0857EasyArrayHash TableSorting

Is the Ticket Roll Standard

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Trains the technique from

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 ticket roll is standard when, for some whole number k of at least 1, the numbers printed on it are exactly the numbers 1 through k each appearing once, plus a second copy of k. So a standard roll holds k + 1 tickets in total.

Return true when the roll nums is standard, and false otherwise.

Examples

Example 1

Input: nums = [2, 1, 3, 3]
Output: true

The roll holds four tickets, so a standard roll would need the numbers 1, 2 and 3 with a second 3. Sorted, the roll reads 1, 2, 3, 3, which is exactly that.

Example 2

Input: nums = [1, 2, 3, 4]
Output: false

Four tickets would need a second copy of 3, but every number here appears once and a 4 is present, so the roll is not standard.

Example 3

Input: nums = [2, 2, 1]
Output: true

Three tickets need the numbers 1 and 2 with a second 2, and sorted the roll reads 1, 2, 2.

Constraints

1 <= nums.length <= 100
1 <= nums[i] <= 200

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 is_standard_roll(nums: list[int]) -> bool:

Java

public boolean isStandardRoll(int[] nums)

The intended approach

Time O(n log n) for n tickets, dominated by the sort.Space O(n) for the sorted copy and the expected roll.

A standard roll of `k + 1` tickets is fixed entirely by its length, so `k` must be one less than the number of tickets. Build the roll that length demands, which is the numbers up to `k` with `k` repeated, and compare it against the sorted tickets. Nothing else can match, so no search is needed.

Pythonentry point: is_standard_roll

def is_standard_roll(nums):
    n = len(nums)
    if n < 2:
        return False
    k = n - 1
    want = list(range(1, k + 1)) + [k]
    want.sort()
    return sorted(nums) == want

The obvious version

Try every candidate k from 1 upwards, build the roll it describes, and compare tallies of how often each number appears. Makes no use of the fact that the length pins k down.

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_is_standard_roll(nums):
    def tally(values):
        out = {}
        for v in values:
            out[v] = out.get(v, 0) + 1
        return out

    seen = tally(nums)
    for k in range(1, 201):
        want = tally(list(range(1, k + 1)) + [k])
        if want == seen:
            return True
    return False

How this problem was checked

Reference solution executed against 12 test cases — 12 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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