1359. Count All Valid Pickup and Delivery Options

Description

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

 

Solutions

Solution: Dynamic Programming

• Time complexity: O(n²)
• Space complexity: O(n²)

 

JavaScript

/**
 * @param {number} n
 * @return {number}
 */
const countOrders = function (n) {
  const MODULO = 10 ** 9 + 7;
  const dp = Array.from({ length: n + 1 }, () => new Array(n + 1).fill(0));

  const releaseOrder = (pickup, delivery) => {
    if (pickup === n && delivery === n) return 1;
    if (dp[pickup][delivery]) return dp[pickup][delivery];

    let result = 0;

    if (pickup < n) {
      const count = releaseOrder(pickup + 1, delivery);

      result = (result + count * (n - pickup)) % MODULO;
    }
    if (pickup > delivery) {
      const count = releaseOrder(pickup, delivery + 1);

      result = (result + count * (pickup - delivery)) % MODULO;
    }
    dp[pickup][delivery] = result;

    return result;
  };

  return releaseOrder(0, 0);
};