1359

1359. Count All Valid Pickup and Delivery Options (Hard)

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

Companies: DoorDash, Apple, Google

Related Topics:
Math, Dynamic Programming, Combinatorics

Solution 1.

Let f(n) be the answer.

It's trivial that f(1) = 1

For n = 2, the single arrangement of f(1) gives us 3 spots to put P2.

• If P2 is put at the first spot, D2 has 3 spots to be put.
• If P2 is put at the 2nd spot, D2 has 2 spots to be put.
• If P2 is put at the 3rd spot, D2 has 1 spot to be put.

So f(2) = 3 + 2 + 1 = 6 or f(2) = (1 + N) * N / 2 * f(1) where N = 3 = 2 * 2 - 1.

By induction, we have f(n) = f(n - 1) * (1 + N) * N / 2 where N = 2n - 1

// OJ: https://leetcode.com/problems/count-all-valid-pickup-and-delivery-options/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    int countOrders(int n) {
        long ans = 1, mod = 1e9 + 7;
        for (int i = 2, cnt = 3; i <= n; ++i, cnt += 2) {
            ans = ans * cnt * (cnt + 1) / 2 % mod;
        }
        return ans;
    }
};