Stack using two queues

Problem Statement - link #

Implement a Stack using two queues q1 and q2

Your Task: Since this is a function problem, you don’t need to take inputs. You are required to complete the two methods push() which takes an integer ‘x’ as input denoting the element to be pushed into the stack and pop() which returns the integer poped out from the stack(-1 if the stack is empty).

Expected Time Complexity: O(1) for push() and O(N) for pop() or O(N) for push() and O(1) for pop()
Expected Auxiliary Space: O(1)

Examples #

Example 1:

Input:
push(2)
push(3)
pop()
push(4)
pop()
Output: 3 4
Explanation:
push(2) the stack will be {2}
push(3) the stack will be {2 3}
pop()   poped element will be 3 the 
        stack will be {2}
push(4) the stack will be {2 4}
pop()   poped element will be 4  

Example 2:

Input:
push(2)
pop()
pop()
push(3)
Output: 2 -1

Constraints #

• 1 <= N <= 100
• 1 <= x <= 100

Solutions #

//Function to push an element into stack using two queues.
void QueueStack :: push(int x)
{
        // Your Code
    q1.push(x);
    while(!q2.empty()) { q1.push(q2.front()); q2.pop(); }
    while(!q1.empty()) { q2.push(q1.front()); q1.pop(); }
    
}

//Function to pop an element from stack using two queues. 
int QueueStack :: pop()
{
    // Your Code   
    int res = -1;
    if(!q2.empty()) { res = q2.front(); q2.pop(); }
    return res;
}