1359. Count All Valid Pickup and Delivery Options

Problem Description

The given problem asks for the number of valid sequences of pickup and delivery operations for n orders. Each order consists of two operations: a pickup (P) and a delivery (D). For each order, the delivery must always occur after the corresponding pickup. The goal is to determine how many different sequences there could be for the n orders, considering this constraint. Finally, because the answer could be very large, the problem instructs us to return the result modulo 10^9 + 7, which is a common practice to keep the numbers within the limits of integer storage in most programming languages.

To illustrate, let's consider there are 2 orders. The valid sequences we can arrange them in, while keeping pickups before deliveries, are:

• P1 P2 D1 D2
• P1 P2 D2 D1
• P1 D1 P2 D2
• P2 P1 D1 D2
• P2 P1 D2 D1
• P2 D2 P1 D1

Notice that as n increases, the number of sequences grows very fast and calculating them one by one might not be feasible, thus a more mathematical or algorithmic approach is required to solve the problem efficiently.

Intuition

The intuition behind the solution comes from considering the possibilities of inserting the i-th pickup and delivery operations into a sequence that has already been arranged for i-1 orders.

At any stage, let's say we have arranged i-1 orders. Now, we intend to insert the i-th order's pickup and delivery operations into this sequence. For the i-th pickup, there are 2 * (i - 1) + 1 different places to put the pickup operation. This is because for each of the previous i-1 orders, there are two operations (a pickup and a delivery), plus one additional place for the start of the sequence. After we've put in the pickup, we need to insert the delivery. Since delivery must come after pickup, there are only i spots where it could go—after each of the delivery and pickup operations that have already been placed.

Mathematically, this means that for the i-th order, there are i spaces for the delivery and 2 * (i - 1) + 1 spaces for the pickup, giving a total of i * (2 * (i - 1) + 1) positions to insert both P and D.

So, the total number of arrangements for all n orders can be calculated by multiplying these numbers for all i from 1 to n. The provided solution uses a for loop to carry out this computation, while also applying the modulo operator at each step to prevent the number from getting too large.

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Solution Approach

The solution provided is a direct application of the mathematical intuition discussed earlier. The solution uses no special data structures—the only structure used is a simple loop, and the approach is entirely iterative and mathematical. Here's the breakdown:

The function countOrders calculates the number of sequences dynamically by using a single integer f to keep track of the result, which is initialized to 1 (since there's only one way to arrange one order).

Within the for loop, for each i from 2 to n inclusive, we calculate the number of possible ways to insert the i-th order into the sequences created for i-1 orders. As explained before, for pickup there are 2 * (i - 1) + 1 options, and for delivery, there are i options afterwards. We then multiply f by these numbers to update the total possible arrangements:

1f = (f * i * (2 * i - 1)) % mod

The mod variable is set to 10**9 + 7, which is a large prime number, and is used here to perform modulo operation to ensure the result stays within integer limits and to avoid overflow.

After the loop ends, f will contain the number of valid sequences modulo 10**9 + 7, and this is exactly what the function returns.

The entire implementation is quite efficient due to its simplicity and directness; it runs in O(n) time complexity because it processes each of the n orders exactly once. Since it only uses a few simple integer variables, its space complexity is O(1).

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

Each iteration computes the possibilities for insertion without building the sequences explicitly, resulting in an elegant and highly efficient solution that scales well with the value of n.

Example Walkthrough

Let's walk through an example using the solution approach for n = 3, meaning we have three orders to sequence.

From the initial explanation, we know we have these constraints:

• For each order, the pickup must occur before the delivery.
• We need to consider all valid combinations of pick up and delivery operations for the given orders.

To begin with the first order (when n = 1), we have no choice other than sequencing it as P1 D1. This is our base case and, intuitively, there's only 1 way to arrange one order.

Now, for the second order (when n = 2), we have to insert P2 and D2 into the existing sequence P1 D1.

• We have three possible spots to insert P2: before P1, between P1 and D1, or after D1.
• Once P2 is inserted, we can insert D2 in two places: anywhere after P2.

Thus, we have 3 (spots for P2) * 2 (spots for D2) = 6 different sequences for two orders.

Now we go to three orders (our example case). Suppose we have already arranged the first two orders (2 pickups and 2 deliveries), making a sequence with four operations. Now we consider inserting P3 and D3.

• We have five spots to insert P3: before the first operation, between any of the two existing operations, or after the last operation in our sequence.
• After placing P3, we have three spots to insert D3, all of them coming after P3.

The total possibilities for n = 3 now are 5 (spots for P3) * 3 (spots for D3) = 15 which we would multiply with the previous possibilities for n = 2 (which were 6).

However, we've missed an important step: taking the modulo to avoid large numbers and potential integer overflow issues. This is done after each multiplication to keep the result manageable.

In code, this is how the process would look:

1mod = 10**9 + 7
2f = 1  # Starting with the case n=1, f starts at 1.
3
4# Now we loop from i=2 to i=3, as we're considering the third order.
5for i in range(2, 4):
6    # Multiply the current number of ways `f` with the possible spots for P and D
7    f = (f * i * (2 * i - 1)) % mod
8
9# By the end of the loop, `f` is our desired output for `n = 3`.

If we run this code, after the iteration for i = 2, f becomes 6, and after the iteration for i = 3, f becomes: f = (6 * 3 * (2 * 3 - 1)) % mod = (6 * 3 * 5) % mod = 90 % mod = 90.

Thus, f is 90, indicating that there are 90 possible sequences for arranging the 3 orders while considering the modulo 10**9 + 7. Since 90 is already less than the modulo, taking the modulo does not change the number. The function would then return 90 as the final result.

Solution Implementation

Time and Space Complexity

Time Complexity: The time complexity of the provided code is O(n) because there is a single loop that iterates from 2 to n. In each iteration, it performs a constant number of mathematical operations which do not depend on n, thus the time complexity is linear with respect to the input size n.

Space Complexity: The space complexity of the code is O(1) as it uses a fixed amount of additional space. There are only a few variables (mod, f, and i) that do not scale with the input size n, hence the space used is constant.

Learn more about how to find time and space complexity quickly using problem constraints.