# How to implement stack using priority queue or heap?

How to Implement stack using a priority queue(using min heap)?.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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.

`pair <int, int> (key, value)`

See Below Image to understand Better

Below is C++ implementation of the idea.

```// 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();
}
}
```

Output:

``` 3
2
1
```

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.

This article is contributed by Mr. Somesh Awasthi. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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