Okay - I think I have an answer that gives you all of these operations in amortized O(1), meaning that any one operation could take up to O(n), but any sequence of n operations takes O(1) time per operation.

The idea is to store your data as a Cartesian tree. This is a binary tree obeying the min-heap property (each node is no bigger than its children) and is ordered in a way such that an inorder traversal of the nodes gives you back the nodes in the same order in which they were added. For example, here's a Cartesian tree for the sequence 2 1 4 3 5:

 1 / \ 2 3 / \ 4 5 

It is possible to insert an element into a Cartesian tree in O(1) amortized time using the following procedure. Look at the right spine of the tree (the path from the root to the rightmost leaf formed by always walking to the right). Starting at rightmost node, scan upward along this path until you find the first node smaller than the node you're inserting.
Change that node so that its right child is this new node, then make that node's former right child the left child of the node you just added. For example, suppose that we want to insert another copy of 2 into the above tree. We walk up the right spine past the 5 and the 3, but stop below the 1 because 1 < 2. We then change the tree to look like this:

 1 / \ 2 2 / 3 / \ 4 5 

Notice that an inorder traversal gives 2 1 4 3 5 2, which is the sequence in which we added the values.

This runs in amortized O(1) because we can create a potential function equal to the number of nodes in the right spine of the tree. The real time required to insert a node is 1 plus the number of nodes in the spine we consider (call this k). Once we find the place to insert the node, the size of the spine shrinks by length k - 1, since each of the k nodes we visited are no longer on the right spine, and the new node is in its place. This gives an amortized cost of 1 + k + (1 - k) = 2 = O(1), for the amortized O(1) insert. As another way of thinking about this, once a node has been moved off the right spine, it's never part of the right spine again, and so we will never have to move it again. Since each of the n nodes can be moved at most once, this means that n insertions can do at most n moves, so the total runtime is at most O(n) for an amortized O(1) per element.

To do a dequeue step, we simply remove the leftmost node from the Cartesian tree. If this node is a leaf, we're done. Otherwise, the node can only have one child (the right child), and so we replace the node with its right child. Provided that we keep track of where the leftmost node is, this step takes O(1) time. However, after removing the leftmost node and replacing it with its right child, we might not know where the new leftmost node is. To fix this, we simply walk down the left spine of the tree starting at the new node we just moved to the leftmost child. I claim that this still runs in O(1) amortized time. To see this, I claim that a node is visited at most once during any one of these passes to find the leftmost node. To see this, note that once a node has been visited this way, the only way that we could ever need to look at it again would be if it were moved from a child of the leftmost node to the leftmost node. But all the nodes visited are parents of the leftmost node, so this can't happen. Consequently, each node is visited at most once during this process, and the pop runs in O(1).

We can do find-min in O(1) because the Cartesian tree gives us access to the smallest element of the tree for free; it's the root of the tree.

Finally, to see that the nodes come back in the same order in which they were inserted, note that a Cartesian tree always stores its elements so that an inorder traversal visits them in sorted order. Since we always remove the leftmost node at each step, and this is the first element of the inorder traversal, we always get the nodes back in the order in which they were inserted.

In short, we get O(1) amortized push and pop, and O(1) worst-case find-min.

If I can come up with a worst-case O(1) implementation, I'll definitely post it. This was a great problem; thanks for posting it!

Ok, here is one solution.

First we need some stuff which provide push_back(),push_front(),pop_back() and pop_front() in 0(1). It's easy to implement with array and 2 iterators. First iterator will point to front, second to back. Let's call such stuff deque.

Here is pseudo-code:

class MyQueue//Our data structure { deque D;//We need 2 deque objects deque Min; push(element)//pushing element to MyQueue { D.push_back(element); while(Min.is_not_empty() and Min.back()>element) Min.pop_back(); Min.push_back(element); } pop()//poping MyQueue { if(Min.front()==D.front() ) Min.pop_front(); D.pop_front(); } min() { return Min.front(); } } 

Explanation:

Example let's push numbers [12,5,10,7,11,19] and to our MyQueue

1)pushing 12

D [12] Min[12] 

2)pushing 5

D[12,5] Min[5] //5>12 so 12 removed 

3)pushing 10

D[12,5,10] Min[5,10] 

4)pushing 7

D[12,5,10,7] Min[5,7] 

6)pushing 11

D[12,5,10,7,11] Min[5,7,11] 

7)pushing 19

D[12,5,10,7,11,19] Min[5,7,11,19] 

Now let's call pop_front()

we got

 D[5,10,7,11,19] Min[5,7,11,19] 

The minimum is 5

Let's call pop_front() again

Explanation: pop_front will remove 5 from D, but it will pop front element of Min too, because it equals to D's front element (5).

 D[10,7,11,19] Min[7,11,19] 

And minimum is 7. :)

Use one deque (A) to store the elements and another deque (B) to store the minimums.

When x is enqueued, push_back it to A and keep pop_backing B until the back of B is smaller than x, then push_back x to B.

when dequeuing A, pop_front A as return value, and if it is equal to the front of B, pop_front B as well.

when getting the minimum of A, use the front of B as return value.

dequeue and getmin are obviously O(1). For the enqueue operation, consider the push_back of n elements. There are n push_back to A, n push_back to B and at most n pop_back of B because each element will either stay in B or being popped out once from B. Over all there are O(3n) operations and therefore the amortized cost is O(1) as well for enqueue.

Lastly the reason this algorithm works is that when you enqueue x to A, if there are elements in B that are larger than x, they will never be minimums now because x will stay in the queue A longer than any elements in B (a queue is FIFO). Therefore we need to pop out elements in B (from the back) that are larger than x before we push x into B.

from collections import deque class MinQueue(deque): def __init__(self): deque.__init__(self) self.minq = deque() def push_rear(self, x): self.append(x) while len(self.minq) > 0 and self.minq[-1] > x: self.minq.pop() self.minq.append(x) def pop_front(self): x = self.popleft() if self.minq[0] == x: self.minq.popleft() return(x) def get_min(self): return(self.minq[0]) 

If you don't mind storing a bit of extra data, it should be trivial to store the minimum value. Push and pop can update the value if the new or removed element is the minimum, and returning the minimum value is as simple as getting the value of the variable.

This is assuming that get_min() does not change the data; if you would rather have something like pop_min() (i.e. remove the minimum element), you can simply store a pointer to the actual element and the element preceding it (if any), and update those accordingly with push_rear() and pop_front() as well.

Edit after comments:

Obviously this leads to O(n) push and pop in the case that the minimum changes on those operations, and so does not strictly satisfy the requirements.

You Can actually use a LinkedList to maintain the Queue.

Each element in LinkedList will be of Type

class LinkedListElement { LinkedListElement next; int currentMin; } 

You can have two pointers One points to the Start and the other points to the End.

If you add an element to the start of the Queue. Examine the Start pointer and the node to insert. If node to insert currentmin is less than start currentmin node to insert currentmin is the minimum. Else update the currentmin with start currentmin.

Repeat the same for enque.

JavaScript implementation

(Credit to adamax's solution for the idea; I loosely based an implementation on it. Jump to the bottom to see fully commented code or read through the general steps below. Note that this finds the maximum value in O(1) constant time rather than the minimum value--easy to change up):

The general idea is to create two Stacks upon construction of the MaxQueue (I used a linked list as the underlying Stack data structure--not included in the code; but any Stack will do as long as it's implemented with O(1) insertion/deletion). One we'll mostly pop from (dqStack) and one we'll mostly push to (eqStack).

For enqueue, if the MaxQueue is empty, we'll push the value to dqStack along with the current max value in a tuple (the same value since it's the only value in the MaxQueue); e.g.:

const m = new MaxQueue(); m.enqueue(6); /* the dqStack now looks like: [6, 6] - [value, max] */ 

If the MaxQueue is not empty, we push just the value to eqStack;

m.enqueue(7); m.enqueue(8); /* dqStack: eqStack: 8 [6, 6] 7 - just the value */ 

then, update the maximum value in the tuple.

/* dqStack: eqStack: 8 [6, 8] 7 */ 

For dequeue we'll pop from dqStack and return the value from the tuple.

m.dequeue(); > 6 // equivalent to: /* const tuple = m.dqStack.pop() // [6, 8] tuple[0]; > 6 */ 

Then, if dqStack is empty, move all values in eqStack to dqStack, e.g.:

// if we build a MaxQueue const maxQ = new MaxQueue(3, 5, 2, 4, 1); /* the stacks will look like: dqStack: eqStack: 1 4 2 [3, 5] 5 */ 

As each value is moved over, we'll check if it's greater than the max so far and store it in each tuple:

maxQ.dequeue(); // pops from dqStack (now empty), so move all from eqStack to dqStack > 3 // as dequeue moves one value over, it checks if it's greater than the ***previous max*** and stores the max at tuple[1], i.e., [data, max]: /* dqStack: [5, 5] => 5 > 4 - update eqStack: [2, 4] => 2 < 4 - no update [4, 4] => 4 > 1 - update [1, 1] => 1st value moved over so max is itself empty */ 

Because each value is moved to dqStack at most once, we can say that dequeue has O(1) amortized time complexity.

Then, at any point in time, we can call getMax to retrieve the current maximum value in O(1) constant time. As long as the MaxQueue is not empty, the maximum value is easily pulled out of the next tuple in dqStack.

maxQ.getMax(); > 5 // equivalent to calling peek on the dqStack and pulling out the maximum value: /* const peekedTuple = maxQ.dqStack.peek(); // [5, 5] peekedTuple[1]; > 5 */ 

Code

class MaxQueue { constructor(...data) { // create a dequeue Stack from which we'll pop this.dqStack = new Stack(); // create an enqueue Stack to which we'll push this.eqStack = new Stack(); // if enqueueing data at construction, iterate through data and enqueue each if (data.length) for (const datum of data) this.enqueue(datum); } enqueue(data) { // O(1) constant insertion time // if the MaxQueue is empty, if (!this.peek()) { // push data to the dequeue Stack and indicate it's the max; this.dqStack.push([data, data]); // e.g., enqueue(8) ==> [data: 8, max: 8] } else { // otherwise, the MaxQueue is not empty; push data to enqueue Stack this.eqStack.push(data); // save a reference to the tuple that's next in line to be dequeued const next = this.dqStack.peek(); // if the enqueueing data is > the max in that tuple, update it if (data > next[1]) next[1] = data; } } moveAllFromEqToDq() { // O(1) amortized as each value will move at most once // start max at -Infinity for comparison with the first value let max = -Infinity; // until enqueue Stack is empty, while (this.eqStack.peek()) { // pop from enqueue Stack and save its data const data = this.eqStack.pop(); // if data is > max, set max to data if (data > max) max = data; // push to dequeue Stack and indicate the current max; e.g., [data: 7: max: 8] this.dqStack.push([data, max]); } } dequeue() { // O(1) amortized deletion due to calling moveAllFromEqToDq from time-to-time // if the MaxQueue is empty, return undefined if (!this.peek()) return; // pop from the dequeue Stack and save it's data const [data] = this.dqStack.pop(); // if there's no data left in dequeue Stack, move all data from enqueue Stack if (!this.dqStack.peek()) this.moveAllFromEqToDq(); // return the data return data; } peek() { // O(1) constant peek time // if the MaxQueue is empty, return undefined if (!this.dqStack.peek()) return; // peek at dequeue Stack and return its data return this.dqStack.peek()[0]; } getMax() { // O(1) constant time to find maximum value // if the MaxQueue is empty, return undefined if (!this.peek()) return; // peek at dequeue Stack and return the current max return this.dqStack.peek()[1]; } }

#include <iostream> #include <queue> #include <deque> using namespace std; queue<int> main_queue; deque<int> min_queue; void clearQueue(deque<int> &q) { while(q.empty() == false) q.pop_front(); } void PushRear(int elem) { main_queue.push(elem); if(min_queue.empty() == false && elem < min_queue.front()) { clearQueue(min_queue); } while(min_queue.empty() == false && elem < min_queue.back()) { min_queue.pop_back(); } min_queue.push_back(elem); } void PopFront() { int elem = main_queue.front(); main_queue.pop(); if (elem == min_queue.front()) { min_queue.pop_front(); } } int GetMin() { return min_queue.front(); } int main() { PushRear(1); PushRear(-1); PushRear(2); cout<<GetMin()<<endl; PopFront(); PopFront(); cout<<GetMin()<<endl; return 0; } 

This solution contains 2 queues:
1. main_q - stores the input numbers.
2. min_q - stores the min numbers by certain rules that we'll described (appear in functions MainQ.enqueue(x), MainQ.dequeue(), MainQ.get_min()).

Here's the code in Python. Queue is implemented using a List.
The main idea lies in the MainQ.enqueue(x), MainQ.dequeue(), MainQ.get_min() functions.
One key assumption is that emptying a queue takes o(0).
A test is provided at the end.

import numbers class EmptyQueueException(Exception): pass class BaseQ(): def __init__(self): self.l = list() def enqueue(self, x): assert isinstance(x, numbers.Number) self.l.append(x) def dequeue(self): return self.l.pop(0) def peek_first(self): return self.l[0] def peek_last(self): return self.l[len(self.l)-1] def empty(self): return self.l==None or len(self.l)==0 def clear(self): self.l=[] class MainQ(BaseQ): def __init__(self, min_q): super().__init__() self.min_q = min_q def enqueue(self, x): super().enqueue(x) if self.min_q.empty(): self.min_q.enqueue(x) elif x > self.min_q.peek_last(): self.min_q.enqueue(x) else: # x <= self.min_q.peek_last(): self.min_q.clear() self.min_q.enqueue(x) def dequeue(self): if self.empty(): raise EmptyQueueException("Queue is empty") x = super().dequeue() if x == self.min_q.peek_first(): self.min_q.dequeue() return x def get_min(self): if self.empty(): raise EmptyQueueException("Queue is empty, NO minimum") return self.min_q.peek_first() INPUT_NUMS = (("+", 5), ("+", 10), ("+", 3), ("+", 6), ("+", 1), ("+", 2), ("+", 4), ("+", -4), ("+", 100), ("+", -40), ("-",None), ("-",None), ("-",None), ("+",-400), ("+",90), ("-",None), ("-",None), ("-",None), ("-",None), ("-",None), ("-",None), ("-",None), ("-",None), ("-",None)) if __name__ == '__main__': min_q = BaseQ() main_q = MainQ(min_q) try: for operator, i in INPUT_NUMS: if operator=="+": main_q.enqueue(i) print("Added {} ; Min is: {}".format(i,main_q.get_min())) print("main_q = {}".format(main_q.l)) print("min_q = {}".format(main_q.min_q.l)) print("==========") else: x = main_q.dequeue() print("Removed {} ; Min is: {}".format(x,main_q.get_min())) print("main_q = {}".format(main_q.l)) print("min_q = {}".format(main_q.min_q.l)) print("==========") except Exception as e: print("exception: {}".format(e)) 

The output of the above test is:

"C:\Program Files\Python35\python.exe" C:/dev/python/py3_pocs/proj1/priority_queue.py Added 5 ; Min is: 5 main_q = [5] min_q = [5] ========== Added 10 ; Min is: 5 main_q = [5, 10] min_q = [5, 10] ========== Added 3 ; Min is: 3 main_q = [5, 10, 3] min_q = [3] ========== Added 6 ; Min is: 3 main_q = [5, 10, 3, 6] min_q = [3, 6] ========== Added 1 ; Min is: 1 main_q = [5, 10, 3, 6, 1] min_q = [1] ========== Added 2 ; Min is: 1 main_q = [5, 10, 3, 6, 1, 2] min_q = [1, 2] ========== Added 4 ; Min is: 1 main_q = [5, 10, 3, 6, 1, 2, 4] min_q = [1, 2, 4] ========== Added -4 ; Min is: -4 main_q = [5, 10, 3, 6, 1, 2, 4, -4] min_q = [-4] ========== Added 100 ; Min is: -4 main_q = [5, 10, 3, 6, 1, 2, 4, -4, 100] min_q = [-4, 100] ========== Added -40 ; Min is: -40 main_q = [5, 10, 3, 6, 1, 2, 4, -4, 100, -40] min_q = [-40] ========== Removed 5 ; Min is: -40 main_q = [10, 3, 6, 1, 2, 4, -4, 100, -40] min_q = [-40] ========== Removed 10 ; Min is: -40 main_q = [3, 6, 1, 2, 4, -4, 100, -40] min_q = [-40] ========== Removed 3 ; Min is: -40 main_q = [6, 1, 2, 4, -4, 100, -40] min_q = [-40] ========== Added -400 ; Min is: -400 main_q = [6, 1, 2, 4, -4, 100, -40, -400] min_q = [-400] ========== Added 90 ; Min is: -400 main_q = [6, 1, 2, 4, -4, 100, -40, -400, 90] min_q = [-400, 90] ========== Removed 6 ; Min is: -400 main_q = [1, 2, 4, -4, 100, -40, -400, 90] min_q = [-400, 90] ========== Removed 1 ; Min is: -400 main_q = [2, 4, -4, 100, -40, -400, 90] min_q = [-400, 90] ========== Removed 2 ; Min is: -400 main_q = [4, -4, 100, -40, -400, 90] min_q = [-400, 90] ========== Removed 4 ; Min is: -400 main_q = [-4, 100, -40, -400, 90] min_q = [-400, 90] ========== Removed -4 ; Min is: -400 main_q = [100, -40, -400, 90] min_q = [-400, 90] ========== Removed 100 ; Min is: -400 main_q = [-40, -400, 90] min_q = [-400, 90] ========== Removed -40 ; Min is: -400 main_q = [-400, 90] min_q = [-400, 90] ========== Removed -400 ; Min is: 90 main_q = [90] min_q = [90] ========== exception: Queue is empty, NO minimum Process finished with exit code 0 

Java Implementation

import java.io.*; import java.util.*; public class queueMin { static class stack { private Node<Integer> head; public void push(int data) { Node<Integer> newNode = new Node<Integer>(data); if(null == head) { head = newNode; } else { Node<Integer> prev = head; head = newNode; head.setNext(prev); } } public int pop() { int data = -1; if(null == head){ System.out.println("Error Nothing to pop"); } else { data = head.getData(); head = head.getNext(); } return data; } public int peek(){ if(null == head){ System.out.println("Error Nothing to pop"); return -1; } else { return head.getData(); } } public boolean isEmpty(){ return null == head; } } static class stackMin extends stack { private stack s2; public stackMin(){ s2 = new stack(); } public void push(int data){ if(data <= getMin()){ s2.push(data); } super.push(data); } public int pop(){ int value = super.pop(); if(value == getMin()) { s2.pop(); } return value; } public int getMin(){ if(s2.isEmpty()) { return Integer.MAX_VALUE; } return s2.peek(); } } static class Queue { private stackMin s1, s2; public Queue(){ s1 = new stackMin(); s2 = new stackMin(); } public void enQueue(int data) { s1.push(data); } public int deQueue() { if(s2.isEmpty()) { while(!s1.isEmpty()) { s2.push(s1.pop()); } } return s2.pop(); } public int getMin(){ return Math.min(s1.isEmpty() ? Integer.MAX_VALUE : s1.getMin(), s2.isEmpty() ? Integer.MAX_VALUE : s2.getMin()); } } static class Node<T> { private T data; private T min; private Node<T> next; public Node(T data){ this.data = data; this.next = null; } public void setNext(Node<T> next){ this.next = next; } public T getData(){ return this.data; } public Node<T> getNext(){ return this.next; } public void setMin(T min){ this.min = min; } public T getMin(){ return this.min; } } public static void main(String args[]){ try { FastScanner in = newInput(); PrintWriter out = newOutput(); // System.out.println(out); Queue q = new Queue(); int t = in.nextInt(); while(t-- > 0) { String[] inp = in.nextLine().split(" "); switch (inp[0]) { case "+": q.enQueue(Integer.parseInt(inp[1])); break; case "-": q.deQueue(); break; case "?": out.println(q.getMin()); default: break; } } out.flush(); out.close(); } catch(IOException e){ e.printStackTrace(); } } static class FastScanner { static BufferedReader br; static StringTokenizer st; FastScanner(File f) { try { br = new BufferedReader(new FileReader(f)); } catch (FileNotFoundException e) { e.printStackTrace(); } } public FastScanner(InputStream f) { br = new BufferedReader(new InputStreamReader(f)); } String next() { while (st == null || !st.hasMoreTokens()) { try { st = new StringTokenizer(br.readLine()); } catch (IOException e) { e.printStackTrace(); } } return st.nextToken(); } String nextLine(){ String str = ""; try { str = br.readLine(); } catch (IOException e) { e.printStackTrace(); } return str; } int nextInt() { return Integer.parseInt(next()); } long nextLong() { return Long.parseLong(next()); } double nextDoulbe() { return Double.parseDouble(next()); } } static FastScanner newInput() throws IOException { if (System.getProperty("JUDGE") != null) { return new FastScanner(new File("input.txt")); } else { return new FastScanner(System.in); } } static PrintWriter newOutput() throws IOException { if (System.getProperty("JUDGE") != null) { return new PrintWriter("output.txt"); } else { return new PrintWriter(System.out); } } } 

We know that push and pop are constant time operations [O(1) to be precise].

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.

This is generally asked very frequently in the interviews, so you must know the trick

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.

Here is the self-descriptive pseudo code based on the above approach mentioned.

 Queue q, minq1, minq2; isMinq1Current=true; void push(int a) { q.push(a); if(isMinq1Current) { if(minq1.empty) minq1.push(a); else { while(!minq1.empty && minq1.top < =a) minq2.push(minq1.pop()); minq2.push(a); while(!minq1.empty) minq1.pop(); isMinq1Current=false; } } else { //mirror if(isMinq1Current) branch. } } int pop() { int a = q.pop(); if(isMinq1Current) { if(a==minq1.top) minq1.pop(); } else { //mirror if(isMinq1Current) branch. } return a; }