# Implementing a queue using two stacks

Question 3.5 of *Cracking the Coding Interview*:

Implement a

`MyQueue`

class which implements a queue using two stacks.

First, let’s try this with one stack,
which we’ll call `queueBack`

.
We’ll treat the top of the stack as the back of the queue:
so enqueueing an element is just pushing it on the stack.
But when we need to to dequeue an element,
we need to access to the bottom of the stack.

One way to access the bottom of a stack
is to reverse the stack.
And one way to reverse a stack
is to repeatedly pop items off of it,
pushing them onto a second stack.
Notice that the top of this second stack
then contains the front of the queue,
in the right order to be dequeued.
So let’s call this second stack `queueFront`

.

So we have two stacks,
`queueBack`

and `queueFront`

,
with easy access to the back and the front of the queue!
The only issue arises when we need to `dequeue`

but `queueFront`

is empty.
In this case, we reverse `queueBack`

onto `queueFront`

before continuing.
This means `dequeue`

is a linear-time operation.

There are better algorithms than this.
In a language with mutation,
you could just keep pointers to the `head`

and `tail`

of a linked list.
This will give you a queue with constant-time `dequeue`

and `enqueue`

,
which is optimal.
In a functional language,
you could use a balanced tree,
which will give you logarithmic-time `dequeue`

and `enqueue`

.

Regardless, here’s an implementation in Haskell:

```
module TwoStackQueue where
import Test.QuickCheck
data Queue a = Queue {
queueBack :: [a], -- head is back of queue
queueFront :: [a] -- head is front of queue
}
emptyQueue :: Queue a
emptyQueue = Queue [] []
enqueue :: a -> Queue a -> Queue a
enqueue x q = q { queueBack = x:(queueBack q) }
dequeue :: Queue a -> Maybe (a, Queue a)
dequeue q =
case queueFront q of
front:rest ->
Just (front, q { queueFront = rest })
[] ->
case reverse (queueBack q) of
front:rest ->
Just (front, Queue [] rest)
[] -> Nothing
-----------------------------------------
----------------- TESTS -----------------
data Op = Enqueue Int | Dequeue deriving (Show)
instance Arbitrary Op where
arbitrary = oneof [Enqueue <$> arbitrary, return Dequeue]
-- returns all elems dequeued when running all ops
runOps :: [Op] -> [Int]
runOps ops = snd $ foldl runOp (emptyQueue, []) ops where
runOp :: (Queue Int, [Int]) -> Op -> (Queue Int, [Int])
runOp (q,out) (Enqueue x) = (enqueue x q, out)
runOp (q,out) Dequeue = case dequeue q of
Just (x, q2) -> (q2, out++[x])
Nothing -> (q, out)
-- model is a plain list
runOpsModel :: [Op] -> [Int]
runOpsModel ops = snd $ foldl runOp ([],[]) ops where
runOp :: ([Int], [Int]) -> Op -> ([Int], [Int])
runOp (q, out) (Enqueue x) = (x:q, out)
runOp (q, out) Dequeue = case reverse q of
[] -> ([], out)
front:rest -> (reverse rest, out ++ [front])
prop_modelcheck :: [Op] -> Bool
prop_modelcheck ops = runOps ops == runOpsModel ops
main = quickCheck prop_modelcheck
```

With Vidrio

With generic competitor

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