Domain: Algorithms & Data Structures

You are given a permutation `p` of size `n` (containing each integer from `1` to `n` exactly once).

For the current permutation, define for each current position `i` (1-indexed):

`c_i = |i - p_i|`

You may repeatedly remove elements from the current permutation. A removal is valid if the chosen element satisfies `c_i <= s`.

After each removal:
- The remaining elements shift left to fill the gap.
- The remaining values are relabeled to again form a permutation of `1..k`:
  subtract `1` from every remaining value that was greater than the removed value.
- Then all `c_i` values are recomputed on the new permutation.

To remove all ambiguity, if more than one position is currently removable, always remove the leftmost removable position.

Goal: find the minimum integer `s` such that this deterministic process removes the entire permutation.

Input:

`n = 5`
`p = [3,1,5,2,4]`

Required output:

Output exactly these 5 lines and nothing else:

`s=<minimum s>`
`positions=<comma-separated removed positions>`
`values=<comma-separated removed values>`
`states=[3,1,5,2,4]->...->[]`
`why_s_minus_1_fails=<one sentence>`

Use the deterministic leftmost-removable rule for every step.


Grading rubric:
criterion: States `s=1` as the minimum integer that allows the deterministic process to remove the entire permutation.
weight: 20
criterion: States `positions=2,1,2,1,1` exactly.
weight: 20
criterion: States `values=1,2,1,2,1` exactly.
weight: 15
criterion: States `states=[3,1,5,2,4]->[2,4,1,3]->[3,1,2]->[2,1]->[1]->[]` exactly.
weight: 20
criterion: Explains that `s-1 = 0` fails because the initial permutation has no position with `i = p_i`, so no removal is valid at the start when `c_i` must equal 0.
weight: 15
criterion: Uses the exact 5-line output structure requested in the prompt with no extra lines or omitted lines.
weight: 10

Answer:
s=1
positions=2,1,2,1,1
values=1,2,1,2,1
states=[3,1,5,2,4]->[2,4,1,3]->[3,1,2]->[2,1]->[1]->[]
why_s_minus_1_fails=s-1=0 fails because the initial permutation [3,1,5,2,4] has no position with i=p_i, so no element has c_i=0 and the process cannot start.

Reasoning

The key detail is that `c_i` must be recomputed after every removal because both positions and values change. The original version treated the initial `c_i` values as if they stayed fixed, which is not correct for this process.

Check `s = 0` first, because the minimum `s` must be the smallest nonnegative integer that works. With `s = 0`, a move is valid only when `|i - p_i| = 0`, which means `i = p_i`. In the initial permutation `[3,1,5,2,4]`, the position-value pairs are `(1,3)`, `(2,1)`, `(3,5)`, `(4,2)`, `(5,4)`. None of them satisfy `i = p_i`, so there is no valid first move. Therefore `s = 0` is impossible.

Now test `s = 1` using the required deterministic rule: always remove the leftmost currently valid position.

Start with `[3,1,5,2,4]`.
- Current `c` values are `[2,1,2,2,1]`.
- Valid positions are 2 and 5, so remove the leftmost one: position 2, value 1.
- After shifting and relabeling values greater than 1 down by 1, the new permutation is `[2,4,1,3]`.

For `[2,4,1,3]`:
- Current `c` values are `[1,2,2,1]`.
- Valid positions are 1 and 4, so remove the leftmost one: position 1, value 2.
- After shifting and relabeling values greater than 2 down by 1, the new permutation is `[3,1,2]`.

For `[3,1,2]`:
- Current `c` values are `[2,1,1]`.
- Valid positions are 2 and 3, so remove the leftmost one: position 2, value 1.
- After shifting and relabeling values greater than 1 down by 1, the new permutation is `[2,1]`.

For `[2,1]`:
- Current `c` values are `[1,1]`.
- Valid positions are 1 and 2, so remove the leftmost one: position 1, value 2.
- After shifting and relabeling, the new permutation is `[1]`.

For `[1]`:
- Current `c` value is `[0]`.
- Remove position 1, value 1, leaving `[]`.

So `s = 1` succeeds, and `s = 0` fails. Therefore the minimum valid `s` is `1`.