 # Implement Queue Using Two Stacks Problem

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Given a sequence of enqueue and dequeue operations, return a result of their execution without using a queue data structure.

Operations are given in the form of a linked list:

● A non-negative integer means “enqueue me”.

● -1 means

○ If the queue is not empty, dequeue the current head and append it to the result.

○ If the queue is empty, append -1 to the result.

Use two stacks as an auxiliary data structure. Using a queue isn’t allowed.

Example One

Input: 1, -1, 2, -1, -1, 3, -1

Output: 1, 2, -1, 3

Here is how we would execute the operations and build the result list:

Example Two

Input: 0, 1, 2, -1, 3

Output: 0

The only dequeue operation results in the first enqueued element, 0, to be appended to the result list.

Notes

Input Parameters: Function has one argument: operations (singly linked list of integers of length N).

Output: Function must return a singly linked list of integers.

Constraints:

● -1

● 1

● There will be at least one dequeue (-1) operation.

Solution

#### 1) brute_force_solution.cpp

Any time we process an operation that puts a number to the queue, we need to store it somewhere. We are only allowed to use stacks to store numbers, so let’s be pushing all enqueued numbers into stack1.

Any time an operation tells us to retrieve a number from the queue, that number would be at the bottom of stack1 then. To access it we’d need to pop all the numbers from there. Luckily we are allowed to use another stack. So we can temporarily push all the numbers into stack2 and access the desired number from the bottom of stack1. After that we can push all the remaining numbers back from stack2 to stack1. They will end up being in the same order as they were there before the dequeue operation, and that works for us.

brute_force_solution.cpp uses this algorithm. What about its complexity?

Time Complexity:

O(N^2) for processing all N operations. Every dequeue operation requires moving around all the numbers so far enqueued, that makes the dequeue operation O(N).

Auxiliary Space Used:

O(N) since we never store more than N numbers.

Space Complexity:

O(N). Input, output and temporary data structures are all O(N) so O(N) + O(N) + O(N) = O(N).

``````
// -------- START --------

{
{
}
else
{
tail->next = node;
tail = tail->next;
}
return tail;
}

int dequeue(stack &s1, stack &s2)
{
if (s1.empty()) return -1;

// Pop all elements from first to second stack.
while (s1.empty() == false)
{
s2.push(s1.top());
s1.pop();
}

// Head of the second stack is the element to dequeue.
int val = s2.top();
s2.pop();

// Shovel all elements back to the first stack.
while (s2.empty() == false)
{
s1.push(s2.top());
s2.pop();
}

return val;
}

{
stack s1, s2;
while (operations != NULL)
{
if (operations->data >= 0)
{
s1.push(operations->data);
}
else
{
tail = insert_node(head, tail, dequeue(s1, s2));
{
}
}
operations = operations->next;
}
}

// -------- END --------
``````

#### 2) optimal_solution.cpp

Let’s imagine that we started with the brute force solution described above and came to this situation:

stack1 = [1, 2, 3], stack2 = [], next operation: -1

To dequeue, we move all numbers from stack1 to stack2:

stack1 = [], stack2 = [3, 2], add append number 1 to the result.

Now we can notice that stack2 has all the remaining numbers in the order that’s perfect for us. For example, if the next operation is -1, we can simply pop and return number 2 from stack2 - a constant time operation.

We can dequeue elements this way if we leave them in stack2, but what about enqueueing new ones? It turns out that we can push them in stack1, and they can remain there until stack2 is empty. Once stack2 is empty and another dequeue operation comes, we can do what was described two paragraphs ago: pop all numbers from stack1 and push them into stack2.

optimal_solution.cpp uses this algorithm.

Time Complexity:

Enqueue operation takes constant time, clearly. Let us examine dequeue operation.

Most dequeue operations will just need to pop one number from stack2, that’s constant time. Some dequeue operations however will need to move some numbers from stack1 to stack2.

Amortized time complexity of the dequeue operation is constant and intuitively we can observe that we never move any given number more than once between stack1 and stack2 (that’s constant time per number).

Overall time complexity of the algorithm would be O(N) since we process N operations, taking constant time per operation (amortized).

Auxiliary Space Used:

O(N) as we never store more than N numbers in our stacks.

Space Complexity:

O(N).

``````
// -------- START --------

{
{
}
else
{
tail->next = node;
tail = tail->next;
}
return tail;
}

int dequeue(stack &s1, stack &s2)
{
if (s1.empty() && s2.empty())
{
return -1;
}

// If second stack isn't empty, pop and return from it.
if (s2.empty() == false)
{
int val = s2.top();
s2.pop();
return val;
}

// Otherwise pop all elements from first to second stack.
while (s1.empty() == false)
{
s2.push(s1.top());
s1.pop();
}

// And then pop and return head of the second stack.
int val = s2.top();
s2.pop();
return val;
}

{
stack s1, s2;
while (operations != NULL)
{
if (operations->data >= 0)
{
s1.push(operations->data);
}
else
{
tail = insert_node(head, tail, dequeue(s1, s2));
{
}
}
operations = operations->next;
}
}

// -------- END --------
``````

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## How To Nail Your Next Tech Interview Hosted By
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Founder, Interview Kickstart Our tried & tested strategy for cracking interviews How FAANG hiring process works The 4 areas you must prepare for How you can accelerate your learnings

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