hard Count All Valid Pickup and Delivery Options

Problem Statement

You are given a positive integer n, representing total orders where each order consists of a pickup and a delivery service.

Count all valid pickup/delivery sequences such that delivery(x) is always after of pickup(x).

Return the answer modulo 10^9 + 7.

Examples

• Example 1:

  • Input: n = 1
  • Expected Output: 1
  • Justification: With just one order, there's only one valid sequence: pickup and then delivery. Thus, the expected output is 1.

• Example 2:

  • Input: n = 2
  • Expected Output: 6
  • Justification: With two orders, there are six valid sequences: P1 D1 P2 D2, P1 P2 D1 D2, P1 P2 D2 D1, P2 D2 P1 D1, P2 P1 D1 D2, P2 P1 D2 D1. Therefore, the expected output is 6.

• Example 3:

  • Input: n = 5
  • Expected Output: 113400
  • Justification: For five orders, the number of valid sequences increases significantly because each additional order adds complexity to how pickups and deliveries can be arranged. The expected output for three orders is 113400.

Solution

To solve this problem, we'll use a dynamic programming approach. This method is chosen because it allows us to build the solution for n orders based on the solutions for fewer orders, thus reducing the overall computational complexity. The key insight is to recognize that for each new order, we can insert its pickup and delivery actions into the sequences generated for the previous orders in a specific number of ways. Specifically, when adding the nth order, we can insert its pickup anywhere in the sequence of the previous n-1 orders, and its delivery can then be placed in any position following its pickup. This results in a pattern where the number of sequences grows significantly with each additional order, but each step can be calculated from the previous one.

The effectiveness of this approach lies in its ability to break down the problem into smaller, manageable pieces, leveraging the fact that the solution for n orders is built upon the solution for n-1 orders. By systematically expanding our calculation with each additional order, we ensure that all valid sequences are counted without redundant computations or overlooking any possibilities.

Step-by-Step Algorithm

1. Initialization:

   • Declare a MOD constant with the value of (10^9 + 7) to handle large numbers during calculations.
   • Initialize a long variable count to 1. This variable will hold the cumulative result of the calculations for the number of valid sequences.

2. Loop Through Orders:

   • Iterate from 1 through n (inclusive), with i representing the current order number. This loop calculates the cumulative number of valid sequences for each order.
     • For each i:
       • First, update count to account for the new combinations introduced by the addition of the ith pickup. Multiply count by i and apply the modulo MOD to ensure the result stays within bounds. This step accounts for the different permutations of the orders themselves.
       • Next, update count again to incorporate the positions where the ith delivery can be inserted. Multiply count by (2 * i - 1) and apply the modulo MOD. This step considers the specific requirement that each delivery follows its corresponding pickup and calculates the number of ways the delivery can be positioned in relation to the pickups and other deliveries already considered.

3. Finalize and Return Result:

   • The loop concludes with count holding the total number of valid sequences, modulo (10^9 + 7).
   • Cast count to an int (as per the method's return type) and return it.

Algorithm Walkthrough

Let's consider the Input: n = 5

• Initialization:

  • MOD is set to (10^9 + 7).
  • count is initialized to 1.

• Loop Through Orders:

  • For i = 1:
    • Multiply count by i (1), count = 1.
    • Multiply count by (2*1 - 1) (1), count = 1.
  • For i = 2:
    • Multiply count by i (2), count = 2.
    • Multiply count by (2*2 - 1) (3), count = 6.
  • For i = 3:
    • Multiply count by i (3), count = 18.
    • Multiply count by (2*3 - 1) (5), count = 90.
  • For i = 4:
    • Multiply count by i (4), count = 360.
    • Multiply count by (2*4 - 1) (7), count = 2520.
  • For i = 5:
    • Multiply count by i (5), count = 12600.
    • Multiply count by (2*5 - 1) (9), count = 113400.

• Finalize and Return Result:

  • At the end of the loop, count equals 113400.

Code

public class Solution {

  private static final int MOD = 1000000007;

  public int countOrders(int n) {
    long count = 1;
    for (int i = 1; i <= n; i++) {
      count = (count * i) % MOD; // Ways to arrange pickups
      count = (count * (2 * i - 1)) % MOD; // Ways to arrange deliveries with respect to pickups
    }
    return (int) count;
  }

  public static void main(String[] args) {
    Solution solution = new Solution();
    // Example 1
    System.out.println(solution.countOrders(1)); // Expected: 1
    // Example 2
    System.out.println(solution.countOrders(2)); // Expected: 6
    // Example 3
    System.out.println(solution.countOrders(5)); // Expected: 113400
  }
}

Complexity Analysis

Time Complexity

• : The primary operation in the algorithm is a loop that runs n times, where n is the number of orders. Within this loop, we perform a constant number of mathematical operations (multiplications and modulo operations), which do not depend on the size of n beyond the loop's iteration. Therefore, the overall time complexity is linear with respect to the input size.

Space Complexity

• : Our algorithm uses a fixed amount of space regardless of the input size. The variables used to calculate the count of valid sequences (including the MOD constant and loop counter) do not scale with n. Thus, the space complexity is constant.