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 JavaScript (judge verdict: Accepted)

class priorityQueue {
  constructor() {
    this.keys = [];
    Number.prototype.parentIdx = function() { return Math.ceil(this/2 - 1); };
    Number.prototype.leftIdx = function() { return this * 2 + 1; };
    Number.prototype.rightIdx = function() { return this * 2 + 2; };
  }

  insert(key) { this.heapIncrease(this.keys.push(key) - 1); }

  heapIncrease(idx) {
    const keys = this.keys;
    for (let i=idx, p=idx.parentIdx(); i > 0 && keys[i] > keys[p]; i=p, p=i.parentIdx()) {
      keys[p] = [keys[i], keys[i] = keys[p]][0];
    }
  }

  extract() {
    const keys = this.keys;
    const res = keys[0];
    keys[0] = keys.pop();
    this.maxHeapify(0);
    return res;
  }

  maxHeapify(idx) {
    const keys = this.keys;
    const n = keys.length;
    const innerMaxHeapify = (i) => {
      let left, right, largest = i;
      left = i.leftIdx();
      right = i.rightIdx();
      if (left < n && Number(keys[left]) > Number(keys[largest])) largest = left;
      if (right < n && Number(keys[right]) > Number(keys[largest])) largest = right;
      if (largest != i) {
        keys[largest] = [keys[i], keys[i] = keys[largest]][0];
        innerMaxHeapify(largest);
      }
    };
    innerMaxHeapify(idx);
  }
}

(function main() {
  const lines = require('fs').readFileSync('/dev/stdin', 'utf8').trim().split('\n');
  lines.pop();
  const queue = new priorityQueue();
  let str = '';
  lines.forEach(function(cmd) {
    switch(cmd.charAt(0)) {
      case 'i': queue.insert(Number(cmd.split(' ')[1])); break;
      case 'e': str += `${queue.extract()|0}\n`; break;
    }
  });
  process.stdout.write(str);
})();