LeetCode 1801: Number of Orders in the Backlog
LeetCode 1801 Solution Explanation
Explanation
To solve this problem, we can use a priority queue for both buy and sell orders. We iterate through the orders and process them accordingly. For each buy order, we match it with the smallest sell order in the sell backlog that has a price less than or equal to the current buy order's price. Similarly, for each sell order, we match it with the largest buy order in the buy backlog that has a price greater than or equal to the current sell order's price.
We maintain two priority queues, one for buy orders and one for sell orders. We process each order and update the backlogs accordingly. Finally, we return the total number of orders left in the backlog modulo 10^9 + 7.
Time Complexity:
- The time complexity of this approach is O(n log n) where n is the number of orders.
Space Complexity:
- The space complexity is O(n) to store the orders in the priority queues.
LeetCode 1801 Solutions in Java, C++, Python
import java.util.PriorityQueue;
class Solution {
public int getNumberOfBacklogOrders(int[][] orders) {
long MOD = 1000000007;
PriorityQueue<int[]> buy = new PriorityQueue<>((a, b) -> b[0] - a[0]); // Max heap for buy orders
PriorityQueue<int[]> sell = new PriorityQueue<>((a, b) -> a[0] - b[0]); // Min heap for sell orders
for (int[] order : orders) {
if (order[2] == 0) { // Buy order
while (!sell.isEmpty() && sell.peek()[0] <= order[0] && order[1] > 0) {
int[] currSell = sell.poll();
int minAmount = Math.min(currSell[1], order[1]);
order[1] -= minAmount;
currSell[1] -= minAmount;
if (currSell[1] > 0) {
sell.offer(currSell);
}
}
if (order[1] > 0) {
buy.offer(order);
}
} else { // Sell order
while (!buy.isEmpty() && buy.peek()[0] >= order[0] && order[1] > 0) {
int[] currBuy = buy.poll();
int minAmount = Math.min(currBuy[1], order[1]);
order[1] -= minAmount;
currBuy[1] -= minAmount;
if (currBuy[1] > 0) {
buy.offer(currBuy);
}
}
if (order[1] > 0) {
sell.offer(order);
}
}
}
long total = 0;
while (!buy.isEmpty()) {
total += buy.poll()[1];
}
while (!sell.isEmpty()) {
total += sell.poll()[1];
}
return (int) (total % MOD);
}
}Interactive Code Editor for LeetCode 1801
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