What is the core pattern of the Count All Valid Pickup and Delivery Options problem?

The core pattern involves dynamic programming with state transitions and combinatorics to compute valid sequences while considering the constraints of each pickup and delivery pair.

How does dynamic programming help in solving this problem?

Dynamic programming is used to build up solutions for smaller subproblems, ensuring that all valid sequences are counted by transitioning states as new (pickup, delivery) pairs are added.

Why is modulo 10^9 + 7 important in this problem?

Modulo 10^9 + 7 is crucial to ensure that the result remains within the problem's constraints, avoiding overflow and keeping computations manageable.

What combinatorial concepts are used in this problem?

Permutation and combination theory is used to calculate the number of valid ways to insert each new (pickup, delivery) pair into an existing sequence.

How can GhostInterview assist in preparing for this dynamic programming problem?

GhostInterview provides step-by-step guidance and feedback on implementing dynamic programming approaches, helping candidates handle transitions, optimizations, and modulo operations effectively.

#1359

Hard

auto_awesome状态·转移·动态规划

LeetCode 题解工作台

有效的快递序列数目

给你 n 笔订单，每笔订单都需要快递服务。计算所有有效的取货 / 交付可能的顺序，使 delivery(i) 总是在 pickup(i) 之后。由于答案可能很大，请返回答案对 10^9 + 7 取余的结果。示例 1：输入： n = 1 输出： 1 解释：只有一种序列 (P1, D1)，…

数学动态规划组合数学

题目描述

给你 n 笔订单，每笔订单都需要快递服务。

计算所有有效的取货 / 交付可能的顺序，使 delivery(i) 总是在 pickup(i) 之后。

由于答案可能很大，请返回答案对 10^9 + 7 取余的结果。

示例 1：

输入：n = 1
输出：1
解释：只有一种序列 (P1, D1)，物品 1 的配送服务（D1）在物品 1 的收件服务（P1）后。

示例 2：

输入：n = 2
输出：6
解释：所有可能的序列包括：
(P1,P2,D1,D2)，(P1,P2,D2,D1)，(P1,D1,P2,D2)，(P2,P1,D1,D2)，(P2,P1,D2,D1) 和 (P2,D2,P1,D1)。
(P1,D2,P2,D1) 是一个无效的序列，因为物品 2 的收件服务（P2）不应在物品 2 的配送服务（D2）之后。

示例 3：

输入：n = 3
输出：90

提示：

1 <= n <= 500

lightbulb

解题思路

方法一：动态规划

我们定义 $f[i]$ 表示 $i$ 个订单的所有有效的收件/配送序列的数目。初始时 $f[1] = 1$ 。

我们可以选择这 $i$ 个订单中的任意一个作为最后一个配送的订单 $D_i$ ，那么它的收件订单 $P_i$ 可以在之前 $2 \times i - 1$ 的任意一个位置，剩下的 $i - 1$ 个订单的配送/收件序列数目为 $f[i - 1]$ ，所以 $f[i]$ 可以表示为：

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

最终的答案即为 $f[n]$ 。

我们注意到 $f[i]$ 的值只与 $f[i - 1]$ 有关，所以可以用一个变量代替数组，降低空间复杂度。

时间复杂度 $O(n)$ ，其中 $n$ 为订单数目。空间复杂度 $O(1)$ 。

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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

speed

复杂度分析

指标	值
时间	Depends on the final approach
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

question_mark
Candidate should demonstrate a strong understanding of dynamic programming and combinatorics.
question_mark
Watch for candidates applying permutations and combinations to efficiently calculate the number of valid sequences.
question_mark
Test if the candidate can explain and handle modulo operations during the computation process.

warning

常见陷阱

外企场景

error
Overcomplicating the problem by generating all sequences explicitly instead of using dynamic programming.
error
Forgetting to account for the modulo 10^9 + 7 operation, leading to incorrect results due to overflow.
error
Misunderstanding the state transition process and failing to correctly compute valid transitions between pickup and delivery pairs.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Extending the problem to account for more than one delivery per pickup, increasing the complexity.
arrow_right_alt
Changing the constraints, such as limiting n to smaller values or increasing the range of possible values for n.
arrow_right_alt
Modifying the problem to work with different numbers of pickups and deliveries, not tied to a one-to-one correspondence.

help

常见问题

外企场景

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