Off-by-one on range boundaries
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Move from brute-force thinking to an efficient approach using array strategy.
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 summary: 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
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array
[[10,5,0],[15,2,1],[25,1,1],[30,4,0]]
[[7,1000000000,1],[15,3,0],[5,999999995,0],[5,1,1]]
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #1801: Number of Orders in the Backlog
class Solution {
public int getNumberOfBacklogOrders(int[][] orders) {
PriorityQueue<int[]> buy = new PriorityQueue<>((a, b) -> b[0] - a[0]);
PriorityQueue<int[]> sell = new PriorityQueue<>((a, b) -> a[0] - b[0]);
for (var e : orders) {
int p = e[0], a = e[1], t = e[2];
if (t == 0) {
while (a > 0 && !sell.isEmpty() && sell.peek()[0] <= p) {
var q = sell.poll();
int x = q[0], y = q[1];
if (a >= y) {
a -= y;
} else {
sell.offer(new int[] {x, y - a});
a = 0;
}
}
if (a > 0) {
buy.offer(new int[] {p, a});
}
} else {
while (a > 0 && !buy.isEmpty() && buy.peek()[0] >= p) {
var q = buy.poll();
int x = q[0], y = q[1];
if (a >= y) {
a -= y;
} else {
buy.offer(new int[] {x, y - a});
a = 0;
}
}
if (a > 0) {
sell.offer(new int[] {p, a});
}
}
}
long ans = 0;
final int mod = (int) 1e9 + 7;
while (!buy.isEmpty()) {
ans += buy.poll()[1];
}
while (!sell.isEmpty()) {
ans += sell.poll()[1];
}
return (int) (ans % mod);
}
}
// Accepted solution for LeetCode #1801: Number of Orders in the Backlog
func getNumberOfBacklogOrders(orders [][]int) (ans int) {
sell := hp{}
buy := hp{}
for _, e := range orders {
p, a, t := e[0], e[1], e[2]
if t == 0 {
for a > 0 && len(sell) > 0 && sell[0].p <= p {
q := heap.Pop(&sell).(pair)
x, y := q.p, q.a
if a >= y {
a -= y
} else {
heap.Push(&sell, pair{x, y - a})
a = 0
}
}
if a > 0 {
heap.Push(&buy, pair{-p, a})
}
} else {
for a > 0 && len(buy) > 0 && -buy[0].p >= p {
q := heap.Pop(&buy).(pair)
x, y := q.p, q.a
if a >= y {
a -= y
} else {
heap.Push(&buy, pair{x, y - a})
a = 0
}
}
if a > 0 {
heap.Push(&sell, pair{p, a})
}
}
}
const mod int = 1e9 + 7
for len(buy) > 0 {
ans += heap.Pop(&buy).(pair).a
}
for len(sell) > 0 {
ans += heap.Pop(&sell).(pair).a
}
return ans % mod
}
type pair struct{ p, a int }
type hp []pair
func (h hp) Len() int { return len(h) }
func (h hp) Less(i, j int) bool { return h[i].p < h[j].p }
func (h hp) Swap(i, j int) { h[i], h[j] = h[j], h[i] }
func (h *hp) Push(v any) { *h = append(*h, v.(pair)) }
func (h *hp) Pop() any { a := *h; v := a[len(a)-1]; *h = a[:len(a)-1]; return v }
# Accepted solution for LeetCode #1801: Number of Orders in the Backlog
class Solution:
def getNumberOfBacklogOrders(self, orders: List[List[int]]) -> int:
buy, sell = [], []
for p, a, t in orders:
if t == 0:
while a and sell and sell[0][0] <= p:
x, y = heappop(sell)
if a >= y:
a -= y
else:
heappush(sell, (x, y - a))
a = 0
if a:
heappush(buy, (-p, a))
else:
while a and buy and -buy[0][0] >= p:
x, y = heappop(buy)
if a >= y:
a -= y
else:
heappush(buy, (x, y - a))
a = 0
if a:
heappush(sell, (p, a))
mod = 10**9 + 7
return sum(v[1] for v in buy + sell) % mod
// Accepted solution for LeetCode #1801: Number of Orders in the Backlog
struct Solution;
use std::cmp::Reverse;
use std::collections::BinaryHeap;
const MOD: i64 = 1_000_000_007;
impl Solution {
fn get_number_of_backlog_orders(orders: Vec<Vec<i32>>) -> i32 {
let mut res: i64 = 0;
let mut buy: BinaryHeap<(i32, i32)> = BinaryHeap::new();
let mut sell: BinaryHeap<(Reverse<i32>, i32)> = BinaryHeap::new();
for order in orders {
let price = order[0];
let mut amount = order[1];
res += amount as i64;
if order[2] == 0 {
while let Some(&(Reverse(p_sell), a_sell)) = sell.peek() {
if p_sell <= price {
sell.pop();
let a_min = a_sell.min(amount);
let a_left = a_sell - a_min;
amount -= a_min;
res -= 2 * a_min as i64;
if a_left > 0 {
sell.push((Reverse(p_sell), a_left));
break;
}
} else {
break;
}
}
if amount > 0 {
buy.push((price, amount));
}
} else {
while let Some(&(p_buy, a_buy)) = buy.peek() {
if p_buy >= price {
buy.pop();
let a_min = a_buy.min(amount);
let a_left = a_buy - a_min;
amount -= a_min;
res -= 2 * a_min as i64;
if a_left > 0 {
buy.push((p_buy, a_left));
break;
}
} else {
break;
}
}
if amount > 0 {
sell.push((Reverse(price), amount));
}
}
}
(res % MOD) as i32
}
}
#[test]
fn test() {
let orders = vec_vec_i32![[10, 5, 0], [15, 2, 1], [25, 1, 1], [30, 4, 0]];
let res = 6;
assert_eq!(Solution::get_number_of_backlog_orders(orders), res);
let orders = vec_vec_i32![[7, 1000000000, 1], [15, 3, 0], [5, 999999995, 0], [5, 1, 1]];
let res = 999999984;
assert_eq!(Solution::get_number_of_backlog_orders(orders), res);
}
// Accepted solution for LeetCode #1801: Number of Orders in the Backlog
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1801: Number of Orders in the Backlog
// class Solution {
// public int getNumberOfBacklogOrders(int[][] orders) {
// PriorityQueue<int[]> buy = new PriorityQueue<>((a, b) -> b[0] - a[0]);
// PriorityQueue<int[]> sell = new PriorityQueue<>((a, b) -> a[0] - b[0]);
// for (var e : orders) {
// int p = e[0], a = e[1], t = e[2];
// if (t == 0) {
// while (a > 0 && !sell.isEmpty() && sell.peek()[0] <= p) {
// var q = sell.poll();
// int x = q[0], y = q[1];
// if (a >= y) {
// a -= y;
// } else {
// sell.offer(new int[] {x, y - a});
// a = 0;
// }
// }
// if (a > 0) {
// buy.offer(new int[] {p, a});
// }
// } else {
// while (a > 0 && !buy.isEmpty() && buy.peek()[0] >= p) {
// var q = buy.poll();
// int x = q[0], y = q[1];
// if (a >= y) {
// a -= y;
// } else {
// buy.offer(new int[] {x, y - a});
// a = 0;
// }
// }
// if (a > 0) {
// sell.offer(new int[] {p, a});
// }
// }
// }
// long ans = 0;
// final int mod = (int) 1e9 + 7;
// while (!buy.isEmpty()) {
// ans += buy.poll()[1];
// }
// while (!sell.isEmpty()) {
// ans += sell.poll()[1];
// }
// return (int) (ans % mod);
// }
// }
Use this to step through a reusable interview workflow for this problem.
Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.
Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.
Review these before coding to avoid predictable interview regressions.
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.