0680EasyArraySimulation

Follow Each Card Twice

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Trains the technique from

LeetCode 1920Build Array from Permutation

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

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 is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def follow_twice(cards: list[int]) -> list[int]:

Java

public int[] followTwice(int[] cards)

The intended approach

Time O(n): one pass with two lookups per slot.Space O(n) for the rack that is returned.

Allocate the new rack up front and fill it slot by slot. For slot `i`, the card in slot `i` gives a slot number, and the card in that slot gives the number to write. Since the answer is written into separate storage, both lookups are guaranteed to see the original cards however far the walk has got.

Pythonentry point: follow_twice

def follow_twice(cards):
    # every lookup reads the original rack, so build the answer into fresh storage
    n = len(cards)
    built = [0] * n
    for i in range(n):
        built[i] = cards[cards[i]]
    return built

The obvious version

For each slot, scan the whole rack to find that slot and note the number on its card, then scan the whole rack again to find the slot that number names and note the number on its card. Collect those numbers in slot order.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_follow_twice(cards):
    out = []
    for i in range(len(cards)):
        first = None
        for slot, written in enumerate(cards):
            if slot == i:
                first = written
                break
        second = None
        for slot, written in enumerate(cards):
            if slot == first:
                second = written
                break
        out.append(second)
    return out

How this problem was checked

Reference solution executed against 10 test cases — 10 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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