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Classic problem #1920EasyArraySimulation

Build Array from Permutation

Real-world version: Follow Each Card Twice
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The statement here is our own: the same inputs, outputs and technique.

A rack has pigeonhole slots numbered 0 to n - 1. Slot i holds a card reading cards[i], which is the number of a slot on the same rack and may be the number of slot i itself. Every slot number appears on exactly one card, so the cards are a rearrangement of the slot numbers.

Fill a second rack of the same size. For each slot i, read the card in slot i of the original rack, go to the slot whose number it gives, and copy the number written on the card there into slot i of the new rack.

Both lookups read the original rack, never the rack being filled. Return the contents of the new rack, in slot order.

Examples

Example 1

Input
cards = [5, 3, 1, 4, 0, 2]
Output
[2, 4, 3, 0, 5, 1]

For slot 1 the card reads 3, and the card in slot 3 reads 4, so slot 1 of the new rack holds 4. For slot 4 the card reads 0, and the card in slot 0 reads 5, so slot 4 of the new rack holds 5.

Example 2

Input
cards = [3, 0, 4, 1, 2]
Output
[1, 3, 2, 0, 4]

For slot 0 the card reads 3 and the card in slot 3 reads 1, so the new rack starts with 1. For slot 2 the card reads 4 and the card in slot 4 reads 2.

Example 3

Input
cards = [2, 1, 0]
Output
[0, 1, 2]

Slot 0 points at slot 2, whose card reads 0. Slot 1 points at itself. Slot 2 points at slot 0, whose card reads 2.

Constraints

  • 1 <= cards.length <= 1000
  • 0 <= cards[i] < cards.length
  • The numbers in cards are all different.

The signature

The editor starts with this. It matches the parent problem's shape, so a solution that works here works on a standard judge unchanged.

Python
def follow_twice(cards: list[int]) -> list[int]:
Java
public int[] followTwice(int[] cards)

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