Problem
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
Solution
class Solution {
public int countOrders(int n) {
long[] dp = new long[n + 1];
dp[1] = 1;
long mod = (long) 1e9 + 7;
for (int i = 2; i <= n; i++) {
long gaps = (i - 1) * 2L + 1;
dp[i] = gaps * (gaps + 1) / 2 * dp[i - 1] % mod;
}
return (int) dp[n];
}
}
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).