Your task is to remove smallest amount elements from a list, so the most elements are on their corresponding place. The element is on it's corresponding place, when it's value is equal to it's position.

Let's look at this example list:

1 5 3 2 5

This list has three elements on their corresponding place (1 on position 1, 5 on position 5, 3 on position 3).


The input is a sequence of decimals in any reasonable format. For example:

1 3 2 3 6


The output is a sequence of decimals in any reasonable format which contains, as its first element, the amount of elements removed from the input then the corresponding, 1-based element positions removed.

For example, for input given above, the (correct) output:

1 2

instructs us to remove one element, the one at index two (the first 3), which will leave us with 1 2 3 6 which has three elements at their index positions.

This output would be incorrect:

2 2 5

Since although removing the two elements at indexes two and five would leave us with 1 2 3 which also has three elements at their index positions, we've removed more than the minimal necessary.

More complicated example

1 2 3 3 4 5 6

In this case you can remove either of 3's, but not both (so 1 3 and 1 4 are both acceptable outputs).

1 3 2 3 4 5 6 7 8

In this case, if you would remove 3 on position 2, it would pass more elements, than if you would remove 3 on position 4 (so 1 4 is incorrect while 1 2 is correct).

1 3 2 3 5 6 7 8 9

Here removing 3 on position 2 is a mistake, because this actually makes the situation worse in obvious way (there are now less correctly placed elements than before). (the correct output is 0 since removing no elements is the best thing to do)

1 7 2 8 1 3 9

In this case we want to remove the three elements at positions 2, 4, and 5 (leaving us 1 2 3 9 for three in the correct location) so the correct output would be 3 2 4 5)


  • This is code golf, so the shortest code wins!
  • Loopholes are forbidden.
  • Assume that input is valid, contains nothing more than digits and spaces, and the input numbers inside the string are decimals in range of 0 <= n <= 99.
  • Please include a link to an online interpreter for your code.
  • If anything is unclear, please let me down in the comments.
  • \$\begingroup\$ Can we use 0 based indices? \$\endgroup\$ – Embodiment of Ignorance Aug 15 at 18:09
  • \$\begingroup\$ @EmbodimentofIgnorance I want to make it uniform, so only 1 based indices are allowed. \$\endgroup\$ – Krzysztof Szewczyk Aug 15 at 18:10

Pyth, 20 bytes


Try it online!

  • \$\begingroup\$ It was stated clearly that you have to. \$\endgroup\$ – Krzysztof Szewczyk Aug 15 at 18:57
  • \$\begingroup\$ Ok, edited the answer. I still don't understand why. \$\endgroup\$ – Mr. Xcoder Aug 15 at 18:58
  • \$\begingroup\$ Because there is such requirement in the answer. \$\endgroup\$ – Krzysztof Szewczyk Aug 15 at 18:59

Jelly, 18 bytes


Try it online!


JŒPṚœPF=J$SʋÞ⁸ṪL;$ - Link: list of numbers, A
J                  - range of length (A) [1,2,3,...,len(A)]
 ŒP                - all partitions      [[1],[2],...,[1,2],[1,3],...,[2,3],...]
   Ṛ               - reversed (so longest to shortest)
            Þ      - sort (the p's in all partitions) by:
             ⁸     -   (using the chain's left argument, A, as the right argument)
           ʋ       -   last four links as a dyad, i.e. f(p, A):
    œP             -     partition A at the indices in p
      F            -     flatten (to give result of dropping the values at indices in p)
         $         -     last two links as a monad:
        J          -       range of length (of the flatten result)
       =           -       equals? (vectorises)
          S        -     sum
              Ṫ    - tail (i.e. the shortest p which yields the maximal sum)
                 $ - last two links as a monad:
               L   -   length (of that result)
                ;  -   concatenate

An alternative for =J$ is ĖE€


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