1359 - Count All Valid Pickup and Delivery Options (Hard)

Problem Link

https://leetcode.com/problems/count-all-valid-pickup-and-delivery-options/

Problem Statement

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

Approach 1: Math

If we just put all $P$ in a row, we would have $N!$ ways. It's a permutation with no repetition. The first choice has $N$ possibilities, and the next choice has $N - 1$ possibilities (as you cannot choose the first choice), and then $N - 2$ (as you cannot choose the previous two choices), $N - 3$ and so on. Hence, we got $N * (N - 1) * (N - 2) * .. * 1 = N!$.

Then we need to think how to put $D_i$in some possible places. We know that$D_i$ must come after $P_i$. It's obvious that there is only one way to put $D$ for the last $P$. For the second $D$, we can put it right after the corresponding $P$ or put it to the left / right of the previous $D$. Hence, we have $3$ possible places to put. If you keep doing the same thing, you should find the number of ways to put $D$ is $1 * 3 * 5 * .. * (2N - 1)$.

Hence, the answer is $N! * \sum_{i=1}^{N} (2 * i - 1)$. Remember to take $MOD$ during the calculation.

C++

Written by @wingkwong

class Solution {
public:
    int countOrders(int n) {
        long long M = 1e9 + 7, ans = 1;
        for (int i = 1; i <= n; i++) {
            // number of ways to put P: N!
            (ans *= i) %= M;
            // number of ways to put D after P: 1 * 3 * 5 * .. * (2N - 1)
            (ans *= (2 * i - 1)) %= M;
        }
        return ans;
    }
};

Java

Written by @wingkwong

class Solution {
    public int countOrders(int n) {
        long M = (long) 1e9 + 7;
        long ans = 1;
        for (int i = 1; i <= n; i++) {
            // number of ways to put P: N!
            ans = (ans * i) % M;
            // number of ways to put D after P: 1 * 3 * 5 * .. * (2N - 1)
            ans = (ans * (2 * i - 1)) % M;
        }
        return (int) ans;
    }
}

Kotlin

Written by @wingkwong

class Solution {
    fun countOrders(n: Int): Int {
        val M = (1e9 + 7).toLong()
        var ans = 1L
        for (i in 1..n) {
            // number of ways to put P: N!
            ans = (ans * i) % M
            // number of ways to put D after P: 1 * 3 * 5 * .. * (2N - 1)
            ans = (ans * (2 * i - 1)) % M
        }
        return ans.toInt()
    }
}