What is the time complexity of Count All Valid Pickup and Delivery Options?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Count All Valid Pickup and Delivery Options?

This problem uses Math, Dynamic Programming, Combinatorics. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Count All Valid Pickup and Delivery Options?

Insight 1: The core idea is to recognize that for each order (Pi, Di), Pi must come before Di. We need to find all possible interleavings of the pickups and deliveries such that this constraint holds. Insight 2: The number of ways to insert the i-th pickup-delivery pair into an existing sequence of (i-1) pairs can be determined combinatorially. There are 2*i-1 possible positions to insert Pi into the sequence. After inserting Pi, there are 2*i possible positions to insert Di.

Which companies ask LeetCode 1359. Count All Valid Pickup and Delivery Options in interviews?

This problem is frequently asked at Acko, DoorDash.

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Count All Valid Pickup and Delivery Options - LeetCode 1359 Solution

Count All Valid Pickup and Delivery Options - Complete Solution Guide

Count All Valid Pickup and Delivery Options is LeetCode problem 1359, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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,

Detailed Explanation

The problem asks us to count the number of valid pickup and delivery sequences for 'n' orders, where each order has a pickup (P) and a delivery (D) service. The constraint is that for each order 'i', delivery(i) must always occur after pickup(i). Since the answer can be very large, we need to return it modulo 10^9 + 7.

Solution Approach

The provided solution uses an iterative approach based on combinatorics. It builds the answer incrementally, starting with one order and adding orders one at a time. For each new order 'i', it calculates the number of ways to insert the pickup and delivery into the existing sequence. The number of ways to place the i-th pickup and delivery is i * (2 * i - 1). We then multiply the number of existing solutions with this value and take the result modulo 10^9 + 7.

Step-by-Step Algorithm

Step 1: Initialize 'ans' to 1. This represents the base case where n=1.
Step 2: Iterate from i = 1 to n (inclusive). Each iteration represents adding a new order.
Step 3: In each iteration, calculate the number of ways to insert the i-th pickup and delivery into the sequence formed by the previous i-1 orders. This is given by i * (2 * i - 1).
Step 4: Multiply 'ans' by i * (2 * i - 1) and take the result modulo 10^9 + 7. This updates 'ans' with the number of valid sequences for 'i' orders.
Step 5: After the loop finishes, return 'ans'. This is the total number of valid sequences for 'n' orders.

Key Insights

Insight 1: The core idea is to recognize that for each order (Pi, Di), Pi must come before Di. We need to find all possible interleavings of the pickups and deliveries such that this constraint holds.
Insight 2: The number of ways to insert the i-th pickup-delivery pair into an existing sequence of (i-1) pairs can be determined combinatorially. There are 2*i-1 possible positions to insert Pi into the sequence. After inserting Pi, there are 2*i possible positions to insert Di.
Insight 3: The total number of valid sequences for 'n' orders can be calculated iteratively. Each time we add a new order, we multiply the number of existing valid sequences by the number of ways to insert the new order's pickup and delivery, modulo 10^9 + 7 to prevent integer overflow.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Math, Dynamic Programming, Combinatorics.

Companies

Asked at: Acko, DoorDash.

Count All Valid Pickup and Delivery Options | LeetCode 1359

Hard

Math Dynamic Programming Combinatorics

Time: O(n)Space: O(1)

Solution Code

Available in:

class Solution {
    public int countOrders(int n) {
        long MOD = 1000000007;
        long ans = 1;
        for (int i = 1; i <= n; i++) {
            // The number of ways to insert the i-th pair (Pi, Di) into a
            // valid sequence of i-1 pairs.
            // There are 2 * (i - 1) items already, so 2 * i - 1 slots for Pi.
            // After placing Pi, there are 2 * i - 1 items, so 2 * i slots for Di.
            // The total number of ways to pick two spots for (Pi, Di) from 2*i spots is C(2i, 2).
            // C(2i, 2) = (2i * (2i - 1)) / 2 = i * (2i - 1).
            // So, f(i) = f(i-1) * i * (2i - 1).
            ans = (ans * i * (2 * i - 1)) % MOD;
        }
        return (int) ans;
    }
}

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Problem Info

Description

Problem 1359: Count All Valid Pickup and Delivery Options - Detailed explanation and solution approaches.

Topics

Math Dynamic Programming Combinatorics

Companies

AckoDoorDash

Stats

Difficulty:Hard

Languages:4

Companies:2

Problem #:1359

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

5 steps

1
Step 1: Initialize 'ans' to 1. This represents the base case where n=1.
2
Step 2: Iterate from i = 1 to n (inclusive). Each iteration represents adding a new order.
3
Step 3: In each iteration, calculate the number of ways to insert the i-th pickup and delivery into the sequence formed by the previous i-1 orders. This is given by i * (2 * i - 1).
4
Step 4: Multiply 'ans' by i * (2 * i - 1) and take the result modulo 10^9 + 7. This updates 'ans' with the number of valid sequences for 'i' orders.
5
Step 5: After the loop finishes, return 'ans'. This is the total number of valid sequences for 'n' orders.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The algorithm iterates from 1 to n, performing a constant amount of work in each iteration. Therefore, the time complexity is O(n). The algorithm uses a constant amount of extra space to store variables such as 'ans', 'i', and 'MOD'. Therefore, the space complexity is O(1).

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Dynamic Programming

Although the given solution is already very efficient, a dynamic programming approach can also solve the problem. A DP table could store the number of valid sequences for i orders, ending with either a pickup or a delivery. However, this would likely result in a more complex implementation for the same O(n) time complexity and larger space complexity.

Time: O(n)Space: O(n) or O(1) with optimization

Trade-offs: DP approach is more verbose and complex for this problem, offering no advantages in time complexity. The iterative approach is more straightforward and efficient.

✨

Key Insights

1Insight 1: The core idea is to recognize that for each order (Pi, Di), Pi must come before Di. We need to find all possible interleavings of the pickups and deliveries such that this constraint holds.
2Insight 2: The number of ways to insert the i-th pickup-delivery pair into an existing sequence of (i-1) pairs can be determined combinatorially. There are 2*i-1 possible positions to insert Pi into the sequence. After inserting Pi, there are 2*i possible positions to insert Di.
3Insight 3: The total number of valid sequences for 'n' orders can be calculated iteratively. Each time we add a new order, we multiply the number of existing valid sequences by the number of ways to insert the new order's pickup and delivery, modulo 10^9 + 7 to prevent integer overflow.

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: n = 1. The solution correctly returns 1 as the only valid sequence is (P1, D1).
Edge case 2: Large n (close to 500). The solution uses modulo arithmetic to prevent integer overflow, ensuring that the answer remains within the required range.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Integer Overflow. Not using modulo arithmetic will lead to incorrect results for larger values of 'n'. The problem specifies returning the answer modulo 10^9 + 7, so it's crucial to apply this operation at each step of the calculation.

✗Common mistake 2: Incorrect Combinatorial Reasoning. Failing to properly understand the relationship between the number of orders and the number of valid ways to insert them leads to incorrect results. Specifically, forgetting that the number of available positions for the delivery depends on whether the pickup has already been inserted.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Task scheduling with dependencies. This problem is analogous to scheduling tasks where some tasks must be completed before others. The pickup and delivery represent starting and completing a task, with the delivery needing to happen after the pickup.

🏦Real-world scenario 2: Resource allocation with release and deallocation. This scenario can be modeled using a similar pattern, where 'pickup' represents allocating a resource, and 'delivery' represents releasing it. A valid allocation must always be followed by a release.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: September 8, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 13

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 6

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJun 29

❓

Frequently Asked Questions

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Count All Valid Pickup and Delivery Options"?

For this hard-level Math problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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