Count All Valid Pickup and Delivery Options

Problem Statement

Given an integer n, we need to count all possible sequences consisting of n pickup services and n delivery services. Each order in our sequence has two components: a pickup labeled as P followed by a delivery labeled as D. The challenge is to ensure that the delivery (Di) for order i always occurs after its corresponding pickup (Pi). The sequence should include exactly n pickup and n delivery operations, strictly adhering to the order constraint for each pair.

Since the total number of valid sequences can be vast, particularly as n grows larger, the final count of sequences should be returned modulo (10^9 + 7) to manage large numbers effectively and handle potential overflow issues.

Examples

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

Approach and Intuition

The essence of this problem revolves around combinatorial arrangements with constraints. Each valid sequence must arrange pickups (P) and deliveries (D) such that no delivery occurs before its corresponding pickup. The following approach can be derived from the examples and constraints provided:

1. Understanding the sequence formation:

   • Observing that for a single order (n=1), the sequence is straightforward: P1, D1. This directly gives us a count of 1.
   • For n=2, the sequences multiply since for every pickup, 2 positions are possible for delivery excluding those violating the order constraint. Analyzing combinations gives a total of 6 possible sequences.

2. Generalizing for larger n:

   • For any given order i, once Pi is picked, the number of available spots for the delivery Di increases as more pickups are added to the sequence before Di is added. Every position for Di has to be after every other Pi.
   • This pattern follows a factorial growth modified by the sequence constraints, often computed using dynamic programming.

3. Efficiency consideration:

   • Given constraints (1 <= n <= 500), a direct combinatorial calculation or brute-force generation of sequences would be computationally infeasible. Dynamic programming or memoization should be used to store intermediate results and build upon smaller sub-problems.

Thus, the solution would involve recognizing patterns in sequence arrangement, leveraging factorial properties, and possibly utilizing dynamic computations to efficiently handle sequences up to a count of 500.

Solutions

class Solution {
public:
    int countOrders(int n) {
        long result = 1;
        int modulo = 1000000007;
            
        for (int i = 1; i <= 2 * n; ++i) {
            result = result * i;
                
            if (i % 2 == 0) {
                result /= 2;
            }
            result %= modulo;
        }
        return result;
    }
};