# 2954. Count the Number of Infection Sequences

## Description

You are given an integer n and a 0-indexed integer array sick which is sorted in increasing order.

There are n children standing in a queue with positions 0 to n - 1 assigned to them. The array sick contains the positions of the children who are infected with an infectious disease. An infected child at position i can spread the disease to either of its immediate neighboring children at positions i - 1 and i + 1 if they exist and are currently not infected. At most one child who was previously not infected can get infected with the disease in one second.

It can be shown that after a finite number of seconds, all the children in the queue will get infected with the disease. An infection sequence is the sequential order of positions in which all of the non-infected children get infected with the disease. Return the total number of possible infection sequences.

Since the answer may be large, return it modulo 109 + 7.

Note that an infection sequence does not contain positions of children who were already infected with the disease in the beginning.

Example 1:

Input: n = 5, sick = [0,4]
Output: 4
Explanation: Children at positions 1, 2, and 3 are not infected in the beginning. There are 4 possible infection sequences:
- The children at positions 1 and 3 can get infected since their positions are adjacent to the infected children 0 and 4. The child at position 1 gets infected first.
Now, the child at position 2 is adjacent to the child at position 1 who is infected and the child at position 3 is adjacent to the child at position 4 who is infected, hence either of them can get infected. The child at position 2 gets infected.
Finally, the child at position 3 gets infected because it is adjacent to children at positions 2 and 4 who are infected. The infection sequence is [1,2,3].
- The children at positions 1 and 3 can get infected because their positions are adjacent to the infected children 0 and 4. The child at position 1 gets infected first.
Now, the child at position 2 is adjacent to the child at position 1 who is infected and the child at position 3 is adjacent to the child at position 4 who is infected, hence either of them can get infected. The child at position 3 gets infected.
Finally, the child at position 2 gets infected because it is adjacent to children at positions 1 and 3 who are infected. The infection sequence is [1,3,2].
- The infection sequence is [3,1,2]. The order of infection of disease in the children can be seen as: [0,1,2,3,4] => [0,1,2,3,4] => [0,1,2,3,4] => [0,1,2,3,4].
- The infection sequence is [3,2,1]. The order of infection of disease in the children can be seen as: [0,1,2,3,4] => [0,1,2,3,4] => [0,1,2,3,4] => [0,1,2,3,4].


Example 2:

Input: n = 4, sick = [1]
Output: 3
Explanation: Children at positions 0, 2, and 3 are not infected in the beginning. There are 3 possible infection sequences:
- The infection sequence is [0,2,3]. The order of infection of disease in the children can be seen as: [0,1,2,3] => [0,1,2,3] => [0,1,2,3] => [0,1,2,3].
- The infection sequence is [2,0,3]. The order of infection of disease in the children can be seen as: [0,1,2,3] => [0,1,2,3] => [0,1,2,3] => [0,1,2,3].
- The infection sequence is [2,3,0]. The order of infection of disease in the children can be seen as: [0,1,2,3] => [0,1,2,3] => [0,1,2,3] => [0,1,2,3].


Constraints:

• 2 <= n <= 105
• 1 <= sick.length <= n - 1
• 0 <= sick[i] <= n - 1
• sick is sorted in increasing order.

## Solutions

Solution 1: Combinatorial Mathematics + Multiplicative Inverse + Fast Power

According to the problem description, the children who have a cold have divided the children who have not yet caught a cold into several continuous segments. We can use an array $nums$ to record the number of children who are not cold in each segment, and there are a total of $s = \sum_{i=0}^{k} nums[k]$ children who are not cold. We can find that the number of cold sequences is the number of permutations of $s$ different elements, that is, $s!$.

Assuming that there is only one transmission scheme for each segment of children who are not cold, there are $\frac{s!}{\prod_{i=0}^{k} nums[k]!}$ cold sequences in total.

Next, we consider the transmission scheme of each segment of children who are not cold. Suppose there are $x$ children in a segment who are not cold, then they have $2^{x-1}$ transmission schemes, because each time you can choose one end from the left and right ends of a segment to transmit, that is: two choices, there are a total of $x-1$ transmissions. However, if it is the first segment or the last segment, there is only one choice.

In summary, the total number of cold sequences is:

$\frac{s!}{\prod_{i=0}^{k} nums[k]!} \prod_{i=1}^{k-1} 2^{nums[i]-1}$

Finally, we need to consider that the answer may be very large and need to be modulo $10^9 + 7$. Therefore, we need to preprocess the factorial and multiplicative inverse.

The time complexity is $O(m)$, where $m$ is the length of the array $sick$. Ignoring the space consumption of the preprocessing array, the space complexity is $O(m)$.

• class Solution {
private static final int MOD = (int) (1e9 + 7);
private static final int MX = 100000;
private static final int[] FAC = new int[MX + 1];

static {
FAC[0] = 1;
for (int i = 1; i <= MX; i++) {
FAC[i] = (int) ((long) FAC[i - 1] * i % MOD);
}
}

public int numberOfSequence(int n, int[] sick) {
int m = sick.length;
int[] nums = new int[m + 1];
nums[0] = sick[0];
nums[m] = n - sick[m - 1] - 1;
for (int i = 1; i < m; i++) {
nums[i] = sick[i] - sick[i - 1] - 1;
}
int s = 0;
for (int x : nums) {
s += x;
}
int ans = FAC[s];
for (int x : nums) {
if (x > 0) {
ans = (int) ((long) ans * qpow(FAC[x], MOD - 2) % MOD);
}
}
for (int i = 1; i < nums.length - 1; ++i) {
if (nums[i] > 1) {
ans = (int) ((long) ans * qpow(2, nums[i] - 1) % MOD);
}
}
return ans;
}

private int qpow(long a, long n) {
long ans = 1;
for (; n > 0; n >>= 1) {
if ((n & 1) == 1) {
ans = ans * a % MOD;
}
a = a * a % MOD;
}
return (int) ans;
}
}

• const int MX = 1e5;
const int MOD = 1e9 + 7;
int fac[MX + 1];

auto init = [] {
fac[0] = 1;
for (int i = 1; i <= MX; ++i) {
fac[i] = 1LL * fac[i - 1] * i % MOD;
}
return 0;
}();

int qpow(long long a, long long n) {
long long ans = 1;
for (; n > 0; n >>= 1) {
if (n & 1) {
ans = (ans * a) % MOD;
}
a = (a * a) % MOD;
}
return ans;
}

class Solution {
public:
int numberOfSequence(int n, vector<int>& sick) {
int m = sick.size();
vector<int> nums(m + 1);

nums[0] = sick[0];
nums[m] = n - sick[m - 1] - 1;
for (int i = 1; i < m; i++) {
nums[i] = sick[i] - sick[i - 1] - 1;
}

int s = accumulate(nums.begin(), nums.end(), 0);
long long ans = fac[s];
for (int x : nums) {
if (x > 0) {
ans = ans * qpow(fac[x], MOD - 2) % MOD;
}
}
for (int i = 1; i < nums.size() - 1; ++i) {
if (nums[i] > 1) {
ans = ans * qpow(2, nums[i] - 1) % MOD;
}
}
return ans;
}
};

• mod = 10**9 + 7
mx = 10**5
fac = [1] * (mx + 1)
for i in range(2, mx + 1):
fac[i] = fac[i - 1] * i % mod

class Solution:
def numberOfSequence(self, n: int, sick: List[int]) -> int:
nums = [b - a - 1 for a, b in pairwise([-1] + sick + [n])]
ans = 1
s = sum(nums)
ans = fac[s]
for x in nums:
if x:
ans = ans * pow(fac[x], mod - 2, mod) % mod
for x in nums[1:-1]:
if x > 1:
ans = ans * pow(2, x - 1, mod) % mod
return ans


• const MX = 1e5
const MOD = 1e9 + 7

var fac [MX + 1]int

func init() {
fac[0] = 1
for i := 1; i <= MX; i++ {
fac[i] = fac[i-1] * i % MOD
}
}

func qpow(a, n int) int {
ans := 1
for n > 0 {
if n&1 == 1 {
ans = (ans * a) % MOD
}
a = (a * a) % MOD
n >>= 1
}
return ans
}

func numberOfSequence(n int, sick []int) int {
m := len(sick)
nums := make([]int, m+1)

nums[0] = sick[0]
nums[m] = n - sick[m-1] - 1
for i := 1; i < m; i++ {
nums[i] = sick[i] - sick[i-1] - 1
}

s := 0
for _, x := range nums {
s += x
}
ans := fac[s]
for _, x := range nums {
if x > 0 {
ans = ans * qpow(fac[x], MOD-2) % MOD
}
}
for i := 1; i < len(nums)-1; i++ {
if nums[i] > 1 {
ans = ans * qpow(2, nums[i]-1) % MOD
}
}
return ans
}

• const MX = 1e5;
const MOD: bigint = BigInt(1e9 + 7);
const fac: bigint[] = Array(MX + 1);

const init = (() => {
fac[0] = 1n;
for (let i = 1; i <= MX; ++i) {
fac[i] = (fac[i - 1] * BigInt(i)) % MOD;
}
return 0;
})();

function qpow(a: bigint, n: number): bigint {
let ans = 1n;
for (; n > 0; n >>= 1) {
if (n & 1) {
ans = (ans * a) % MOD;
}
a = (a * a) % MOD;
}
return ans;
}

function numberOfSequence(n: number, sick: number[]): number {
const m = sick.length;
const nums: number[] = Array(m + 1);
nums[0] = sick[0];
nums[m] = n - sick[m - 1] - 1;
for (let i = 1; i < m; i++) {
nums[i] = sick[i] - sick[i - 1] - 1;
}

const s = nums.reduce((acc, x) => acc + x, 0);
let ans = fac[s];
for (let x of nums) {
if (x > 0) {
ans = (ans * qpow(fac[x], Number(MOD) - 2)) % MOD;
}
}
for (let i = 1; i < nums.length - 1; ++i) {
if (nums[i] > 1) {
ans = (ans * qpow(2n, nums[i] - 1)) % MOD;
}
}
return Number(ans);
}