Count All Valid Pickup and Delivery Options

• Create a 2D array memo of size (n+1) x (n+1) and initialize it with a value indicating that the state has not been computed (e.g., 0 or -1).
• Define a recursive function solve(unpicked, undelivered):
  • If unpicked == 0 and undelivered == 0, it means we have successfully placed all items, so we return 1 (base case).
  • If unpicked < 0 or undelivered < 0, it's an invalid state, so return 0.
  • If memo[unpicked][undelivered] is already computed, return the stored value.
  • Calculate the number of ways by placing a pickup: res = (long)unpicked * solve(unpicked - 1, undelivered + 1). We have unpicked choices for the pickup.
  • Add the number of ways by placing a delivery: res += (long)undelivered * solve(unpicked, undelivered - 1). We have undelivered choices for the delivery.
  • Store the result modulo 10^9 + 7 in memo[unpicked][undelivered] and return it.
• The main function initiates the process by calling solve(n, 0).

We can think of building the sequence of 2n events from left to right. At each step, we have two choices: place a pickup or place a delivery.

Let's define a state by (unpicked, undelivered):

• unpicked: The number of orders for which we haven't placed the pickup yet.
• undelivered: The number of orders for which we have placed the pickup but not the delivery.

At any point, we have unpicked pickups we can place and undelivered deliveries we can place.

• Choice 1: Place a Pickup.

  • We have unpicked choices for which pickup to place (e.g., if P1, P2 are unpicked, we can choose either).
  • After placing a pickup, the number of unpicked orders decreases by 1, and the number of undelivered orders increases by 1.
  • The number of ways for this choice is unpicked * solve(unpicked - 1, undelivered + 1).

• Choice 2: Place a Delivery.

  • We can only place a delivery for an order whose pickup has already been placed. There are undelivered such orders.
  • After placing a delivery, the number of undelivered orders decreases by 1. The number of unpicked orders remains the same.
  • The number of ways for this choice is undelivered * solve(unpicked, undelivered - 1).

This leads to the following recurrence relation:

1solve(unpicked, undelivered) = (unpicked * solve(unpicked - 1, undelivered + 1)) + (undelivered * solve(unpicked, undelivered - 1))

The base case is solve(0, 0) = 1, as there is one way to arrange zero items. We start with the initial call solve(n, 0). To avoid recomputing the same states, we use a 2D array memo[n+1][n+1] for memoization.

1class Solution {2    long[][] memo;3    int MOD = 1_000_000_007;4 5    public int countOrders(int n) {6        memo = new long[n + 1][n + 1];7        return (int) solve(n, 0);8    }9 10    private long solve(int unpicked, int undelivered) {11        if (unpicked == 0 && undelivered == 0) {12            return 1;13        }14        if (unpicked < 0 || undelivered < 0) {15            return 0;16        }17        if (memo[unpicked][undelivered] != 0) {18            return memo[unpicked][undelivered];19        }20 21        long ans = 0;22 23        // Option 1: Pick up an order.24        // We have 'unpicked' choices.25        ans += (long) unpicked * solve(unpicked - 1, undelivered + 1);26        ans %= MOD;27 28        // Option 2: Deliver an order.29        // We have 'undelivered' choices.30        ans += (long) undelivered * solve(unpicked, undelivered - 1);31        ans %= MOD;32 33        return memo[unpicked][undelivered] = ans;34    }35}

• Let f(n) be the number of valid sequences for n orders.
• Base Case: f(1) = 1.
• Recurrence: f(n) = f(n-1) * n * (2n - 1).
• Iterative Calculation:
  • Initialize ans = 1 and MOD = 10^9 + 7.
  • Loop with a variable i from 1 to n.
  • In each iteration, update the answer by multiplying with the ways to place the i-th pair. The number of ways to place the i-th pickup and delivery among 2i total slots is i * (2i - 1).
  • Update the answer: ans = (ans * i * (2L * i - 1)) % MOD.
  • After the loop, ans will hold the result for f(n).

Let f(n) be the number of valid pickup and delivery sequences for n orders.

Base Case: For n=1, we have one sequence (P1, D1), so f(1) = 1.

Recurrence Relation: Let's find a formula for f(n) based on f(n-1). Assume we have a valid sequence for n-1 orders, which has 2(n-1) positions. Now, we need to insert the new pair, Pn and Dn. There are 2n total slots in the final sequence.

Consider the 2n available slots. We need to choose 2 slots for Pn and Dn. The number of ways to choose 2 slots is C(2n, 2). Once the slots are chosen, Pn must be in the first and Dn in the second to maintain validity. So there is only 1 way to place them for each pair of chosen slots. The remaining 2n-2 slots must be filled by the other n-1 pairs, which can be done in f(n-1) ways.

This gives the recurrence: f(n) = C(2n, 2) * f(n-1) f(n) = (2n * (2n-1) / 2) * f(n-1) f(n) = n * (2n-1) * f(n-1)

We can compute f(n) iteratively:

• f(1) = 1
• f(2) = f(1) * 2 * (2*2 - 1) = 1 * 2 * 3 = 6
• f(3) = f(2) * 3 * (2*3 - 1) = 6 * 3 * 5 = 90

This can be implemented with a simple loop.

Alternative Derivation The total number of permutations of 2n distinct items is (2n)!. For each of the n pairs (Pi, Di), Pi must come before Di. By symmetry, in exactly half of the permutations, Pi is before Di. Since there are n such independent constraints, we divide by 2 for each pair. Total valid sequences = (2n)! / 2^n. This can be calculated as (1 * 3 * 5 * ... * (2n-1)) * n!, which is equivalent to the iterative calculation from the recurrence.

1class Solution {2    public int countOrders(int n) {3        long ans = 1;4        int MOD = 1_000_000_007;5 6        for (int i = 1; i <= n; i++) {7            // For the i-th pair (Pi, Di), we have 2i-1 previous items.8            // Total slots available are 2i.9            // Ways to choose 2 slots for Pi and Di is C(2i, 2) = i * (2i - 1).10            // We multiply this with the arrangements of previous i-1 pairs.11            // ans for i = (ans for i-1) * i * (2i-1)12            ans = ans * i;13            ans %= MOD;14            ans = ans * (2L * i - 1);15            ans %= MOD;16        }17        18        return (int) ans;19    }20}