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 <= 500The key challenge in #1359 Count All Valid Pickup and Delivery Options is counting valid sequences where each delivery must appear after its corresponding pickup. Instead of generating permutations, an efficient approach uses combinatorics and dynamic counting.
Consider adding orders one by one. When introducing the i-th order, there are multiple valid positions to insert its pickup and delivery while preserving the rule that the pickup occurs before the delivery. The number of valid placements grows based on previously placed pairs. This leads to a mathematical recurrence where each step multiplies the previous count by the number of new valid placements.
This approach avoids brute-force permutation generation and reduces the problem to a simple iterative computation using modular arithmetic. It can also be interpreted using dynamic programming, where each state represents the number of valid sequences for i orders.
The resulting algorithm runs in O(n) time with O(1) extra space, making it efficient even for large inputs.
| Approach | Time Complexity | Space Complexity |
|---|---|---|
| Combinatorics / Mathematical Counting | O(n) | O(1) |
| Dynamic Programming Interpretation | O(n) | O(n) |
NeetCodeIO
Use these hints if you're stuck. Try solving on your own first.
Use the permutation and combination theory to add one (P, D) pair each time until n pairs.
This approach leverages recursion with memoization to calculate the number of valid sequences. The main idea is to determine how many ways we can arrange pickups and deliveries by handling one order at a time.
Time Complexity: O(N^2), where N is the number of orders, due to overlapping subproblems.
Space Complexity: O(N^2) due to storing results in the memoization dictionary.
1MOD = 10**9 + 7
2
3def countOrders(n):
4 memo = {}
5
6 def dp(p, d):
7 if p == 0 and d == 0:
8 return 1
9 if (p, d) in memo:
10 return memo[(p, d)]
11
12 ways = 0
13 if p > 0:
14 ways += p * dp(p - 1, d + 1)
15 if d > 0:
16 ways += d * dp(p, d - 1)
17
18 memo[(p, d)] = ways % MOD
19 return ways % MOD
20
21 return dp(n, 0)This Python function uses a recursive helper function dp(p, d) that tracks how many pickups and deliveries are remaining. If both pickups and deliveries are zero, it returns 1 as a valid sequence is formed. Otherwise, it calculates the possible sequences either by picking up an order or delivering an order, avoiding violating the sequence rule.
This approach uses a combinatorial method by calculating factorial operations iteratively to compute the number of valid sequences. It simplifies the problem by deriving a general formula.
Time Complexity: O(N), where N is the number of orders.
Space Complexity: O(1), as we only maintain a constant amount of extra space.
1#include <vector>
2using namespace std;
3
4const int MOD = 1e9 + 7;
5
6class Solution {
public:
int countOrders(int n) {
long long res = 1;
for(int i = 1; i <= n; ++i){
res = res * i % MOD;
res = res * (2 * i - 1) % MOD;
}
return res;
}
};Watch expert explanations and walkthroughs
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Explore More ProblemsJot down your thoughts, approach, and key learnings
Generating all permutations of pickup and delivery events would lead to factorial complexity, which becomes infeasible for larger values of n. The combinatorial insight allows us to count valid sequences directly in linear time instead.
Yes, this problem or similar variations appear in FAANG-style interviews. It tests a candidate's understanding of combinatorics, counting techniques, and recognizing patterns that avoid brute-force permutations.
No special data structure is required for the optimal solution. The problem can be solved using simple arithmetic with integers, though it can also be interpreted through dynamic programming states representing the number of valid sequences for a given number of orders.
The optimal approach uses combinatorics to count valid placements when adding each new pickup–delivery pair. Instead of generating permutations, you iteratively compute the number of valid arrangements using a mathematical recurrence and apply modulo arithmetic.
This C++ solution iteratively applies a combinatorial formula that derives from possible positioning of deliveries after pickups. It computes using factorial increases corresponding to the number of current orders.