Number of Orders in the Backlog

Problem

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, or
1 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:

If the order is a 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.
Vice versa, if the order is a 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 totalamount of orders in the backlog after placing all the orders from the input. Since this number can be large, return it modulo 10^9 + 7.

Examples

Example 1

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![](https://assets.leetcode.com/uploads/2021/03/11/ex1.png)

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

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![](https://assets.leetcode.com/uploads/2021/03/11/ex2.png)

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 % (10^9 + 7).

Constraints

1 <= orders.length <= 10^5
orders[i].length == 3
1 <= pricei, amounti <= 10^9
orderTypei is either 0 or 1.

Solution

Method 1 – Heap Simulation (Priority Queue)

Intuition

We need to efficiently match buy and sell orders according to their prices. Buy orders match with the lowest-priced sell orders, and sell orders match with the highest-priced buy orders. We use two heaps: a max-heap for buy orders and a min-heap for sell orders, simulating the backlog and matching process.

Approach

Use a max-heap for buy orders (price, amount) and a min-heap for sell orders (price, amount).
For each order:
- If buy, try to match with lowest sell orders (price <= buy price).
- If sell, try to match with highest buy orders (price >= sell price).
- If not fully matched, add remaining to backlog.
After all orders, sum the amounts in both heaps for the answer.
Return the answer modulo 1e9+7.

Code

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#include <vector>
#include <queue>
using namespace std;
class Solution {
public:
    int getNumberOfBacklogOrders(vector<vector<int>>& orders) {
        const int MOD = 1e9+7;
        priority_queue<pair<int,int>> buy; // max-heap
        priority_queue<pair<int,int>, vector<pair<int,int>>, greater<>> sell; // min-heap
        for (auto& o : orders) {
            int price = o[0], amount = o[1], type = o[2];
            if (type == 0) { // buy
                while (amount && !sell.empty() && sell.top().first <= price) {
                    auto [sp, sa] = sell.top(); sell.pop();
                    int match = min(amount, sa);
                    amount -= match; sa -= match;
                    if (sa) sell.push({sp, sa});
                }
                if (amount) buy.push({price, amount});
            } else { // sell
                while (amount && !buy.empty() && buy.top().first >= price) {
                    auto [bp, ba] = buy.top(); buy.pop();
                    int match = min(amount, ba);
                    amount -= match; ba -= match;
                    if (ba) buy.push({bp, ba});
                }
                if (amount) sell.push({price, amount});
            }
        }
        long long ans = 0;
        while (!buy.empty()) { ans = (ans + buy.top().second) % MOD; buy.pop(); }
        while (!sell.empty()) { ans = (ans + sell.top().second) % MOD; sell.pop(); }
        return ans;
    }
};

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import "container/heap"
type Order struct{ price, amount int }
type BuyHeap []Order
func (h BuyHeap) Len() int { return len(h) }
func (h BuyHeap) Less(i, j int) bool { return h[i].price > h[j].price }
func (h BuyHeap) Swap(i, j int) { h[i], h[j] = h[j], h[i] }
func (h *BuyHeap) Push(x interface{}) { *h = append(*h, x.(Order)) }
func (h *BuyHeap) Pop() interface{} { old := *h; x := old[len(old)-1]; *h = old[:len(old)-1]; return x }
type SellHeap []Order
func (h SellHeap) Len() int { return len(h) }
func (h SellHeap) Less(i, j int) bool { return h[i].price < h[j].price }
func (h SellHeap) Swap(i, j int) { h[i], h[j] = h[j], h[i] }
func (h *SellHeap) Push(x interface{}) { *h = append(*h, x.(Order)) }
func (h *SellHeap) Pop() interface{} { old := *h; x := old[len(old)-1]; *h = old[:len(old)-1]; return x }
func getNumberOfBacklogOrders(orders [][]int) int {
    const mod = 1e9+7
    buy := &BuyHeap{}; sell := &SellHeap{}
    heap.Init(buy); heap.Init(sell)
    for _, o := range orders {
        price, amount, typ := o[0], o[1], o[2]
        if typ == 0 {
            for amount > 0 && sell.Len() > 0 && (*sell)[0].price <= price {
                sa := (*sell)[0].amount
                match := min(amount, sa)
                amount -= match; (*sell)[0].amount -= match
                if (*sell)[0].amount == 0 { heap.Pop(sell) }
            }
            if amount > 0 { heap.Push(buy, Order{price, amount}) }
        } else {
            for amount > 0 && buy.Len() > 0 && (*buy)[0].price >= price {
                ba := (*buy)[0].amount
                match := min(amount, ba)
                amount -= match; (*buy)[0].amount -= match
                if (*buy)[0].amount == 0 { heap.Pop(buy) }
            }
            if amount > 0 { heap.Push(sell, Order{price, amount}) }
        }
    }
    ans := 0
    for buy.Len() > 0 { ans = (ans + heap.Pop(buy).(Order).amount) % mod }
    for sell.Len() > 0 { ans = (ans + heap.Pop(sell).(Order).amount) % mod }
    return ans
}
func min(a, b int) int { if a < b { return a } else { return b } }

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import java.util.*;
class Solution {
    public int getNumberOfBacklogOrders(int[][] orders) {
        int MOD = (int)1e9+7;
        PriorityQueue<int[]> buy = new PriorityQueue<>((a,b)->b[0]-a[0]);
        PriorityQueue<int[]> sell = new PriorityQueue<>(Comparator.comparingInt(a->a[0]));
        for (int[] o : orders) {
            int price = o[0], amount = o[1], type = o[2];
            if (type == 0) {
                while (amount > 0 && !sell.isEmpty() && sell.peek()[0] <= price) {
                    int[] s = sell.poll();
                    int match = Math.min(amount, s[1]);
                    amount -= match; s[1] -= match;
                    if (s[1] > 0) sell.offer(s);
                }
                if (amount > 0) buy.offer(new int[]{price, amount});
            } else {
                while (amount > 0 && !buy.isEmpty() && buy.peek()[0] >= price) {
                    int[] b = buy.poll();
                    int match = Math.min(amount, b[1]);
                    amount -= match; b[1] -= match;
                    if (b[1] > 0) buy.offer(b);
                }
                if (amount > 0) sell.offer(new int[]{price, amount});
            }
        }
        long ans = 0;
        while (!buy.isEmpty()) ans = (ans + buy.poll()[1]) % MOD;
        while (!sell.isEmpty()) ans = (ans + sell.poll()[1]) % MOD;
        return (int)ans;
    }
}

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import java.util.PriorityQueue
class Solution {
    fun getNumberOfBacklogOrders(orders: Array<IntArray>): Int {
        val MOD = 1_000_000_007
        val buy = PriorityQueue<IntArray> { a, b -> b[0] - a[0] }
        val sell = PriorityQueue<IntArray> { a, b -> a[0] - b[0] }
        for (o in orders) {
            var price = o[0]; var amount = o[1]; val type = o[2]
            if (type == 0) {
                while (amount > 0 && sell.isNotEmpty() && sell.peek()[0] <= price) {
                    val s = sell.poll()
                    val match = minOf(amount, s[1])
                    amount -= match; s[1] -= match
                    if (s[1] > 0) sell.offer(s)
                }
                if (amount > 0) buy.offer(intArrayOf(price, amount))
            } else {
                while (amount > 0 && buy.isNotEmpty() && buy.peek()[0] >= price) {
                    val b = buy.poll()
                    val match = minOf(amount, b[1])
                    amount -= match; b[1] -= match
                    if (b[1] > 0) buy.offer(b)
                }
                if (amount > 0) sell.offer(intArrayOf(price, amount))
            }
        }
        var ans = 0L
        while (buy.isNotEmpty()) ans = (ans + buy.poll()[1]) % MOD
        while (sell.isNotEmpty()) ans = (ans + sell.poll()[1]) % MOD
        return ans.toInt()
    }
}

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import heapq
class Solution:
    def getNumberOfBacklogOrders(self, orders: list[list[int]]) -> int:
        MOD = 10**9+7
        buy = [] # max-heap
        sell = [] # min-heap
        for price, amount, typ in orders:
            if typ == 0:
                while amount and sell and sell[0][0] <= price:
                    sp, sa = heapq.heappop(sell)
                    match = min(amount, sa)
                    amount -= match; sa -= match
                    if sa: heapq.heappush(sell, (sp, sa))
                if amount: heapq.heappush(buy, (-price, amount))
            else:
                while amount and buy and -buy[0][0] >= price:
                    bp, ba = heapq.heappop(buy)
                    match = min(amount, ba)
                    amount -= match; ba -= match
                    if ba: heapq.heappush(buy, (bp, ba))
                if amount: heapq.heappush(sell, (price, amount))
        ans = 0
        for _, a in buy: ans = (ans + a) % MOD
        for _, a in sell: ans = (ans + a) % MOD
        return ans

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use std::collections::BinaryHeap;
impl Solution {
    pub fn get_number_of_backlog_orders(orders: Vec<Vec<i32>>) -> i32 {
        let mut buy = BinaryHeap::new(); // max-heap
        let mut sell = std::collections::BinaryHeap::new_by(|a: &(i32, i32), b: &(i32, i32)| a.0.cmp(&b.0).reverse()); // min-heap
        let m = 1_000_000_007;
        for o in orders {
            let (price, mut amount, typ) = (o[0], o[1], o[2]);
            if typ == 0 {
                while amount > 0 && !sell.is_empty() && sell.peek().unwrap().0 <= price {
                    let (sp, sa) = sell.pop().unwrap();
                    let match_amt = amount.min(sa);
                    amount -= match_amt;
                    let sa = sa - match_amt;
                    if sa > 0 { sell.push((sp, sa)); }
                }
                if amount > 0 { buy.push((price, amount)); }
            } else {
                while amount > 0 && !buy.is_empty() && buy.peek().unwrap().0 >= price {
                    let (bp, ba) = buy.pop().unwrap();
                    let match_amt = amount.min(ba);
                    amount -= match_amt;
                    let ba = ba - match_amt;
                    if ba > 0 { buy.push((bp, ba)); }
                }
                if amount > 0 { sell.push((price, amount)); }
            }
        }
        let mut ans = 0i64;
        while let Some((_, a)) = buy.pop() { ans = (ans + a as i64) % m; }
        while let Some((_, a)) = sell.pop() { ans = (ans + a as i64) % m; }
        ans as i32
    }
}

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class Solution {
    getNumberOfBacklogOrders(orders: number[][]): number {
        const MOD = 1e9+7;
        const buy: [number, number][] = [];
        const sell: [number, number][] = [];
        function buySort() { buy.sort((a,b)=>b[0]-a[0]); }
        function sellSort() { sell.sort((a,b)=>a[0]-b[0]); }
        for (const [price, amount, typ] of orders) {
            if (typ === 0) {
                sellSort();
                let i = 0;
                while (amount > 0 && i < sell.length && sell[i][0] <= price) {
                    const match = Math.min(amount, sell[i][1]);
                    amount -= match; sell[i][1] -= match;
                    if (sell[i][1] === 0) { sell.splice(i,1); } else { i++; }
                }
                if (amount > 0) buy.push([price, amount]);
            } else {
                buySort();
                let i = 0;
                while (amount > 0 && i < buy.length && buy[i][0] >= price) {
                    const match = Math.min(amount, buy[i][1]);
                    amount -= match; buy[i][1] -= match;
                    if (buy[i][1] === 0) { buy.splice(i,1); } else { i++; }
                }
                if (amount > 0) sell.push([price, amount]);
            }
        }
        let ans = 0;
        for (const [, a] of buy) ans = (ans + a) % MOD;
        for (const [, a] of sell) ans = (ans + a) % MOD;
        return ans;
    }
}

Complexity

⏰ Time complexity: O(n log n), where n is the number of orders (heap operations).
🧺 Space complexity: O(n), for the heaps/backlog.

Problem#

Examples#

Example 1#

Example 2#

Constraints#

Solution#

Method 1 – Heap Simulation (Priority Queue)#

Intuition#

Approach#

Code#

Complexity#