# 1359. Count All Valid Pickup and Delivery Options

## Description

Given n orders, each order consists of a pickup and a delivery service.

Count all valid pickup/delivery possible sequences such that delivery(i) is always after of pickup(i).

Since the answer may be too large, return it modulo 10^9 + 7.

Example 1:

Input: n = 1
Output: 1
Explanation: Unique order (P1, D1), Delivery 1 always is after of Pickup 1.


Example 2:

Input: n = 2
Output: 6
Explanation: All possible orders:
(P1,P2,D1,D2), (P1,P2,D2,D1), (P1,D1,P2,D2), (P2,P1,D1,D2), (P2,P1,D2,D1) and (P2,D2,P1,D1).
This is an invalid order (P1,D2,P2,D1) because Pickup 2 is after of Delivery 2.


Example 3:

Input: n = 3
Output: 90


Constraints:

• 1 <= n <= 500

## Solutions

Solution 1: Dynamic Programming

We define $f[i]$ as the number of all valid pickup/delivery sequences for $i$ orders. Initially, $f[1] = 1$.

We can choose any of these $i$ orders as the last delivery order $D_i$, then its pickup order $P_i$ can be at any position in the previous $2 \times i - 1$, and the number of pickup/delivery sequences for the remaining $i - 1$ orders is $f[i - 1]$, so $f[i]$ can be expressed as:

$f[i] = i \times (2 \times i - 1) \times f[i - 1]$

The final answer is $f[n]$.

We notice that the value of $f[i]$ is only related to $f[i - 1]$, so we can use a variable instead of an array to reduce the space complexity.

The time complexity is $O(n)$, where $n$ is the number of orders. The space complexity is $O(1)$.

• class Solution {
public int countOrders(int n) {
final int mod = (int) 1e9 + 7;
long f = 1;
for (int i = 2; i <= n; ++i) {
f = f * i * (2 * i - 1) % mod;
}
return (int) f;
}
}

• class Solution {
public:
int countOrders(int n) {
const int mod = 1e9 + 7;
long long f = 1;
for (int i = 2; i <= n; ++i) {
f = f * i * (2 * i - 1) % mod;
}
return f;
}
};

• class Solution:
def countOrders(self, n: int) -> int:
mod = 10**9 + 7
f = 1
for i in range(2, n + 1):
f = (f * i * (2 * i - 1)) % mod
return f


• func countOrders(n int) int {
const mod = 1e9 + 7
f := 1
for i := 2; i <= n; i++ {
f = f * i * (2*i - 1) % mod
}
return f
}

• const MOD: i64 = (1e9 as i64) + 7;

impl Solution {
pub fn count_orders(n: i32) -> i32 {
let mut f = 1;
for i in 2..=n as i64 {
f = (i * (2 * i - 1) * f) % MOD;
}
f as i32
}
}