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
/**
* @param {number} n
* @return {number}
*/
var countOrders = function(n) {
return helper(n, 0, 0, {});
};
var helper = function(n, x, y, dp) {
if (x === n && y === n) return 1;
var mod = Math.pow(10, 9) + 7;
var key = `${x}-${y}`;
if (dp[key] === undefined) {
var choosePickup = x < n ? ((n - x) * helper(n, x + 1, y, dp) % mod) : 0;
var chooseDelivery = y < n && x > y ? ((x - y) * helper(n, x, y + 1, dp) % mod) : 0;
dp[key] = (choosePickup + chooseDelivery) % mod;
}
return dp[key];
};
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).