A queue with constant time operations

Implement a queue in which push_rear(), pop_front() and get_min() are all constant time operations.

We know that push and pop are constant time operations But when we think of get_min()[i.e to find the current minimum number in the queue] generally the first thing that comes to mind is searching the whole queue every time the request for the minimum element is made. But this will never give the constant time operation, which is the main aim of the problem.

To do this we have to use two more queues which will keep the track of minimum element and we have to go on modifying these 2 queues as we do push and pop operations on the queue so that minimum element is obtained in O(1) time.

import java.util.LinkedList;
import java.util.Queue;

public class MinQueue {

	Queue<Integer> q = new LinkedList<Integer>();
	Queue<Integer> minq1 = new LinkedList<Integer>();
	Queue<Integer> minq2 = new LinkedList<Integer>();
	boolean isMinq1Current = true;

	public void push(int a) {
		q.offer(a);
		if (isMinq1Current) {
			if (minq1.isEmpty())
				minq1.offer(a);
			else {
				while (!minq1.isEmpty() && minq1.peek() <= a)
					minq2.offer(minq1.poll());
				minq2.offer(a);
				while (!minq1.isEmpty())
					minq1.poll();
				isMinq1Current = false;
			}
		} else {
			if (minq2.isEmpty())
				minq2.offer(a);
			else {
				while (!minq2.isEmpty() && minq2.peek() <= a)
					minq1.offer(minq2.poll());
				minq1.offer(a);
				while (!minq2.isEmpty())
					minq2.poll();
				isMinq1Current = true;
			}
		}
	}

	public int pop() {
		int a = q.poll();
		if (isMinq1Current) {
			if (a == minq1.peek())
				minq1.poll();
		} else {
			if (a == minq2.peek())
				minq2.poll();
		}
		return a;
	}

	public int min() {

		if (isMinq1Current) {
			return minq1.peek();
		} else {
			return minq2.peek();
		}
	}

}