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: Accepted)

import java.util.Scanner;
public class Main {
private static Scanner s = new Scanner(System.in);
static int count = 0;
static int[] A = new int[2000001];
public static void main(String[] args) {
	int k;
	while(true) {
		String com = s.next();
		if(com.equals("insert")) {
			k = s.nextInt();
			insert(k);
		}else if(com.equals("extract")) {
			System.out.println(extract());
		}else {
			return;
		}
	}
}
static void insert(int n) {
	count++;
	A[count] = n;
	int i = count;
	while(i > 1 && A[i/2] < A[i]) {
		int tmp = A[i/2];
		A[i/2] = A[i];
		A[i] = tmp;
		i /= 2;
	}
}

static int extract() {
	if(count < 1) return -1;
	int set = A[1];
	A[1] = A[count];
	count--;
	maxHeapify(1);
	return set;
}

static void maxHeapify(int i) {
	int l = i * 2;
	int r = i * 2 + 1;
	int max;
	if(l <= count && A[l] > A[i]) {
		max = l;
	} else {
		max = i;
	}
	if(r <= count && A[r] > A[max]) {
		max = r;
	}
	if(max != i) {
		int tmp = A[i];
		A[i] = A[max];
		A[max] = tmp;
		maxHeapify(max);
	}
}
}