How to remove repeated lists and get only one of them? like: [[3,7], [7,3]]

Learning Questions & Discussions Homework Help
1

This is output
[[3, 2, 2, 3], [5, 2, 3], [3, 4, 3], [7, 3], [5, 2, 3], [3, 5, 2], [3, 7], [3, 5, 2]]

This i want
[[3, 7], [3, 2, 5], [3, 4, 3], [3, 2, 2, 3]]

like: [[3,7], [7,3]] → sum of the element in both subsets is 10 but i want only one of them [3,7] or [7,3] in my output.*

 defmodule Test do
      @moduledoc false
    
      def possible_subsets(list, sum) do
        list
        |> subsets()
        |> Enum.filter(&(Enum.sum(&1) == sum))
      end
    
      defp subsets([]), do: [[]]
      defp subsets([h | t]) do
        t
        |> subsets()
        |> Enum.flat_map(&[[h | &1], &1])
      end
    
    end
2

One common approach for handling situations where you want to compare lists without considering order is to use MapSet - but that also ignores duplicates, so something like MapSet.new([3, 4, 3]) isn’t distinguishable from MapSet.new([3, 4]).

Another duplicate-friendly approach is to normalize the lists so that only one of [3,7] and [7,3] appear in the output. For instance, you could sort the values in each list into increasing order.

1 Like
3

this perhaps?

[
  [3, 2, 2, 3],
  [5, 2, 3],
  [3, 4, 3],
  [7, 3],
  [5, 2, 3],
  [3, 5, 2],
  [3, 7],
  [3, 5, 2]
]
|> Enum.uniq_by(& Enum.sort(&1))

[[3, 2, 2, 3], [5, 2, 3], [3, 4, 3], [7, 3]]
3 Likes
4

This is a poorly described problem. You talk about the sum of the sublists, but they are all 10.

Enum.uniq_by(input, &Enum.sum/1) will remove all but the first element because they all sum to 10.

Trisolaran’s solution of Enum.uniq_by(input, &Enum.sort/1) matches what you say you want the output to be, so go with that.

2 Likes
5

When sum = 0 then possible_subsets is returning
output : [0], []
but Expected output : []
How can i do that??

6

Can you describe why [0] is returned in the output?
If you can understand why it’s returned, you might be able to find a solution yourself.
If you cannot find, please describe what you thought about, tried out,…

4 Likes
7

I spent lot of time on it but it not fix.

iex(12)> list = [3, 5, 2, 7, 4, 2, 3]
[3, 5, 2, 7, 4, 2, 3]
iex(13)> possible_subsets(list, 0)         
[[]]

mix test result

 1) test check the expected sum of subsets of matrix for sum as 0 (SumOfSubsetsTest)
     test/sum_of_subsets_test.exs:98
     Assertion with == failed
     code:  assert SumOfSubsets.possible_subsets(list, 0) == expected_possible_subsets_matrix
     left:  [[]]
     right: []
     stacktrace:
       test/sum_of_subsets_test.exs:101: (test)

........

Finished in 0.2 seconds (0.00s async, 0.2s sync)
9 tests, 1 failure
8

You should use MapSet, as @al2o3cr told You :slight_smile:

It is a personal assignment, people are unwilling to give You a direct answer, but hints to get You on the right track.

2 Likes
9

ok i got it why [0] is coming in output but now i don’t understand how to get this as output [] not this [[]] when i run this possible_subsets(list, 0)??

10

Your code literally has that as output:

defp subsets([]), do: [[]]

So if subsets is called with an empty list. You return a list with one element: an empty list.

3 Likes
11

yeah…so how to fix it , i tried if & else loop also but it is not working…Can u tell me how to do it??

12

Look at the function…

What is the return value?

You can do it… just return what You want

3 Likes
13

Yeah…but if i did this

defp subsets([]), do: []

Then this happens

iex(8)> sum_of_one(array_of_digits, 0)
[]
iex(9)> sum_of_one(array_of_digits, 10)
[]
14

Make sure you clearly understand what subsets is doing.
Do you understand what happens if you pass in an empty list, a list with one element, 2 elements, more?

If you understand that and it appears correct to you, then continue to possible_subsets and check how it reacts to getting [[]] as output from subsets.
If that is still correct but you don’t want to return it you could add an other filter at the end.

I suggest you try to go through what happens in your code on paper and validate if your understanding is correct. You only have a small amount of code so it shouldn’t take much time.

2 Likes
15

Done Guys, Thankyou so much all of us.

16

Hello sir please help me sum of all

17

how did this happen please understand

18

Hello and welcome,

What have You tried so far?

19 
def sum_of_all(list, matrix_2d, list_sanitizer \\ &valid_sum/1) do
         calculated_integer_sum = list_sanitizer.(matrix_2d) |> Enum.sum()
         sum_of_one(list, calculated_integer_sum)
       end
20

I formatted your last post… code sample should be wrapped inside ```as in Markdown