1359. Count All Valid Pickup and Delivery Options

Hard

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Given n orders, each order consist in pickup and delivery services.

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

class Solution:
    def countOrders(self, n: int) -> int:
        MOD = 10**9+7
        pickup_pos = math.factorial(n) % MOD
        delivery_pos = 1
        for i in range(1, n+1):
            delivery_pos *= (2*i-1)
        return pickup_pos*delivery_pos % MOD