LeetCode 1359. Count All Valid Pickup and Delivery Options

原题链接在这里：https://leetcode.com/problems/count-all-valid-pickup-and-delivery-options/description/

题目：

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

题解：

Consider we already have n - 1 pairs sequence, for the nth pair, pickup we have 2 * n - 1 positions. delivery we have 2 * n positions.

e.g. n - 1 = 1, we have p1, d1 sequence. For n = 2, we have (2 * 2 -1) = 3 positions to place p2.

_, p1, d1,

p1, _, d1,

p1, d1, _,

For d2, assume, we already placed p2, we have 2 * 2 = 4 positions to place d2.

assume, we put p2, p1, d1. there are *, p2, * p1, *, d1, * options. * represent the possible places to place d2.

So, we have (2 * n - 1) * 2 * n permutations. But d2 must come after p2, so need to divide by 2 and we get (2 * n - 1) * n.

Time Complexity: O(n).

Space: O(1).

AC Java:

 1 class Solution {
 2     public int countOrders(int n) {
 3         long mod = (long)1e9 + 7;
 4         long res = 1;
 5         for(int i = 2; i <= n; i++){
 6             res = res * (i * 2 - 1) * i % mod;
 7         }
 8 
 9         return (int)res;
10     }
11 }