The problem is opposite of this post. We are given a Queue data structure that supports standard operations like enqueue() and dequeue(). We need to implement a Stack data structure using only instances of Queue and queue operations allowed on the instances.
A stack can be implemented using two queues. Let stack to be implemented be ‘s’ and queues used to implement be ‘q1’ and ‘q2’. Stack ‘s’ can be implemented in two ways:
Method 1 (By making push operation costly)
This method makes sure that newly entered element is always at the front of ‘q1’, so that pop operation just dequeues from ‘q1’. ‘q2’ is used to put every new element at front of ‘q1’.
push(s, x) // x is the element to be pushed and s is stack 1) Enqueue x to q2 2) One by one dequeue everything from q1 and enqueue to q2. 3) Swap the names of q1 and q2 // Swapping of names is done to avoid one more movement of all elements // from q2 to q1. pop(s) 1) Dequeue an item from q1 and return it.
C++
/* Program to implement a stack using two queue */#include<bits/stdc++.h> using namespace std; class Stack { // Two inbuilt queues queue<int> q1, q2; // To maintain current number of // elements int curr_size; public: Stack() { curr_size = 0; } void push(int x) { curr_size++; // Push x first in empty q2 q2.push(x); // Push all the remaining // elements in q1 to q2. while (!q1.empty()) { q2.push(q1.front()); q1.pop(); } // swap the names of two queues queue<int> q = q1; q1 = q2; q2 = q; } void pop(){ // if no elements are there in q1 if (q1.empty()) return ; q1.pop(); curr_size--; } int top() { if (q1.empty()) return -1; return q1.front(); } int size() { return curr_size; } }; // driver code int main() { Stack s; s.push(1); s.push(2); s.push(3); cout << "current size: " << s.size() << endl; cout << s.top() << endl; s.pop(); cout << s.top() << endl; s.pop(); cout << s.top() << endl; cout << "current size: " << s.size() << endl; return 0; } // This code is contributed by Chhavi |
Java
/* Java Program to implement a stack using two queue */import java.util.*; class GfG { static class Stack { // Two inbuilt queues static Queue<Integer> q1 = new LinkedList<Integer>(); static Queue<Integer> q2 = new LinkedList<Integer>(); // To maintain current number of // elements static int curr_size; Stack() { curr_size = 0; } static void push(int x) { curr_size++; // Push x first in empty q2 q2.add(x); // Push all the remaining // elements in q1 to q2. while (!q1.isEmpty()) { q2.add(q1.peek()); q1.remove(); } // swap the names of two queues Queue<Integer> q = q1; q1 = q2; q2 = q; } static void pop(){ // if no elements are there in q1 if (q1.isEmpty()) return ; q1.remove(); curr_size--; } static int top() { if (q1.isEmpty()) return -1; return q1.peek(); } static int size() { return curr_size; } }; // driver code public static void main(String[] args) { Stack s = new Stack(); s.push(1); s.push(2); s.push(3); System.out.println("current size: " + s.size()); System.out.println(s.top()); s.pop(); System.out.println(s.top()); s.pop(); System.out.println(s.top()); System.out.println("current size: " + s.size()); } } // This code is contributed by Prerna |
Output :
current size: 3 3 2 1 current size: 1
Method 2 (By making pop operation costly)
In push operation, the new element is always enqueued to q1. In pop() operation, if q2 is empty then all the elements except the last, are moved to q2. Finally the last element is dequeued from q1 and returned.
push(s, x) 1) Enqueue x to q1 (assuming size of q1 is unlimited). pop(s) 1) One by one dequeue everything except the last element from q1 and enqueue to q2. 2) Dequeue the last item of q1, the dequeued item is result, store it. 3) Swap the names of q1 and q2 4) Return the item stored in step 2. // Swapping of names is done to avoid one more movement of all elements // from q2 to q1.
/* Program to implement a stack using two queue */#include<bits/stdc++.h> using namespace std; class Stack { queue<int> q1, q2; int curr_size; public: Stack() { curr_size = 0; } void pop() { if (q1.empty()) return; // Leave one element in q1 and // push others in q2. while (q1.size() != 1) { q2.push(q1.front()); q1.pop(); } // Pop the only left element // from q1 q1.pop(); curr_size--; // swap the names of two queues queue<int> q = q1; q1 = q2; q2 = q; } void push(int x) { q1.push(x); curr_size++; } int top() { if (q1.empty()) return -1; while( q1.size() != 1 ) { q2.push(q1.front()); q1.pop(); } // last pushed element int temp = q1.front(); // to empty the auxiliary queue after // last operation q1.pop(); // push last element to q2 q2.push(temp); // swap the two queues names queue<int> q = q1; q1 = q2; q2 = q; return temp; } int size() { return curr_size; } }; // driver code int main() { Stack s; s.push(1); s.push(2); s.push(3); s.push(4); cout << "current size: " << s.size() << endl; cout << s.top() << endl; s.pop(); cout << s.top() << endl; s.pop(); cout << s.top() << endl; cout << "current size: " << s.size() << endl; return 0; } // This code is contributed by Chhavi |
Output :
current size: 4 4 3 2 current size: 2
References:
Implement Stack using Two Queues
This article is compiled by Sumit Jain and reviewed by GeeksforGeeks team. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
Improved By : prerna saini