How to implement stack using priority queue or heap?

How to Implement stack using a priority queue(using min heap)?. Asked In: Microsoft, Adobe.

Solution: In priority queue, we assign priority to the elements that are being pushed. A stack requires elements to be processed in Last in First Out manner. The idea is to associate a count that determines when it was pushed. This count works as a key for the priority queue. So the implementation of stack uses a priority queue of pairs, with the first element serving as the key.

CPP

pair& lt;
int, int& gt;
(key, value)

See Below Image to understand Better 

 Below is C++ implementation of the idea. 

C++

// C++ program to implement a stack using
// Priority queue(min heap)
#include<bits/stdc++.h>
using namespace std;

typedef pair<int, int> pi;

// User defined stack class
class Stack{

    // cnt is used to keep track of the number of
    //elements in the stack and also serves as key
    //for the priority queue.
    int cnt;
    priority_queue<pair<int, int> > pq;
public:
    Stack():cnt(0){}
    void push(int n);
    void pop();
    int top();
    bool isEmpty();
};

// push function increases cnt by 1 and
// inserts this cnt with the original value.
void Stack::push(int n){
    cnt++;
    pq.push(pi(cnt, n));
}

// pops element and reduces count.
void Stack::pop(){
    if(pq.empty()){ cout<<"Nothing to pop!!!";}
    cnt--;
    pq.pop();
}

// returns the top element in the stack using
// cnt as key to determine top(highest priority),
// default comparator for pairs works fine in this case
int Stack::top(){
    pi temp=pq.top();
    return temp.second;
}

// return true if stack is empty
bool Stack::isEmpty(){
    return pq.empty();
}

// Driver code
int main()
{
    Stack* s=new Stack();
    s->push(1);
    s->push(2);
    s->push(3);
    while(!s->isEmpty()){
        cout<<s->top()<<endl;
        s->pop();
    }
}

Java

// Java program to implement a stack using
// Priority queue(min heap)
import java.util.PriorityQueue;

class Stack
{

  // cnt is used to keep track of the number of
  //elements in the stack and also serves as key
  //for the priority queue.
  int cnt;
  PriorityQueue<int[]> pq = new PriorityQueue<>((a, b) -> a[0] - b[0]);

  public Stack() {
    cnt = 0;
  }

  public void push(int n) {
    cnt++;
    pq.offer(new int[]{cnt, n});
  }

  public void pop() {
    if (pq.isEmpty()) {
      System.out.println("Nothing to pop!!!");
      return;
    }
    cnt--;
    pq.poll();
  }

  public int top() {
    int[] temp = pq.peek();
    return temp[1];
  }

  public boolean isEmpty() {
    return pq.isEmpty();
  }

  public static void main(String[] args) {
    Stack s = new Stack();
    s.push(3);
    s.push(2);
    s.push(1);
    while (!s.isEmpty()) {
      System.out.println(s.top());
      s.pop();
    }
  }
}

// This code is contributed by adityamaharshi21

Python3

import heapq

# User defined stack class
class Stack:
    # cnt is used to keep track of the number of
    # elements in the stack and also serves as key
    # for the priority queue.
    def __init__(self):
        self.cnt = 0
        self.pq = []

    def push(self, n):
        # push function increases cnt by 1 and
        # inserts this cnt with the original value.
        self.cnt += 1
        heapq.heappush(self.pq, (-self.cnt, n))

    def pop(self):
        # pops element and reduces count.
        if not self.pq:
            print("Nothing to pop!!!")
        self.cnt -= 1
        return heapq.heappop(self.pq)[1]

    def top(self):
        # returns the top element in the stack using
        # cnt as key to determine top(highest priority),
        # default comparator for pairs works fine in this case
        return self.pq[0][1]

    def isEmpty(self):
        # return true if stack is empty
        return not bool(self.pq)

# Driver code
s = Stack()
s.push(1)
s.push(2)
s.push(3)
while not s.isEmpty():
    print(s.top())
    s.pop()

C#

// C# program to implement a stack using
// Priority queue(min heap)
using System;
using System.Collections.Generic;

class Stack
{
    // cnt is used to keep track of the number of
   //elements in the stack and also serves as key
  //for the priority queue.
    List<int> stack = new List<int>();

    public void Push(int n)
    {
        stack.Add(n);
    }

    public int Pop()
    {
        if (stack.Count == 0)
        {
            Console.WriteLine("Nothing to pop!!!");
            return -1;
        }
        int lastIndex = stack.Count - 1;
        int last = stack[lastIndex];
        stack.RemoveAt(lastIndex);
        return last;
    }

    public int Top()
    {
        if (stack.Count == 0)
        {
            Console.WriteLine("Nothing to get the top!!!");
            return -1;
        }
        return stack[stack.Count - 1];
    }

    public bool IsEmpty()
    {
        return stack.Count == 0;
    }
}

class Program
{
    static void Main(string[] args)
    {
        Stack s = new Stack();
        s.Push(1);
        s.Push(2);
        s.Push(3);
        while (!s.IsEmpty())
        {
            Console.WriteLine(s.Top());
            s.Pop();
        }
    }
}

Javascript

class Stack {
    constructor() {
        this.stack = [];
    }

    push(n) {
        this.stack.push(n);
    }

    pop() {
        if (this.stack.length === 0) {
            console.log("Nothing to pop!!!");
        }
        return this.stack.pop();
    }

    top() {
        return this.stack[this.stack.length - 1];
    }

    isEmpty() {
        return this.stack.length === 0;
    }
}

// Driver code
let s = new Stack();
s.push(1);
s.push(2);
s.push(3);
while (!s.isEmpty()) {
    console.log(s.top());
    s.pop();
}

Output

3
2
1

Time Complexity: O(logn)

Now, as we can see this implementation takes O(log n) time for both push and pop operations. This can be slightly optimized by using fibonacci heap implementation of priority queue which would give us O(1) time complexity for push operation, but pop still requires O(log n) time. 

Auxiliary Space: O(n) where n is size of priority queue

This article is contributed by Mr. Somesh Awasthi. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.