0597MediumArrayHash TableGreedySorting

Filing Journal Issues Into Display Runs

Tracked in this browser only

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 trolley holds loose journal booklets. You are given hand, where hand[i] is the issue number printed on the i-th booklet; the trolley may well hold several booklets bearing the same issue number.

The booklets have to go out on display in runs. A run is a shelf holding exactly groupSize booklets whose issue numbers are groupSize consecutive whole numbers, with each of those numbers appearing on exactly one booklet of that run.

Return true if every booklet on the trolley can be placed on some run, each booklet used once and no booklet left over, and false otherwise.

Examples

Example 1

Input: hand = [8, 6, 7, 9, 10], groupSize = 5
Output: true

The five booklets carry the consecutive issue numbers 6 through 10, which is exactly one run of five.

Example 2

Input: hand = [4, 4, 5, 5, 6, 6], groupSize = 3
Output: true

Two runs of three can be filled, each holding issues 4, 5 and 6, and that uses all six booklets.

Example 3

Input: hand = [4, 4, 5, 6], groupSize = 2
Output: false

Two runs of two are needed. One of them must hold the pair 4 and 5, which leaves the booklets 4 and 6, and those issue numbers are not consecutive.

Constraints

1 <= hand.length <= 10^4
0 <= hand[i] <= 10^9
1 <= groupSize <= hand.length

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_n_straight_hand(hand: list[int], groupSize: int) -> bool:

Java

public boolean isNStraightHand(int[] hand, int groupSize)

The intended approach

Time O(n log n): sorting the distinct issue numbers dominates, and the subtraction loop removes each booklet at most once.Space O(n) for the tally.

Tally how many booklets carry each issue number, then walk the distinct issue numbers upward. When the walk reaches an issue number whose tally is still positive, no smaller number is left, so each remaining copy of it has to be the lowest booklet of its own run; that fixes exactly how many runs start there and which numbers they claim, so the same count is subtracted from the following `groupSize - 1` numbers and any shortfall means failure. A length not divisible by `groupSize` is rejected up front.

Pythonentry point: is_n_straight_hand

def is_n_straight_hand(hand, groupSize):
    # O(n log n) time from sorting the distinct issue numbers; the nested loop
    # removes each booklet from the tally at most once, so it adds only O(n).
    # O(n) space for the tally.
    if len(hand) % groupSize:
        return False
    left = {}
    for issue in hand:
        left[issue] = left.get(issue, 0) + 1
    for start in sorted(left):
        need = left[start]
        if need <= 0:
            continue
        # Nothing below start remains, so all copies of start must open runs.
        for step in range(groupSize):
            issue = start + step
            if left.get(issue, 0) < need:
                return False
            left[issue] -= need
    return True

The obvious version

Exhaustive search over shelvings. Take the highest issue number still on the trolley and try each of the groupSize runs of consecutive numbers that could contain it; whenever the trolley supplies all the numbers of that run, remove them and recurse on what is left. Results are cached against the remaining multiset.

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_n_straight_hand(hand, groupSize):
    if len(hand) % groupSize:
        return False
    cache = {}

    def solve(state):
        if not state:
            return True
        if state in cache:
            return cache[state]
        items = list(state)
        top = items[-1]
        answer = False
        for offset in range(groupSize):
            begin = top - groupSize + 1 + offset
            rest = items[:]
            fits = True
            for step in range(groupSize):
                issue = begin + step
                if issue in rest:
                    rest.remove(issue)
                else:
                    fits = False
                    break
            if fits and solve(tuple(rest)):
                answer = True
                break
        cache[state] = answer
        return answer

    return solve(tuple(sorted(hand)))

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

Create a free account to unlock the solution

See the step-by-step animation, the intuition, and clean code in every language — free, no credit card.

Create a free account Already have an account? Sign in