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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 1: Dynamic Programming

We define \(f[i]\) as the number of all valid pickup/delivery sequences for \(i\) orders. Initially, \(f[1] = 1\).

We can choose any of these \(i\) orders as the last delivery order \(D_i\), then its pickup order \(P_i\) can be at any position in the previous \(2 \times i - 1\), and the number of pickup/delivery sequences for the remaining \(i - 1\) orders is \(f[i - 1]\), so \(f[i]\) can be expressed as:

\[ f[i] = i \times (2 \times i - 1) \times f[i - 1] \]

The final answer is \(f[n]\).

We notice that the value of \(f[i]\) is only related to \(f[i - 1]\), so we can use a variable instead of an array to reduce the space complexity.

The time complexity is \(O(n)\), where \(n\) is the number of orders. The space complexity is \(O(1)\).

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class Solution:
    def countOrders(self, n: int) -> int:
        mod = 10**9 + 7
        f = 1
        for i in range(2, n + 1):
            f = (f * i * (2 * i - 1)) % mod
        return f

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class Solution {
    public int countOrders(int n) {
        final int mod = (int) 1e9 + 7;
        long f = 1;
        for (int i = 2; i <= n; ++i) {
            f = f * i * (2 * i - 1) % mod;
        }
        return (int) f;
    }
}

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class Solution {
public:
    int countOrders(int n) {
        const int mod = 1e9 + 7;
        long long f = 1;
        for (int i = 2; i <= n; ++i) {
            f = f * i * (2 * i - 1) % mod;
        }
        return f;
    }
};

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func countOrders(n int) int {
    const mod = 1e9 + 7
    f := 1
    for i := 2; i <= n; i++ {
        f = f * i * (2*i - 1) % mod
    }
    return f
}

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const MOD: i64 = (1e9 as i64) + 7;

impl Solution {
    #[allow(dead_code)]
    pub fn count_orders(n: i32) -> i32 {
        let mut f = 1;
        for i in 2..=n as i64 {
            f = (i * (2 * i - 1) * f) % MOD;
        }
        f as i32
    }
}

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