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: Time Limit Exceeded)

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Main {
	private static int[] que;
	private static int count;
	public static void main(String[] args)throws IOException{
		que = new int[10000000];
		count =0;
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		while(true){
			String[] str=br.readLine().split(" ");
			if(str[0].equals("end")) break;
			else if(str[0].equals("extract")) extract();
			else if(str[0].equals("insert")){
				int n=Integer.parseInt(str[1]);
				insert(n);
			}
		}
	}
	
	private static void extract(){
		int max=que[1];
		System.out.println(max);
		que[1]=que[count];
		count--;
		maxHeap(1);
	}
	
	private static void insert(int n){
		count++;
		que[count]=n;
		for(int i=n/2;i>=1;i--) maxHeap(i);
	}
	
	private static void maxHeap(int i){
		int l=2*i;
		int r=2*i+1;
		int max;
		if(l<=count && que[l]>que[i]) max=l;
		else max=i;
		if(r<=count && que[r]>que[max]) max=r;
		
		if(max!=i){
			int tmp=que[i];
			que[i]=que[max];
			que[max]=tmp;
			maxHeap(max);
		}
	}
	
}