0618HardMathEnumeration

Rows of Increasing Seats

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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 hall has exactly n seats and they all have to be used. The seats are laid out in one or more rows, read from the front of the hall to the back, under two rules: every row holds at least one seat, and each row after the first holds exactly one seat more than the row directly in front of it.

Two layouts are different when they use a different number of rows, or when some row holds a different number of seats. A single row holding all n seats is a valid layout.

Return how many different layouts the hall allows.

Examples

Example 1

Input: n = 12
Output: 2

Two layouts work: one row of 12, and the three rows 3, 4, 5 which also total 12.

Example 2

Input: n = 27
Output: 4

The four layouts are the single row 27, the two rows 13 and 14, the three rows 8, 9 and 10, and the six rows 2 through 7.

Example 3

Input: n = 4
Output: 1

Only the single row of 4 works. A two-row layout would need a front row and a back row one larger that add to 4, and no whole number does that, while the smallest three-row layout already totals 1 + 2 + 3 = 6.

Constraints

1 <= n <= 10^9

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 consecutive_numbers_sum(n: int) -> int:

Java

public int consecutiveNumbersSum(int n)

The intended approach

Time O(sqrt(n)) — the loop runs while the k-th triangular number is at most n.Space O(1)

Fix the number of rows `k` and the layout is pinned down by the front row alone, since the rows then read as `k` values each one larger than the last. Their sum is `k` copies of the front row plus the fixed amount `0 + 1 + ... + (k - 1)`, so a whole-number front row exists exactly when `n` minus that fixed amount is a multiple of `k`. Requiring the front row to hold at least one seat means `1 + 2 + ... + k` cannot exceed `n`, which caps `k` at roughly the square root of twice `n` and bounds the loop. Each `k` that passes the divisibility test contributes one layout.

Pythonentry point: consecutive_numbers_sum

def consecutive_numbers_sum(n):
    # O(sqrt(n)) time, O(1) space. An arrangement is fixed once the number of
    # rows k is fixed, because the rows then read a, a+1, ..., a+k-1 for a
    # single unknown front row a. Their total is k*a plus the fixed amount
    # 0+1+...+(k-1), so a whole number a exists exactly when n minus that fixed
    # amount divides by k, and a is at least 1 exactly while 1+2+...+k is at
    # most n. Every intermediate here stays below about 10^9.
    ways = 0
    k = 1
    while k * (k + 1) // 2 <= n:
        if (n - k * (k + 1) // 2) % k == 0:
            ways += 1
        k += 1
    return ways

The obvious version

Try every possible front row from 1 upwards. For each one keep adding the next row, one seat larger each time, until the running total reaches or passes `n`, and count the front rows whose running total lands exactly on `n`.

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_consecutive_numbers_sum(n):
    ways = 0
    for front in range(1, n + 1):
        total = 0
        row = front
        while total < n:
            total += row
            row += 1
        if total == n:
            ways += 1
    return ways

How this problem was checked

Reference solution executed against 15 test cases — 15 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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