You are given a 2D integer array orders, where each orders[i] = [pricei, amounti, orderTypei] denotes that amounti orders have been placed of type orderTypei at the price pricei. The orderTypei is:
0 if it is a batch of buy orders, or1 if it is a batch of sell orders.Note that orders[i] represents a batch of amounti 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.
There is a backlog that consists of orders that have not been executed. The backlog is initially empty. When an order is placed, the following happens:
buy order, you look at the sell order with the smallest price in the backlog. If that sell order's price is smaller than or equal to the current buy order's price, they will match and be executed, and that sell order will be removed from the backlog. Else, the buy order is added to the backlog.sell order, you look at the buy order with the largest price in the backlog. If that buy order's price is larger than or equal to the current sell order's price, they will match and be executed, and that buy order will be removed from the backlog. Else, the sell order is added to the backlog.Return the total amount of orders in the backlog after placing all the orders from the input. Since this number can be large, return it modulo 109 + 7.
Example 1:
Input: orders = [[10,5,0],[15,2,1],[25,1,1],[30,4,0]] Output: 6 Explanation: Here is what happens with the orders: - 5 orders of type buy with price 10 are placed. There are no sell orders, so the 5 orders are added to the backlog. - 2 orders of type sell with price 15 are placed. There are no buy orders with prices larger than or equal to 15, so the 2 orders are added to the backlog. - 1 order of type sell with price 25 is placed. There are no buy orders with prices larger than or equal to 25 in the backlog, so this order is added to the backlog. - 4 orders of type buy with price 30 are placed. The first 2 orders are matched with the 2 sell orders of the least price, which is 15 and these 2 sell orders are removed from the backlog. The 3rd order is matched with the sell order of the least price, which is 25 and this sell order is removed from the backlog. Then, there are no more sell orders in the backlog, so the 4th order is added to the backlog. Finally, the backlog has 5 buy orders with price 10, and 1 buy order with price 30. So the total number of orders in the backlog is 6.
Example 2:
Input: orders = [[7,1000000000,1],[15,3,0],[5,999999995,0],[5,1,1]] Output: 999999984 Explanation: Here is what happens with the orders: - 109 orders of type sell with price 7 are placed. There are no buy orders, so the 109 orders are added to the backlog. - 3 orders of type buy with price 15 are placed. They are matched with the 3 sell orders with the least price which is 7, and those 3 sell orders are removed from the backlog. - 999999995 orders of type buy with price 5 are placed. The least price of a sell order is 7, so the 999999995 orders are added to the backlog. - 1 order of type sell with price 5 is placed. It is matched with the buy order of the highest price, which is 5, and that buy order is removed from the backlog. Finally, the backlog has (1000000000-3) sell orders with price 7, and (999999995-1) buy orders with price 5. So the total number of orders = 1999999991, which is equal to 999999984 % (109 + 7).
Constraints:
1 <= orders.length <= 105orders[i].length == 31 <= pricei, amounti <= 109orderTypei is either 0 or 1.Problem Overview: You receive a stream of buy and sell orders for a stock. Each order has a price, amount, and type (buy or sell). When possible, orders should match with existing opposite orders in the backlog. After processing all orders, return the total remaining amount in the backlog modulo 1e9 + 7.
Approach 1: Using Two Priority Queues (O(n log n) time, O(n) space)
This approach simulates a real trading order book using two heaps. Maintain a max‑heap for buy orders (highest price first) and a min‑heap for sell orders (lowest price first). When a new buy order arrives, repeatedly match it with the cheapest sell order while the sell price is less than or equal to the buy price. Similarly, when a sell order arrives, match it with the highest buy order while the buy price is greater than or equal to the sell price. Each match reduces quantities until one order is exhausted. Remaining quantities are pushed back into the appropriate heap. Heap operations like push and pop take O(log n), so processing all orders costs O(n log n). This approach directly models the trading system and is the standard solution using a heap (priority queue).
Approach 2: Sorting and Two Pointers (O(n log n) time, O(n) space)
Another way to reason about the problem is to process orders after grouping and sorting them by price. Buy orders are sorted descending by price while sell orders are sorted ascending. Two pointers then iterate through both lists and simulate matching between the current highest buy and lowest sell order. Whenever the buy price is at least the sell price, reduce the smaller amount and move the pointer when an order is exhausted. If prices no longer match, the unmatched orders remain in the backlog. Sorting dominates the runtime at O(n log n), while pointer traversal is linear. This approach avoids heaps but requires preprocessing and additional arrays, making it less natural for streaming input. It still relies on understanding the matching process using arrays and careful simulation of order execution.
Recommended for interviews: The two‑priority‑queue approach is what interviewers expect. It mirrors how real order books work and demonstrates strong knowledge of heaps and simulation problems. The sorting method shows the matching logic clearly but is less flexible because the input is processed sequentially. Showing awareness of both approaches is useful, but implementing the heap-based solution signals solid problem‑solving skill.
This approach involves using two priority queues (heaps) to represent the buy and sell orders in the backlog:
This Python solution uses two heaps to manage the buy and sell orders:
Python
C++
Java
C#
JavaScript
Time Complexity: O(n log n), where n is the number of orders since each order insertion/deletion into the heap takes O(log n), and there can be at most n such operations.
Space Complexity: O(n), storing all unmatched orders in heaps.
This approach leverages sorting to handle and manage orders efficiently:
This approach can be efficient, as sorting can be done in O(n log n), allowing a straightforward linear scan for matching orders.
This solution involves sorting both buy and sell orders to efficiently match them using two pointers:
Python
JavaScript
Time Complexity: O(n log n) due to sorting, followed by O(n) scanning.
Space Complexity: O(n) to store orders separately.
We can use a priority queue (max-min heap) to maintain the current backlog of orders, where the max heap buy maintains the backlog of purchase orders, and the min heap sell maintains the backlog of sales orders. Each element in the heap is a tuple (price, amount), indicating that the number of orders at price price is amount.
Next, we traverse the order array orders, and simulate according to the problem's requirements.
After the traversal, we add the order quantities in buy and sell, which is the final backlog of orders. Note that the answer may be very large, so we need to take the modulus of 10^9 + 7.
The time complexity is O(n times log n), and the space complexity is O(n). Here, n is the length of orders.
| Approach | Complexity |
|---|---|
| Using Two Priority Queues | Time Complexity: O(n log n), where n is the number of orders since each order insertion/deletion into the heap takes O(log n), and there can be at most n such operations. |
| Using Sorting and Two Pointers | Time Complexity: O(n log n) due to sorting, followed by O(n) scanning. |
| Priority Queue (Max-Min Heap) + Simulation | — |
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Two Priority Queues (Order Book Simulation) | O(n log n) | O(n) | General case. Best for streaming order processing and typical interview expectations. |
| Sorting + Two Pointers | O(n log n) | O(n) | Useful when orders can be preprocessed and sorted before matching. |
LeetCode 1801. Number of Orders in the Backlog | 🏆 Weekly Contest 233 | Medium | Algorithm Explained • Cherry Coding [IIT-G] • 1,455 views views
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