What is the time complexity of Number of Orders in the Backlog?

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

What data structures are used to solve Number of Orders in the Backlog?

This problem uses Array, Heap (Priority Queue), Simulation. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Number of Orders in the Backlog?

Insight 1: The key insight is to use priority queues (heaps) to efficiently maintain the buy and sell order backlogs, allowing for quick access to the minimum sell price and maximum buy price. Insight 2: Using a min-heap for sell orders and a max-heap for buy orders ensures that the best matching orders are always at the top of the heaps.

Which companies ask LeetCode 1801. Number of Orders in the Backlog in interviews?

This problem is frequently asked at Coinbase, Jane Street, Robinhood.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Number of Orders in the Backlog - LeetCode 1801 Solution

Number of Orders in the Backlog - Complete Solution Guide

Number of Orders in the Backlog is LeetCode problem 1801, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given a 2D integer array orders , where each orders[i] = [price i , amount i , orderType i ] denotes that amount i orders have been placed of type orderType i at the price price i . The orderType i is: 0 if it is a batch of buy orders, or 1 if it is a batch of sell orders. Note that orders[i] represents a batch of amount i independent orders with the same price and order type. All orders represented by orders[i] will be placed before all orders represented by orders[i+1] for all valid i

Detailed Explanation

The problem simulates an order book where buy and sell orders are matched based on price. Given a series of orders, each with a price, amount, and type (buy or sell), the goal is to determine the total amount of orders remaining in the backlog after processing all orders, modulo 10^9 + 7. The backlog consists of unmatched buy and sell orders. Buy orders are matched with the lowest-priced sell orders, and sell orders are matched with the highest-priced buy orders. Orders are processed in the order they are given in the input array.

Solution Approach

The solution uses two priority queues, `sell_backlog` (min-heap) and `buy_backlog` (max-heap), to simulate the order book. It iterates through the given `orders` array. For each order, it checks if it's a buy or sell order. If it's a buy order, it tries to match it with the lowest-priced sell orders in `sell_backlog`. If it's a sell order, it tries to match it with the highest-priced buy orders in `buy_backlog`. If there are remaining amounts after matching, the remaining order is added to its respective backlog. Finally, the total amount of orders in both backlogs is calculated and returned modulo 10^9 + 7.

Step-by-Step Algorithm

Step 1: Initialize a min-heap (`sell_backlog`) to store sell orders (price, amount) and a max-heap (`buy_backlog`) to store buy orders (price, amount).
Step 2: Iterate through the input `orders` array.
Step 3: For each order, extract the price, amount, and order type.
Step 4: If the order is a buy order (order_type == 0):
Step 5: While there are sell orders in the `sell_backlog`, the current buy amount is greater than 0, and the lowest sell price is less than or equal to the buy price, match the orders:
Step 6: Pop the lowest sell order from `sell_backlog`.
Step 7: If the sell amount is less than or equal to the buy amount, reduce the buy amount by the sell amount.
Step 8: Otherwise, reduce the sell amount by the buy amount and push the remaining sell order back into `sell_backlog`, then set the buy amount to 0.
Step 9: If there's any remaining buy amount, push the buy order into `buy_backlog`.
Step 10: If the order is a sell order (order_type == 1):
Step 11: While there are buy orders in the `buy_backlog`, the current sell amount is greater than 0, and the highest buy price is greater than or equal to the sell price, match the orders:
Step 12: Pop the highest buy order from `buy_backlog`.
Step 13: If the buy amount is less than or equal to the sell amount, reduce the sell amount by the buy amount.
Step 14: Otherwise, reduce the buy amount by the sell amount and push the remaining buy order back into `buy_backlog`, then set the sell amount to 0.
Step 15: If there's any remaining sell amount, push the sell order into `sell_backlog`.
Step 16: After processing all orders, calculate the total amount of orders in both `sell_backlog` and `buy_backlog`.
Step 17: Return the total amount modulo 10^9 + 7.

Key Insights

Insight 1: The key insight is to use priority queues (heaps) to efficiently maintain the buy and sell order backlogs, allowing for quick access to the minimum sell price and maximum buy price.
Insight 2: Using a min-heap for sell orders and a max-heap for buy orders ensures that the best matching orders are always at the top of the heaps.
Insight 3: The modulo operation (10^9 + 7) is crucial to prevent integer overflow because the amount of orders can be very large (up to 10^9 per order).

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(n)

Topics

This problem involves: Array, Heap (Priority Queue), Simulation.

Companies

Asked at: Coinbase, Jane Street, Robinhood.

Number of Orders in the Backlog | LeetCode 1801

Medium

Array Heap (Priority Queue)Simulation

Time: O(n log n)Space: O(n)

Solution Code

Available in:

import java.util.PriorityQueue;

class Solution {
    public int getNumberOfBacklogOrders(int[][] orders) {
        PriorityQueue<int[]> sell_backlog = new PriorityQueue<>((a, b) -> a[0] - b[0]); // min-heap for sell orders [price, amount]
        PriorityQueue<int[]> buy_backlog = new PriorityQueue<>((a, b) -> b[0] - a[0]);   // max-heap for buy orders [-price, amount]
        int MOD = 1000000007;

        for (int[] order : orders) {
            int price = order[0];
            int amount = order[1];
            int order_type = order[2];

            if (order_type == 0) {  // Buy order
                while (amount > 0 && !sell_backlog.isEmpty() && sell_backlog.peek()[0] <= price) {
                    int[] sell_order = sell_backlog.poll();
                    int sell_p = sell_order[0];
                    int sell_a = sell_order[1];
                    if (sell_a <= amount) {
                        amount -= sell_a;
                    } else {
                        sell_backlog.offer(new int[]{sell_p, sell_a - amount});
                        amount = 0;
                    }
                }
                if (amount > 0) {
                    buy_backlog.offer(new int[]{price, amount});
                }
            } else {  // Sell order
                while (amount > 0 && !buy_backlog.isEmpty() && buy_backlog.peek()[0] >= price) {
                    int[] buy_order = buy_backlog.poll();
                    int buy_p = buy_order[0];
                    int buy_a = buy_order[1];
                    if (buy_a <= amount) {
                        amount -= buy_a;
                    } else {
                        buy_backlog.offer(new int[]{buy_p, buy_a - amount});
                        amount = 0;
                    }
                }
                if (amount > 0) {
                    sell_backlog.offer(new int[]{price, amount});
                }
            }
        }

        long total_amount = 0;
        for (int[] order : sell_backlog) {
            total_amount += order[1];
        }
        for (int[] order : buy_backlog) {
            total_amount += order[1];
        }

        return (int) (total_amount % MOD);
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 1801: Number of Orders in the Backlog - Detailed explanation and solution approaches.

Topics

Array Heap (Priority Queue)Simulation

Companies

CoinbaseJane StreetRobinhood

Stats

Difficulty:Medium

Languages:4

Companies:3

Problem #:1801

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

17 steps

1
Step 1: Initialize a min-heap (`sell_backlog`) to store sell orders (price, amount) and a max-heap (`buy_backlog`) to store buy orders (price, amount).
2
Step 2: Iterate through the input `orders` array.
3
Step 3: For each order, extract the price, amount, and order type.
4
Step 4: If the order is a buy order (order_type == 0):
5
Step 5: While there are sell orders in the `sell_backlog`, the current buy amount is greater than 0, and the lowest sell price is less than or equal to the buy price, match the orders:
6
Step 6: Pop the lowest sell order from `sell_backlog`.
7
Step 7: If the sell amount is less than or equal to the buy amount, reduce the buy amount by the sell amount.
8
Step 8: Otherwise, reduce the sell amount by the buy amount and push the remaining sell order back into `sell_backlog`, then set the buy amount to 0.
9
Step 9: If there's any remaining buy amount, push the buy order into `buy_backlog`.
10
Step 10: If the order is a sell order (order_type == 1):
11
Step 11: While there are buy orders in the `buy_backlog`, the current sell amount is greater than 0, and the highest buy price is greater than or equal to the sell price, match the orders:
12
Step 12: Pop the highest buy order from `buy_backlog`.
13
Step 13: If the buy amount is less than or equal to the sell amount, reduce the sell amount by the buy amount.
14
Step 14: Otherwise, reduce the buy amount by the sell amount and push the remaining buy order back into `buy_backlog`, then set the sell amount to 0.
15
Step 15: If there's any remaining sell amount, push the sell order into `sell_backlog`.
16
Step 16: After processing all orders, calculate the total amount of orders in both `sell_backlog` and `buy_backlog`.
17
Step 17: Return the total amount modulo 10^9 + 7.

💡

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 log n)

✨ Log-linear - Great for sorting algorithms!

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is determined by the heap operations (push and pop), each costing O(log n), performed for each of the n orders. The space complexity is determined by the size of the heaps, which in the worst case can store all n orders.

📖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

Using balanced trees

Instead of heaps, balanced binary search trees (e.g., AVL trees, Red-Black trees) could be used to store the buy and sell orders. Insertion and deletion would still be O(log n), but implementation would be more complex.

Time: O(n log n)Space: O(n)

Trade-offs: Balanced trees provide similar time complexity, but the implementation overhead is higher than using heaps, making heaps the preferred choice here due to simplicity and efficiency.

✨

Key Insights

1Insight 1: The key insight is to use priority queues (heaps) to efficiently maintain the buy and sell order backlogs, allowing for quick access to the minimum sell price and maximum buy price.
2Insight 2: Using a min-heap for sell orders and a max-heap for buy orders ensures that the best matching orders are always at the top of the heaps.
3Insight 3: The modulo operation (10^9 + 7) is crucial to prevent integer overflow because the amount of orders can be very large (up to 10^9 per order).

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Edge Cases

Edge case 1: Large amount of orders (up to 10^5) with large order sizes (up to 10^9). The modulo operation handles the potential integer overflow when calculating the total amount.
Edge case 2: Orders arriving in a sequence that creates large backlogs. The heap data structures efficiently manage these backlogs ensuring the best matching orders are prioritized according to price.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not using priority queues (heaps) and instead using linear search or sorting to find the best match, resulting in a significantly slower O(n^2) time complexity.

✗Common mistake 2: Failing to apply the modulo operation when calculating the total amount, which can lead to integer overflow and incorrect results.

✅

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: This algorithm is directly applicable to stock market order matching engines, where buy and sell orders are continuously matched based on price and time priority.

🏦Real-world scenario 2: Could be used in any auction system or marketplace where prices are dynamically determined by matching bids (buy orders) and asks (sell orders).

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

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: June 2, 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

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

@AlgoMasterJul 13

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 6

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

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 "Number of Orders in the Backlog"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether heap (priority queue) techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?