Introduction

In this blog post, we will explore how to solve the LeetCode problem "1359" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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

Explanation

Here's the breakdown of the solution:

• Combinatorial Approach: For n orders, we have 2n slots to fill. We can choose 2 slots for the first order's pickup and delivery in (2n * (2n - 1)) / 2 ways, but since delivery must come after pickup, we divide by 2 (resulting in 2n * (2n - 1) / 2). After placing the first order, we have 2n - 2 slots remaining, and we repeat the process for the second order, and so on.

• Iterative Calculation: Instead of directly calculating combinations, we can express the number of ways as a product. For each order i from 1 to n, we have 2 * i - 1 available slots to place the pickup and delivery. The number of ways to arrange these is (2 * i) * (2 * i - 1) / 2, where we pick two slots. We multiply these values for each order to get the final result.

• Modulo Arithmetic: Since the result can be very large, we perform modulo operations at each step to prevent overflow.

• Runtime & Storage Complexity: The solution has a time complexity of O(n) because it iterates from 1 to n. The space complexity is O(1) because it uses a constant amount of extra space.

Code

    def countOrders(n: int) -> int:
    """
    Counts the number of valid pickup/delivery sequences such that delivery(i) is always after pickup(i).

    Args:
        n: The number of orders.

    Returns:
        The number of valid sequences modulo 10^9 + 7.
    """

    MOD = 10**9 + 7
    result = 1
    for i in range(1, n + 1):
        result = (result * (2 * i - 1) * i) % MOD
    return result