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What is the optimal algorithm to find all the paths between 2 nodes of an adjacency list
Example input:
source = 1
destination = 5
list = {1: [2], 2: [4], 3: [4, 5], 4: [5]}
def paths(adj, st, en)
return [] unless adj.key?(st)
adj[st].each_with_object([]) do |nxt,arr|
nxt == en ? arr << [st, en] :
paths(adj, nxt, en).each { |a| arr << [st, *a] }
end
end
adj = { 1=>[2,3], 2=>[4,7], 3=>[4,8], 4=>[5,6], 5=>[7], 6=>[7] }
Note that I added an isolated node 8.
paths(adj, 1, 7)
#=> [[1, 2, 4, 5, 7],
# [1, 2, 4, 6, 7],
# [1, 2, 7],
# [1, 3, 4, 5, 7],
# [1, 3, 4, 6, 7]]
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The instructions:
You have to write a method, that return the length of the missing array.
Example:
[[1, 2], [4, 5, 1, 1], [1], [5, 6, 7, 8, 9]] --> 3
If the array of arrays is null/nil or empty, the method should return 0.
When an array in the array is null or empty, the method should return 0 too!
There will always be a missing element and its length will be always between the given arrays.
If you do array_of_arrays.sort_by! {|x| x.size} you'll get this:
[[1], [1, 2], [4, 5, 1, 1], [5, 6, 7, 8, 9]] it's missing an array with three elements inside the main array.
Also the def name: def getLengthOfMissingArray(array_of_arrays)
a = [[1, 2], [4, 5, 1, 1], [1], [5, 6, 7, 8, 9]]
new_arr = a.map(&:length)
missing_num_arr = (new_arr.min..new_arr.max).to_a - new_arr
# => [3]
You can map given array by length
Take minimum & maximum from mapped array
Create new array from minimum & maximum and subtract mapped array
You will get result
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I want to add array in this manner
arr1=[1,2,3,4]
arr2=[4,5,6,7]
adding should be like
arr1[0]+arr2[0]
arr1[0]+arr2[1]
arr1[0]+arr2[2]
and so on similarly with other second and other elements of arr1
Try a combination of map over both arrays:
p [1,2,3,4].map { |e| [4,5,6,7].map { |f| f + e } }
# => [[5, 6, 7, 8], [6, 7, 8, 9], [7, 8, 9, 10], [8, 9, 10, 11]]
arr1.product(arr2).map { |a,b| a + b }
#=> [5, 6, 7, 8, 6, 7, 8, 9, 7, 8, 9, 10, 8, 9, 10, 11]
Use sum maybe if you have Ruby > 2.4, otherwise you can use Enumerable#inject.
arr1.product(arr2).map { |a| a.sum }
See Array#product and Array#sum for further information.
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I want to search through a hash for keys with only one value, and divide that value by 2. How do I go about this?
Example
hash = { a => [4], b => [2, 4, 5], c => [3, 5] }
Result sought
hash = { a => [2], b => [2, 4, 5], c => [3, 5] }
hash = {:a=>[4], :b=>[2, 4, 5], :c=>[3, 5]}
You can use transform_values:
hash.transform_values { |v| v.one? ? [v[0]/2.0] : v }
#=> {:a=>[2.0], :b=>[2, 4, 5], :c=>[3, 5]}
Or map
hash.map { |k,v| v.one? ? [k,[v[0]/2.0]] : [k,v] }.to_h
#=> {:a=>[2.0], :b=>[2, 4, 5], :c=>[3, 5]}
This should do the trick, just iterate over each key/value pair and check the length of the values, if it only has 1 item, set the key in the hash to be equal to half the original value.
hash = { a: [4], b: [2, 4, 5], c: [3, 5] }
# => {:a=>[4], :b=>[2, 4, 5], :c=>[3, 5]}
hash.each do |key, values|
if 1 == values.length
hash[key] = [values.first / 2]
end
end
# => {:a=>[2], :b=>[2, 4, 5], :c=>[3, 5]}
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If I have a hash that looks like this:
C = [[1, 1, 1, 1],
[1, 2, 1, 1],
[1, 3, 1, 7],
[1, 1, 4, 1]]
What is a fast way to sum the columns and produce the following result:
C = [4, 7, 7, 10]
Edit: The way I was doing it coming from a C background was to parse thru the result and summing manually, that's why I asked. didn't know where else to look for.
arr = [[1, 1, 1, 1],
[1, 2, 1, 1],
[1, 3, 1, 7],
[1, 1, 4, 1]]
arr.transpose.map{|e| e.inject(:+)}
# => [4, 7, 7, 10]
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I am new to ruby on rails .
I was programming ruby and want to try checking if 2 numbers of array add up to Input number in ruby.
eg,
array A[]= {3, 1, 8, 11, 5, 7}
given integer say N = 6
answer will be 1,5.
I know how to program it in java,C++ but i am stuck in ruby coding,
Can anyone please help me.Thanks in advance
You can use Array#combination:
ary = [3, 1, 8, 11, 5, 7]
n = 6
ary.combination(2).detect { |a, b| a + b == n }
#=> [1, 5]
combination(2) creates an array of all combinations of length 2, i.e. [3,1], [3,8], [3,11] etc.
detect { |a, b| a + b == n } returns the first pair with sum n
You can use find_all instead of detect to return all pairs with sum n.
a = [3, 1, 8, 11, 4, 5, 7, 2]
> a.combination(2).select {|i| i.inject(:+) == 6 }
#=> [[1, 5], [4, 2]]
a = [3, 1, 8, 11, 5, 7]
p a.combination(2).find{|i| i.inject(:+) == 6}
# >> [1, 5]