
Problem Statement
Given an integer n
, we need to count all possible sequences consisting of n
pickup services and n
delivery services. Each order in our sequence has two components: a pickup labeled as P
followed by a delivery labeled as D
. The challenge is to ensure that the delivery (Di
) for order i
always occurs after its corresponding pickup (Pi
). The sequence should include exactly n
pickup and n
delivery operations, strictly adhering to the order constraint for each pair.
Since the total number of valid sequences can be vast, particularly as n
grows larger, the final count of sequences should be returned modulo (10^9 + 7) to manage large numbers effectively and handle potential overflow issues.
Examples
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 and Intuition
The essence of this problem revolves around combinatorial arrangements with constraints. Each valid sequence must arrange pickups (P
) and deliveries (D
) such that no delivery occurs before its corresponding pickup. The following approach can be derived from the examples and constraints provided:
Understanding the sequence formation:
- Observing that for a single order (
n=1
), the sequence is straightforward:P1, D1
. This directly gives us a count of 1. - For
n=2
, the sequences multiply since for every pickup, 2 positions are possible for delivery excluding those violating the order constraint. Analyzing combinations gives a total of 6 possible sequences.
- Observing that for a single order (
Generalizing for larger
n
:- For any given order
i
, oncePi
is picked, the number of available spots for the deliveryDi
increases as more pickups are added to the sequence beforeDi
is added. Every position forDi
has to be after every otherPi
. - This pattern follows a factorial growth modified by the sequence constraints, often computed using dynamic programming.
- For any given order
Efficiency consideration:
- Given constraints (
1 <= n <= 500
), a direct combinatorial calculation or brute-force generation of sequences would be computationally infeasible. Dynamic programming or memoization should be used to store intermediate results and build upon smaller sub-problems.
- Given constraints (
Thus, the solution would involve recognizing patterns in sequence arrangement, leveraging factorial properties, and possibly utilizing dynamic computations to efficiently handle sequences up to a count of 500.
Solutions
- C++
- Java
- JavaScript
- Python
class Solution {
public:
int countOrders(int n) {
long result = 1;
int modulo = 1000000007;
for (int i = 1; i <= 2 * n; ++i) {
result = result * i;
if (i % 2 == 0) {
result /= 2;
}
result %= modulo;
}
return result;
}
};
The solution addresses the problem of counting all valid pickup and delivery options using C++. The given implementation efficiently calculates the number of ways to organize pickups and deliveries with the constraints that a delivery cannot happen before its corresponding pickup.
- Initialize a variable
result
set to 1 to hold the final number of permutations. - Define
modulo
with the value1000000007
which is commonly used to prevent overflow in problems requiring large number computations. - Loop through numbers from 1 to
2 * n
, wheren
is the number of pickup/delivery pairs:- Multiply
result
by the current loop variablei
to account for additional permutations as more orders are considered. - Check if
i
is even which corresponds to the point where both a pickup and its delivery have been accounted for. At this point, divideresult
by 2 to correct for the overcounting of arrangements where order is not a factor. - Apply the
modulo
operation to keep the result manageable and within the bounds of typical integer sizes.
- Multiply
- Return the computed value of
result
.
This solution takes full advantage of modular arithmetic and combinatorial logic to efficiently handle large values and returns the number of valid sequences of pickup and delivery events modulo 1000000007
.
class Solution {
public int calculateOrders(int total) {
long result = 1;
final int MODULO = 1_000_000_007;
for (int j = 1; j <= 2 * total; ++j) {
result *= j;
// Adjust result by dividing by 2 every time j is even.
if (j % 2 == 0) {
result /= 2;
}
result %= MODULO;
}
return (int)result;
}
}
The Java solution provided calculates the number of valid pickup and delivery options where orders must be picked up before they can be delivered. Follow the steps and overview below to understand the implementation details:
Declare and initialize a
long
variableresult
to 1, which will hold the final computed number of ways.Define a constant
MODULO
with the value1_000_000_007
(a large prime number) to ensure that the result remains within manageable limits by taking modulo at each step.Loop from 1 through
2 * total
to simulate the factorial calculation of (2n)! which is necessary since each order has two actions (pickup and delivery).- Within the loop:
- Multiply
result
by the loop variablej
. - Every time
j
is even, divideresult
by 2. This division handles the pairing of pickup and delivery actions correctly without overcounting. - Apply modulo operation to keep
result
within the limits of integer values.
- Multiply
- Within the loop:
At the end of the loop,
result
holds the count of valid sequences mod1_000_000_007
.Finally, cast
result
to anint
and return it.
This methodology leverages combinatorial logic and properties of permutations, treating every order as contributing two actions that are interleaved among other orders' actions while maintaining the order-specific constraint. By careful application of division and the modulo operation, the solution efficiently handles the large numbers involved.
var calculateOrders = function(n) {
let result = 1;
let MODULO = 1_000_000_007;
for (let i = 1; i <= 2 * n; ++i) {
result = result * i;
// We only need to divide the result by 2 n-times.
// To prevent decimal results we divide after multiplying an even number.
if (i % 2 == 0) {
result = result / 2;
}
result %= MODULO;
}
return result;
};
The code provided is a solution for calculating all valid pickup and delivery options using a specified number of pairs n
. This JavaScript function operates by determining the number of distinct sequences in which n
pickup and n
delivery actions could be legally executed, adhering to the condition that each delivery follows its corresponding pickup.
Here's an overview of how the function works:
- Initialize
result
to store the cumulative product of numbers. - Define
MODULO
to keep the results manageable and avoid overflow by applying modulo1_000_000_007
. - Loop through numbers from 1 to
2n
(every pair requires two actions: a pickup and a delivery):- Multiply the current number with the
result
. - After multiplying by each even number, divide the result by 2. This approach is taken to achieve the factorial division required by the problem's combinatorial nature only after an even product, which ensures the result remains an integer and not a floating point number.
- Apply modulo operation to avoid overflow and keep the numbers manageable.
- Multiply the current number with the
- Return the resulting calculation from the loop which gives the total valid sequences for the given
n
.
The implementation effectively uses modular arithmetic to handle large numbers resulting from factorial calculations, ensuring the solution is optimized for performance and can handle scenarios where n
is relatively large. This function is precise for calculating the number of possible valid orders of operations for pairs of pickup and delivery requests.
class Solution:
def calculateOrders(self, total: int) -> int:
CONSTANT_MOD = 1_000_000_007
result = 1
for value in range(1, 2 * total + 1):
result = result * value
# Divide result by 2 on even indices to maintain integer results
if value % 2 == 0:
result = result // 2
result %= CONSTANT_MOD
return result
Given a problem that necessitates counting all valid pickup and delivery options, the provided Python solution implements an efficient approach to calculate permutations of pickup and delivery pairs using modulo arithmetic for large numbers. Follow this explanation to understand the mechanics of the solution:
- The function,
calculateOrders
, accepts an integertotal
which represents the number of pickup and delivery pairs. - A constant
CONSTANT_MOD
is defined to prevent integer overflow and keep calculations manageable, typical in problems with potential large outputs. - Initialize
result
to 1, which will accumulate the product of sequence values. - Iterate from 1 up to
2 * total + 1
(inclusive of all pickups and deliveries). Multiplyresult
by each value in the loop. - If the current value in the sequence is even, divide
result
by 2 immediately to handle the pairing of pickups and deliveries correctly. - After each multiplication (and optional division), apply modulo operation with
CONSTANT_MOD
. - Finally, return the computed
result
.
The approach ensures that each pairing is uniquely accounted for and that large number outputs are handled efficiently. By sequentially multiplying values and selectively dividing by 2 on even indices, the solution correctly constructs the count of valid sequences without redundancy.
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