Problem:
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Priority Queue

A priority queue is a data structure which maintains a set $S$ of elements, each of with an associated value (key), and supports the following operations:

$insert(S, k)$: insert an element $k$ into the set $S$

$extractMax(S)$: remove and return the element of $S$ with the largest key

Write a program which performs the $insert(S, k)$ and $extractMax(S)$ operations to a priority queue $S$.
The priority queue manages a set of integers, which are also keys for the priority.

Input

Multiple operations to the priority queue $S$ are given. Each operation is given by "insert $k$", "extract" or "end" in a line. Here, $k$ represents an integer element to be inserted to the priority queue.

The input ends with "end" operation.

Output

For each "extract" operation, print the element extracted from the priority queue $S$ in a line.

Constraints

The number of operations $\leq 2,000,000$

$0 \leq k \leq 2,000,000,000$

Sample Input 1

insert 8
insert 2
extract
insert 10
extract
insert 11
extract
extract
end

Sample Output 1

8
10
11
2

Solution in JAVA (judge verdict: Wrong Answer)

import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        List<Integer> a = new ArrayList<>();
        while (true) {
            String cmd = scanner.next();
            if ("insert".equals(cmd)) {
                a.add(scanner.nextInt());
                int i = a.size() - 1;
                while (i > 0 && a.get(i / 2) < a.get(i)) {
                    int t = a.get(i);
                    a.set(i, a.get(i / 2));
                    a.set(i / 2, t);
                    i /= 2;
                }
            } else if ("extract".equals(cmd)) {
                if (a.size() < 1) {
                    continue;
                }
                System.out.println(a.get(0));
                a.set(0, a.get(a.size() - 1));
                a.remove(a.size() - 1);
                maxHeapify(a, 0);
            } else {
                break;
            }
        }
    }

    private static void maxHeapify(List<Integer> a, int i) {
        int l = (i + 1) * 2 - 1;
        int r = l + 1;
        int largest = i;
        if (l <= a.size() - 1 && a.get(l) > a.get(i)) {
            largest = l;
        }
        if (r <= a.size() - 1 && a.get(r) > a.get(largest)) {
            largest = r;
        }
        if (largest != i) {
            int t = a.get(i);
            a.set(i, a.get(largest));
            a.set(largest, t);
            maxHeapify(a, largest);
        }
    }

    private static void buildMaxHeap(List<Integer> a) {
        for (int i = a.size() / 2; i >= 0; i--) {
            maxHeapify(a, i);
        }
    }
}