How do you solve Number of Orders in the Backlog on LeetCode?

Number of Orders in the Backlog (LeetCode #1801) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Using heaps allows for efficient order matching, reducing the time complexity significantly.

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

The optimal solution for Number of Orders in the Backlog runs in O(n log n) time and uses O(n) space. The time complexity is O(n log n) due to the heap operations (insert and remove) being logarithmic in nature, while we process each order once.

What is the brute force solution for Number of Orders in the Backlog?

The brute force approach for Number of Orders in the Backlog is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we simulate the order processing by iterating through each order and checking if it can be executed against existing orders in the backlog. This method is straightforward but inefficient due to repeated checks. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Number of Orders in the Backlog?

Number of Orders in the Backlog uses the following concepts: Array, Heap (Priority Queue), Simulation. The recommended patterns to study are: Heap (Priority Queue), Simulation.

What are the interview tips for Number of Orders in the Backlog?

Explain your thought process clearly while coding, especially when using data structures like heaps. Consider edge cases, such as when all buy orders are higher than sell orders or vice versa. Practice implementing heaps and understand their operations thoroughly.

What are common mistakes when solving Number of Orders in the Backlog?

Failing to handle the case where orders cannot be matched correctly. Not using heaps effectively, leading to unnecessary iterations.

#1801

Number of Orders in the Backlog

Medium

ArrayHeap (Priority Queue)SimulationHeap (Priority Queue)Simulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

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Intuition

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

In the optimal approach, we use two heaps (priority queues) to efficiently manage buy and sell orders. This allows us to quickly match orders without needing to iterate through the entire backlog.

⚙️

Algorithm

4 steps

1Step 1: Initialize a max heap for buy orders and a min heap for sell orders.
2Step 2: For each order, process it based on its type, matching it with the appropriate orders from the heaps.
3Step 3: If an order cannot be matched, add it to the respective heap.
4Step 4: At the end, sum the amounts in both heaps to get the total backlog.

solution.py29 lines

1# Full working Python code
2import heapq
3class Solution:
4    def getNumberOfBacklogOrders(self, orders):
5        buy_orders = []  # max heap
6        sell_orders = []  # min heap
7        for price, amount, orderType in orders:
8            if orderType == 0:  # Buy order
9                while amount > 0 and sell_orders and sell_orders[0][0] <= price:
10                    sell_price, sell_amount = heapq.heappop(sell_orders)
11                    if sell_amount <= amount:
12                        amount -= sell_amount
13                    else:
14                        heapq.heappush(sell_orders, (sell_price, sell_amount - amount))
15                        amount = 0
16                if amount > 0:
17                    heapq.heappush(buy_orders, (-price, amount))  # store as negative for max heap
18            else:  # Sell order
19                while amount > 0 and buy_orders and -buy_orders[0][0] >= price:
20                    buy_price, buy_amount = heapq.heappop(buy_orders)
21                    if buy_amount <= amount:
22                        amount -= buy_amount
23                    else:
24                        heapq.heappush(buy_orders, (buy_price, buy_amount - amount))
25                        amount = 0
26                if amount > 0:
27                    heapq.heappush(sell_orders, (price, amount))
28        total = sum(amount for _, amount in buy_orders) + sum(amount for _, amount in sell_orders)
29        return total % (10**9 + 7)

ℹ

Complexity note: The time complexity is O(n log n) due to the heap operations (insert and remove) being logarithmic in nature, while we process each order once.

1Using heaps allows for efficient order matching, reducing the time complexity significantly.
2Understanding the nature of buy and sell orders helps in structuring the heaps correctly.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.