0581HardArrayMathCombinatoricsFermat's Little Theorem

Ways the Blight Can Spread

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Trains the technique from

LeetCode 2954Count the Number of Infection Sequences

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 nursery bench holds n saplings in one straight row, numbered 0 through n - 1. Two saplings are neighbours when their numbers differ by one.

The positions listed in sick, given in increasing order, already carry a blight. Every other sapling is healthy for now. Time passes in rounds, and in each round exactly one healthy sapling that has at least one blighted neighbour catches the blight. When several healthy saplings qualify in a round, any one of them may be the one that catches it. Rounds continue until nothing healthy is left.

A spread order is the list of positions of the initially healthy saplings, written in the order they caught the blight. Return how many different spread orders are possible, taken modulo 1000000007.

Examples

Example 1

Input: n = 5, sick = [0, 4]
Output: 4

Saplings 1, 2 and 3 start healthy. One possible spread order is [3, 2, 1]: sapling 3 touches the blighted sapling 4, then sapling 2 touches sapling 3, then sapling 1. Four spread orders are possible altogether.

Example 2

Input: n = 6, sick = [2, 3]
Output: 6

Saplings 0, 1, 4 and 5 start healthy. One possible spread order is [1, 4, 5, 0]. An order beginning with sapling 5 is not possible, because in the first round neither neighbour of sapling 5 carries the blight. Six spread orders are possible altogether.

Example 3

Input: n = 9, sick = [4]
Output: 70

The blight sits in the middle of the bench with four healthy saplings on either side, and 70 spread orders are possible. One of them is [3, 5, 6, 2, 1, 7, 8, 0].

Constraints

2 <= n <= 10^5
1 <= sick.length <= n - 1
0 <= sick[i] <= n - 1
sick is given in strictly increasing order.
Return the count taken modulo 1000000007, so the answer is below 1000000007.

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 number_of_sequence(n: int, sick: list[int]) -> int:

Java

public int numberOfSequence(int n, int[] sick)

The intended approach

Time O(n + g log MOD) — one factorial table plus a modular power per run.Space O(n) for the factorial table.

The healthy saplings fall into runs separated by the blighted ones, and a sapling in one run never depends on another run, so a spread order is fixed by two independent choices: how the rounds of the different runs are interleaved, and in what order each run is eaten through. The interleavings number as the multinomial coefficient of the run lengths, that is the factorial of the total healthy count divided by the factorial of each run length. Inside a run pinned between two blighted saplings, every round except the last offers exactly two choices, its left end or its right end, which gives two to the power of one less than its length. A run lying against either end of the bench can only be eaten inward, so it has a single internal order. Factorials are taken modulo 1000000007 and the divisions are done by raising to the power `MOD - 2`, which is the modular inverse because the modulus is prime.

Pythonentry point: number_of_sequence

def number_of_sequence(n, sick):
    # O(n + g log MOD) time, O(n) space, where g is the number of gaps: the
    # healthy runs between blighted saplings are independent apart from the
    # interleaving, so the count is the number of ways to interleave the runs
    # times the number of internal orders of each run. A run pinned between two
    # blighted saplings can be taken from either end at every step except the
    # last, and a run at an end of the row has only one possible internal order.
    MOD = 1000000007
    healthy = n - len(sick)
    if healthy == 0:
        return 1
    fact = [1] * (healthy + 1)
    for i in range(1, healthy + 1):
        fact[i] = fact[i - 1] * i % MOD
    answer = fact[healthy]
    runs = []
    if sick[0] > 0:
        runs.append((sick[0], False))
    for i in range(1, len(sick)):
        span = sick[i] - sick[i - 1] - 1
        if span > 0:
            runs.append((span, True))
    tail = n - 1 - sick[-1]
    if tail > 0:
        runs.append((tail, False))
    for length, pinned in runs:
        answer = answer * pow(fact[length], MOD - 2, MOD) % MOD
        if pinned:
            answer = answer * pow(2, length - 1, MOD) % MOD
    return answer

The obvious version

Grow the spread one round at a time. At each step, try every healthy sapling that touches a blighted one, mark it, recurse, and unmark it. Each complete branch of the search is one spread order, so the number of branches that reach the end is the count.

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_number_of_sequence(n, sick):
    MOD = 1000000007
    blighted = [False] * n
    for i in sick:
        blighted[i] = True
    left = n - len(sick)

    def walk(left):
        if left == 0:
            return 1
        total = 0
        for i in range(n):
            if blighted[i]:
                continue
            touching = (i > 0 and blighted[i - 1]) or (i + 1 < n and blighted[i + 1])
            if touching:
                blighted[i] = True
                total += walk(left - 1)
                blighted[i] = False
        return total

    return walk(left) % MOD

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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