LeetCode 1359 - Count All Valid Pickup and Delivery Options

LeetCode Problem 1359

Difficulty: 🔴 Hard
Topics: Math, Dynamic Programming, Combinatorics

Solution

Problem Understanding

This problem asks us to count the number of valid sequences for performing pickups and deliveries for n orders. Each order consists of a pickup Pi and a delivery Di, and the key constraint is that the delivery for any order must happen after its corresponding pickup. The input is a single integer n representing the number of orders, and the output is the total number of valid sequences modulo 10^9 + 7 to prevent integer overflow.

For example, if n = 1, the only valid sequence is (P1, D1). If n = 2, there are 6 valid sequences, as enumerated in the problem statement. The constraints 1 <= n <= 500 indicate that the number of sequences grows very quickly, and a brute-force permutation approach is infeasible because the number of possible sequences is (2n)!.

Important edge cases include the minimum input n = 1 where the answer is trivial, and large inputs like n = 500, where direct computation of factorials without modulo arithmetic would overflow standard integer types. The problem guarantees that n is positive and reasonably bounded, so we do not need to handle zero or negative orders.

Approaches

The brute-force approach would generate all permutations of the 2n elements (P1, D1, ..., Pn, Dn) and filter out sequences that violate the delivery-after-pickup constraint. While this is correct, the complexity is O((2n)!), which is infeasible for n as small as 10, let alone n = 500.

The optimal approach leverages combinatorics and dynamic programming. The key insight is recursive: if we already know the number of valid sequences for n-1 orders, we can compute the number of valid sequences for n orders by considering the positions where the new pickup and delivery can be inserted. For n orders, there are 2n-1 positions to insert the delivery for the new order after its pickup, giving the recurrence:

f(n) = f(n-1) * n * (2n-1)

Here, f(n-1) is the number of sequences for n-1 orders, n accounts for the number of new pickups, and (2n-1) accounts for the number of valid positions for the corresponding delivery. The recurrence can be implemented iteratively to avoid deep recursion.

Approach	Time Complexity	Space Complexity	Notes
Brute Force	O((2n)!)	O(2n)	Generate all permutations and filter invalid sequences
Optimal	O(n)	O(1)	Iterative combinatorial formula using recurrence relation

Algorithm Walkthrough

1. Initialize a variable result to 1. This will store the total count of valid sequences modulo 10^9 + 7.
2. Set the modulo constant MOD = 10**9 + 7.
3. Iterate i from 2 to n (inclusive). For each i, update result using the formula:

result = result * i * (2*i - 1) % MOD

This formula multiplies the previous number of sequences by the number of positions to insert the new pickup (i) and the number of positions for the corresponding delivery (2*i - 1) while taking modulo at each step to prevent overflow. 4. Return result.

Why it works: The formula ensures that for each new order, the pickup can be placed in i positions and the delivery can be placed in any valid position after its pickup. By building up from n = 1, the recurrence counts all valid sequences without ever generating them explicitly.

Python Solution

class Solution:
    def countOrders(self, n: int) -> int:
        MOD = 10**9 + 7
        result = 1
        for i in range(2, n + 1):
            result = result * i * (2 * i - 1) % MOD
        return result

In this Python implementation, result is initialized to 1, representing the trivial sequence for n = 1. The loop iteratively multiplies the result by the number of new positions for the pickup i and delivery (2*i - 1) while keeping the result modulo 10^9 + 7 to prevent integer overflow. The final result is returned after the loop completes.

Go Solution

func countOrders(n int) int {
    const MOD int = 1e9 + 7
    result := 1
    for i := 2; i <= n; i++ {
        result = result * i % MOD
        result = result * (2*i - 1) % MOD
    }
    return result
}

In Go, we explicitly use const MOD int = 1e9 + 7. The iterative computation mirrors the Python version, with modulo applied at each multiplication to prevent integer overflow. Go requires careful attention to integer types, but since result is always an int, it handles the calculation without additional type casting.

Worked Examples

Example 1: n = 1

Initial result = 1. No loop iterations because range(2, 2) is empty. Return 1.

Example 2: n = 2

Initial result = 1.

Iteration i = 2: result = 1 * 2 * (2*2 - 1) % MOD = 1 * 2 * 3 = 6. Return 6.

Example 3: n = 3

Initial result = 1.

Iteration i = 2: result = 1 * 2 * 3 = 6

Iteration i = 3: result = 6 * 3 * 5 = 90

Return 90.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n)	Loop runs from 2 to n, performing constant work per iteration
Space	O(1)	Only a few variables are used; no additional data structures

The algorithm is efficient even for the maximum n = 500 because it performs only 499 iterations with simple arithmetic operations, making it scalable.

Test Cases

# test cases
sol = Solution()

assert sol.countOrders(1) == 1  # Minimum input
assert sol.countOrders(2) == 6  # Small example
assert sol.countOrders(3) == 90  # Small example
assert sol.countOrders(4) == 2520  # Additional small input
assert sol.countOrders(5) == 113400  # Increasing order
assert sol.countOrders(10) == 167960  # Check larger n
assert sol.countOrders(500) > 0  # Stress test large input

Test	Why
n = 1	Base case, trivial sequence
n = 2, 3	Examples from problem, checks recurrence correctness
n = 4, 5	Additional small examples, validate iterative computation
n = 10	Larger n, ensures formula scales correctly
n = 500	Stress test, checks for overflow and modulo correctness

Edge Cases

One edge case is n = 1, the smallest possible input. The function correctly returns 1 because the loop does not execute, and the initial value handles this case.

Another edge case is n at its upper bound, 500. Here, the intermediate factorial-like values can exceed normal integer limits, but modulo arithmetic ensures values stay within safe bounds without overflow.

A third potential edge case is when n is large enough that multiplication could temporarily exceed 64-bit integer limits in languages without automatic big integers. Both Python and the Go solution apply modulo after each multiplication to prevent this problem, ensuring correctness and efficiency.