0813MediumArraySortingHeap (Priority Queue)

Ragged Hall Call Order

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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 lecture hall is described by seats, a list of rows. Row r holds seats[r].length seats and seats[r][c] is the badge number of the seat in position c of that row, counting from 0. Rows may hold very different numbers of seats, so seats is ragged: nothing guarantees that two rows are the same length.

An usher calls the badges in groups. The group of a seat is r + c, its row index plus its position in the row. Groups are called in increasing order of that number, and inside one group the seats are called from the largest row index down to the smallest.

Return the badge numbers in the order they are called.

Examples

Example 1

Input: seats = [[1, 2, 3, 4, 5, 6, 7, 8], [9], [10, 11], [12, 13, 14, 15]]
Output: [1, 9, 2, 10, 3, 12, 11, 4, 13, 5, 14, 6, 15, 7, 8]

Group 0 holds only badge 1. Group 1 holds badge 2 from row 0 and badge 9 from row 1, and the higher row is called first, so badge 9 comes before badge 2. Group 2 holds badge 3 from row 0, badge 10 from row 2 and nothing from row 1, since row 1 has no seat in position 1, so the order there is 10 then 3. The remaining groups follow the same rule.

Example 2

Input: seats = [[9, 8, 7]]
Output: [9, 8, 7]

With a single row, the group numbers are 0, 1 and 2, one seat each, so the badges come out in row order.

Example 3

Input: seats = [[4], [5], [6]]
Output: [4, 5, 6]

Every row holds one seat, so again each group holds a single badge: group 0 is badge 4, group 1 is badge 5 and group 2 is badge 6.

Constraints

1 <= seats.length <= 10^5
1 <= seats[i].length <= 10^5
1 <= seats[i][j] <= 10^5
The hall holds at most 10^5 seats in total.

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 diagonal_call_order(seats: list[list[int]]) -> list[int]:

Java

public int[] diagonalCallOrder(List<List<Integer>> seats)

The intended approach

Time O(N) for a hall of N seats.Space O(N) for the buckets, plus the returned list of N badges.

A seat's group number is its row index plus its position, and that can never exceed the number of seats in the hall, so the groups go into buckets indexed by that number instead of being sorted. Reading the rows from the top down fills each bucket in increasing row order, which is the reverse of the calling order, so each bucket is reversed as it is appended to the answer. Ragged rows need no special handling because each row is walked over its own length.

Pythonentry point: diagonal_call_order

def diagonal_call_order(seats):
    # Every seat belongs to the group numbered row + seat, and no group number
    # can exceed the total number of seats, so the groups are collected in
    # buckets rather than sorted. Rows are read in increasing order, which drops
    # each group's seats in increasing row order, so every bucket is reversed on
    # the way out to get the highest row first.
    #
    # Time: O(N) for N seats in total. Space: O(N) for the buckets and the answer.
    buckets = []
    for r, row in enumerate(seats):
        for c, badge in enumerate(row):
            k = r + c
            while len(buckets) <= k:
                buckets.append([])
            buckets[k].append(badge)
    out = []
    for bucket in buckets:
        bucket.reverse()
        out.extend(bucket)
    return out

The obvious version

Tag every seat with the pair (row plus position, negated row index) and hand the whole list to a general sort, then read the badges off in sorted order. Uses no bucketing at all, so it checks the ordering rule independently.

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_diagonal_call_order(seats):
    """Tag every seat with its sort key and let a general sort do the ordering."""
    tagged = []
    for r, row in enumerate(seats):
        for c, badge in enumerate(row):
            tagged.append((r + c, -r, badge))
    tagged.sort()
    return [badge for _k, _neg, badge in tagged]

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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