx and num_hash[num], which is also idx. So what you want to do is:
return [num_hash[target - num], idx]
and then to 'fix' all the elements being returned when no result is found, just return [] at the end of the method:
def two_sum(nums, target)
num_hash = Hash.new(0)
nums.each_with_index do |num,idx|
num_hash[num] = idx
if num_hash.key?(target - num) && target % num != 0
return [num_hash[target - num], idx]
end
end
[]
end
and now:
two_sum([2, 7, 11, 15], 14) # => []
two_sum([2, 7, 11, 15], 9) # => [0, 1]
Note: you also have a problem with the code where, if you have the same number twice, it doesn't find the answer:
two_sum([2, 7, 11, 7, 15], 14) # => []
left for you to figure out, just wanted to point this out to you.
You can use the method Array#combination to advantage here.
def two_sum(nums, target)
nums.each_index.to_a.combination(2).select { |i,j| nums[i] + nums[j] == target }
end
two_sum([2, 7, 11, 15], 14)
#=> []
two_sum([2, 7, 11, 15], 9)
#=> [[0, 1]]
two_sum([2, 4, 7, 5], 9)
#=> [[0, 2], [1, 3]]
two_sum([2, 2, 2, 2], 4)
#=> [[0, 1], [0, 2], [0, 3], [1, 2], [1, 3], [2, 3]]
two_sum([2, 4, 7, 5], 8)
#=> []
For
nums = [2, 4, 7, 5]
target = 9
the steps are as follows.
a = nums.each_index
#=> #<Enumerator: [2, 4, 7, 5]:each_index>
We can see the elements that will be generated by this enumerator by converting it to an array.
b = a.to_a
#=> [0, 1, 2, 3]
Next,
c = b.combination(2)
#=> #<Enumerator: [0, 1, 2, 3]:combination(2)>
c.to_a
#=> [[0, 1], [0, 2], [0, 3], [1, 2], [1, 3], [2, 3]]
The rest is straightforward as select merely selects those pairs of indices passed to it (i,j) whose corresponding values, num[i] and num[j], sum to target.
I think what you want is ...
return [num_hash[target-num], idx]
I want to write a function that takes in a nested array and return the size of the longest array and itself.
max_with_size([]) # [0, []]
max_with_size([2,3,4]) # [3, [2, 3, 4]]
max_with_size([1,[2,3,4]]) # [3, [2, 3, 4]]
max_with_size([[5,[6],[7,8,9],10,11]]) # [5, [5, [6], [7, 8, 9], 10, 11]]
max_with_size([[1,[2,3,4]],[[[5,[6],[7,8,9],10,11]]]]) # [5, [5, [6], [7, 8, 9], 10, 11]]
So far I've got this
def max_with_size (ary)
max_size = ary.size
max_ary = ary
ary.each { |elem|
if elem.is_a? Array
if elem.size > max_size
max_size = max_with_size(elem)[0]
max_ary = max_with_size(elem)[1]
end
end
}
[max_size, max_ary]
end
It works fine for the first 4 cases, but the 5th fails and only delivers this
max_with_size([[1,[2,3,4]],[[[5,[6],[7,8,9],10,11]]]]) # [2, [[1, [2, 3, 4]], [[[5, [6], [7, 8, 9], 10, 11]]]]]
How can I achieve the wanted result?
The following code should print the desired result. I explained code with Inline comments.
#Initialize #max to empty array, #max is an array with two elements, like this: [max_array_size, max_array]
#max = []
def max_with_size(array)
# when #max is empty or when array size is greater than what is store in #max, store array size and array contents in #max
(#max = [array.size, array]) if #max.empty? || (#max[0] < array.size)
#Iterate through each element in array
array.each do |x|
#Skip to next element if x is not an array
next unless x.is_a? Array
#Recursively find max of array x
max_with_size(x)
end
#max
end
Code
def max_arr(arr)
[arr, *arr.each_with_object([]) {|e,a| a << max_arr(e) if e.is_a?(Array) && e.any?}].
max_by(&:size)
end
Examples
examples = [[],
[2,3,4],
[1,[2,3,4]],
[[5,[6],[7,8,9],10,11]],
[[1,[2,3,4]],[[[5,[6],[7,8,9],10,11]]]],
[1, [2, [3, 4, [6, 7, 8, 9, 10], [11, 12]], 13]]]
examples.each do |arr|
a = max_arr(arr)
puts "\n#{arr}\n \#=> #{a.size}, #{a}"
end·
[]
#=> 0, []
[2, 3, 4]
#=> 3, [2, 3, 4]
[1, [2, 3, 4]]
#=> 3, [2, 3, 4]
[[5, [6], [7, 8, 9], 10, 11]]
#=> 5, [5, [6], [7, 8, 9], 10, 11]
[[1, [2, 3, 4]], [[[5, [6], [7, 8, 9], 10, 11]]]]
#=> 5, [5, [6], [7, 8, 9], 10, 11]
[1, [2, [3, 4, [6, 7, 8, 9, 10], [11, 12]], 13]]
#=> 5, [6, 7, 8, 9, 10]
I have a find_num method that returns the index of a specified number in an ordered array, e.g.
find_num(6, [1, 4, 6, 9, 13]) #=> 2
however my spec also requires that if the number is not available it finds the position of the next highest number so ...
find_num(8, [1, 4, 6, 9, 13]) #=> 3
as 9 is the next available number.
Having trouble implementing this... I have thought of iterating through the whole array but I am told to take into account the array could be large...
You can pass a block to index and it ...
[...] returns the index of the first object for which the block returns true. Returns nil if no match is found.
Examples:
[1, 4, 6, 9, 13].index { |n| n >= 6 } #=> 2
[1, 4, 6, 9, 13].index { |n| n >= 8 } #=> 3
[1, 4, 6, 9, 13].index { |n| n >= 15 } #=> nil
Because this requires the array to be ordered, you can also use bsearch_index which performs a binary search.
you can also find a index of any element in array like this.
2.1.8 :040 > [1, 4, 6, 9, 13].index(6)
=> 2
2.1.8 :041 > [1, 4, 6, 9, 13].index(15)
=> nil
def find_num(n,a)
a.each_with_index.to_a.sort_by(&:first).find { |nbr,_| nbr >= n }.last
end
find_num(6, [1, 4, 6, 9, 13])
#=> 2
find_num(8, [1, 4, 6, 9, 13]) #=> 3
#=> 3
The steps for
n = 8
a = [1, 4, 6, 9, 13]
are as follows.
b = a.each_with_index.to_a
#=> [[1, 0], [4, 1], [6, 2], [9, 3], [13, 4]]
c = b.sort_by(&:first)
#=> [[1, 0], [4, 1], [6, 2], [9, 3], [13, 4]]
d = c.find { |nbr,_| nbr >= n }
#=> [9, 3]
d.last
#=> 3
I am looking for a way to perform a certain operation (for instance delete_if) on an array and return both the deleted elements, and the remaining elements.
For example
a = [1,2,3,4,5,6,7,8,9,10]
a.delete_if {|x| x.even? } #=> [[1, 3, 5, 7, 9]]
But what I am looking for is something like
a = [1,2,3,4,5,6,7,8,9,10]
a.some_operation #=> [[1,3,5,7,9],[2,4,6,8,10]]
How would I go about doing this?
Using Enumerable#partition:
a = [1,2,3,4,5,6,7,8,9,10]
a.partition &:even?
# => [[2, 4, 6, 8, 10], [1, 3, 5, 7, 9]]
The first element of the Enumerable#partition return value contains the elements that are evaluated to true in the block. So you need to use odd? to get what you want.
a.partition &:odd?
# => [[1, 3, 5, 7, 9], [2, 4, 6, 8, 10]]
You might be looking for something like this:
a = [1,2,3,4,5,6,7,8,9,10]
a.group_by { |x| x.even? }.values
#=> [[1, 3, 5, 7, 9], [2, 4, 6, 8, 10]]
I have an array of arrays like the following:
=> [[1,2,3,4,5],[6,7,8,9,10],[11,12,13,14,15]]
I want to rearrange it by order of elements in the inner array, e.g.:
=> [[1,6,11],[2,7,12],[3,8,13],[4,9,14],[5,10,15]]
How can I achieve this?
I know I can iterate an array of arrays like
array1.each do |bla,blo|
#do anything
end
But the side of inner arrays isn't fixed.
p [[1,2,3,4,5],[6,7,8,9,10],[11,12,13,14,15]].transpose
#=> [[1, 6, 11], [2, 7, 12], [3, 8, 13], [4, 9, 14], [5, 10, 15]]
use transpose method on Array
a = [[1,2,3,4,5],[6,7,8,9,10],[11,12,13,14,15]]
a.transpose
#=> [[1, 6, 11], [2, 7, 12], [3, 8, 13], [4, 9, 14], [5, 10, 15]]
Note that this only works if the arrays are of all the same length.
If you want to handle transposing arrays that have different lengths to each other, something like this should do it
class Array
def safe_transpose
max_size = self.map(&:size).max
self.dup.map{|r| r << nil while r.size < max_size; r}.transpose
end
end
and will yield the following
a = [[1,2,3,4,5],[6,7,8,9,10],[11,12,13,14,15,16]]
a.safe_transpose
#=> [[1, 6, 11], [2, 7, 12], [3, 8, 13], [4, 9, 14], [5, 10, 15], [nil, nil, 16]]