I have defined the Stack and Queue class already so I can use any of the Queue methods - Queue(), enqueue(), dequeue(), peek(), size() and is_empty() and any of the Stack ADT methods: Stack(), push(), pop(), peek(), size() and is_empty().
What I basically need to do is modify the parameter Queue object so that the original queue items appear in their original order followed by a copy of the queue items in reverse order.
Here is what I have so far. But this is only giving me the reversed queue. Can someone help how I can modify the queue to both the original and reversed versions.
The Stack class:
class Stack:
def __init__(self):
self.items = []
def is_empty(self):
return self.items == []
def push(self, item):
self.items.append(item)
def pop(self):
if Stack.is_empty(self) == True:
return None
else:
return self.items.pop()
def peek(self):
if Stack.is_empty(self) == True:
return None
else:
return self.items[-1]
def size(self):
return len(self.items)
Queue class:
class Queue:
def __init__(self):
self.items = []
def is_empty(self):
return self.items == []
def enqueue(self, item):
self.items.insert(0,item)
def dequeue(self):
if len(self.items)==0:
raise IndexError("ERROR: The queue is empty!")
return self.items.pop()
def size(self):
return len(self.items)
def peek(self):
if Queue.is_empty(self) == True:
raise IndexError("ERROR: The queue is empty!")
else:
return self.items[-1]
def __eq__(self, other):
self.other = other
if self.items == self.other:
return True
else:
return False
def clear(self):
del self.items[:]
def __str__(self):
return "{}".format(self.items[::-1])
Function I am required to define:
def mirror_queue(q):
stack = Stack()
while not q.is_empty():
stack.push(q.dequeue())
while not stack.is_empty():
q.enqueue(stack.pop())
---test
q1 = Queue()
q1.enqueue(1)
q1.enqueue(2)
q1.enqueue(3)
print(q1)
mirror_queue(q1)
print(q1)
---Expected Output
Queue: [1, 2, 3]
Queue: [1, 2, 3, 3, 2, 1]
---Gotten Ouput
Queue: [1, 2, 3]
Queue: [3, 2, 1]
mirror_queue seems to just reverse your queue.
My approach:
Don’t change the original q but reverse a copy of q1 called q1r.
Afterwards concat those two which then is your result (q1 + q1r).
When you use the function dequeue, it takes out one item from your queue, so when the first loop ends you have an empty queue and the stack with 3 elements. When you pop from your stack it enqueue back to your queue in reverse order.
You should make a copy of the elements of the queue without dequeue it in order to archive what you want.
Related
How do you remove and return an item at the index of the queue for Python? I'm trying to do this problem.
When you send a file to be printed on a shared printer, it is put onto a print queue with other jobs. Anytime before your job prints, you can access the queue to remove it. Thus, some queues support a remove operation. The remove() method should expect an item as an argument. It should remove the given item in the queue or raise an exception if the item is not found.
class Queue:
def __init__(self): # Constructor function
self.items = []
def isEmpty(self): # Returns True if the queue is empty or False otherwise
return self.items == []
def len(self): # Returns the number of items in the queue
return len(self.items)
def peek(self): # Returns the item at the front of the queue
if len(self.items) == 0:
return None
else:
return self.items[(len(self.items)-1)]
def add(self, item): # Adds item to the rear of the queue
return self.items.append(item)
def pop(self): # Removes and returns the item at the front of the queue
return self.items.pop()
def remove(self):
return.self.index.pop()
if __name__ == "__main__":
queue = Queue()
To remove an item you can use:
def remove(self, item):
return self.items.remove(item)
which will generate a ValueError if item is not in the items List. If required you could use a try: except: block to handle the error in some other way.
import collections
class MyStack:
def __init__(self):
self._data = []
def push(self, value):
self._data.append(value)
def size(self):
#Return the number of elements in the stack
return len(self._data)
def toString(self):
#Return a string representing the content of this stack
return str(self._data)
class MyQueue:
def __init__(self):
self._data = collections.deque([])
def enqueue(self, value):
self._data.append(value)
def dequeue(self):
return self._data.popleft()
def size(self):
#return the number of elements in the queue
return len(self._data)
queue1 = MyQueue()
dq = MyStack()
stack1 = MyStack()
stack1.push(['stone', 'atone'])
print "size of stack is now :" ,stack1.size()
queue1.enqueue(stack1)
print "size of queue is now :", queue1.size()
print "size of stack is now :" ,stack1.size()
stack1.push(['stone', 'shone'])
stack1.push(['stone', 'scone'])
dq = queue1.dequeue() # i would like dq to be ['stone','atone']
print dq.toString()
This is a school assignment, I'm not allowed to modify codes for the functions enqueue and dequeue and def__init__(self).
I'm trying to use popleft as a means to dequeue an item from a queue. However, the compiler returns the whole queue rather than only the first stack of the queue while I'm using popleft. That being said , the size of the queue decreases by one. Any explanation for this?
The compiler returns [['stone', 'atone'], ['stone', 'shone'], ['stone', 'scone']] when I only want it to return ['stone', 'atone'].
Just like the above comment, you are passing a Stack object not the elements in the stack. Try this one
import collections
class MyStack:
def __init__(self):
self._data = []
def push(self, value):
self._data.append(value)
def size(self):
#Return the number of elements in the stack
return len(self._data)
def toString(self):
#Return a string representing the content of this stack
return str(self._data)
class MyQueue:
def __init__(self):
self._data = collections.deque([])
def enqueue(self, value):
self._data.append(value)
def dequeue(self):
return self._data.popleft()
def size(self):
#return the number of elements in the queue
return len(self._data)
queue1 = MyQueue()
dq = MyStack()
stack1 = MyStack()
stack1.push(['stone', 'atone'])
print "size of stack is now :" ,stack1.size()
queue1.enqueue(stack1.toString())
print "size of queue is now :", queue1.size()
print "size of stack is now :" ,stack1.size()
stack1.push(['stone', 'shone'])
stack1.push(['stone', 'scone'])
dq = queue1.dequeue() # i would like dq to be ['stone','atone']
print dq
I've been going over some of the many coding interview questions.
I was wondering about implementing a queue using two stacks in Python. I'm working on algorithm question to implement a queue with two stacks for purposes of understanding both data structures.
I have the below:
class QueueTwoStacks:
def __init__(self):
self.in_stack = []
self.out_stack = []
def enqueue(self, item):
self.in_stack.append(item)
def dequeue(self):
if len(self.out_stack) == 0:
# Move items from in_stack to out_stack, reversing order
while len(self.in_stack) > 0:
newest_in_stack_item = self.in_stack.pop()
self.out_stack.append(newest_in_stack_item)
# If out_stack is still empty, raise an error
if len(self.out_stack) == 0:
raise IndexError("Can't dequeue from empty queue!")
return self.out_stack.pop()
What is the runtime analysis for this one?
Why is it true that we can get O(m)O(m) runtime for mm function calls.
Am I assuming have a stack implementation and it gives O(1)O(1) time push and pop?
I appreciate your explanation for this. thank you.
Yes. We can Optimize for the time cost of m function calls on your queue. This optimization can be any mix of enqueue and dequeue calls.
Assume you already have a stack implementation and it gives O(1)O(1) time push and pop.
#
#
class Stack():
def __init__(self):
self.stk = []
def pop(self):
"""raises IndexError if you pop when it's empty"""
return self.stk.pop()
def push(self, elt):
self.stk.append(elt)
def is_empty(self):
return len(self.stk) == 0
def peek(self):
if not self.stk.is_empty():
return self.stk[-1]
class Queue():
def __init__(self):
self.q = Stack() # the primary queue
self.b = Stack() # the reverse, opposite q (a joke: q vs b)
self.front = None
def is_empty(self):
return self.q.is_empty()
def peek(self):
if self.q.is_empty():
return None
else:
return self.front
def enqueue(self, elt):
self.front = elt
self.q.push(elt)
def dequeue(self):
"""raises IndexError if you dequeue from an empty queue"""
while not self.q.is_empty() > 0:
elt = self.q.pop()
self.b.push(elt)
val = self.b.pop()
elt = None
while not self.b.is_empty() > 0:
elt = self.b.pop()
self.q.push(elt)
self.front = elt
return val
# Now let's test
class TestQueueTwoStacks(unittest.TestCase):
def setUp(self):
self.q = Queue()
def test_queuedequue(self):
"""queue up 5 integers, check they are in there, dequeue them, check for emptiness, perform other blackbox and whitebox tests"""
self.assertTrue(self.q.is_empty())
self.assertTrue(self.q.q.is_empty())
self.assertTrue(self.q.b.is_empty())
l = range(5)
for i in l:
self.q.enqueue(i)
self.assertEqual(4, self.q.peek())
self.assertEqual(l, self.q.q.stk)
s = []
l.reverse()
for i in l:
elt = self.q.dequeue()
s.append(elt)
self.assertTrue(self.q.is_empty())
self.assertTrue(self.q.q.is_empty())
self.assertTrue(self.q.b.is_empty())
l.reverse()
self.assertEqual(s, l)
self.assertEqual([], self.q.b.stk)
self.assertEqual([], self.q.q.stk)
if __name__ == "__main__":
# unittest.main()
suite = unittest.TestLoader().loadTestsFromTestCase(TestQueueTwoStacks)
unittest.TextTestRunner(verbosity=2).run(suite)
I've been going over some of the many coding interview questions. I was wondering how you would go about implementing a queue using two stacks in Python? Python is not my strongest language so I need all the help I can get.
Like the enqueue, dequeue, and front functions.
class Queue(object):
def __init__(self):
self.instack=[]
self.outstack=[]
def enqueue(self,element):
self.instack.append(element)
def dequeue(self):
if not self.outstack:
while self.instack:
self.outstack.append(self.instack.pop())
return self.outstack.pop()
q=Queue()
for i in range(10):
q.enqueue(i)
for i in xrange(10):
print q.dequeue(),
class MyQueue(object):
def __init__(self):
self.first = []
self.second = []
def peek(self):
if not self.second:
while self.first:
self.second.append(self.first.pop());
return self.second[len(self.second)-1];
def pop(self):
if not self.second:
while self.first:
self.second.append(self.first.pop());
return self.second.pop();
def put(self, value):
self.first.append(value);
queue = MyQueue()
t = int(raw_input())
for line in xrange(t):
values = map(int, raw_input().split())
if values[0] == 1:
queue.put(values[1])
elif values[0] == 2:
queue.pop()
else:
print queue.peek()
class Stack:
def __init__(self):
self.elements = []
def push(self, item):
self.elements.append(item)
def pop(self):
return self.elements.pop()
def size(self):
return len(self.elements)
def is_empty(self):
return self.size() == 0
class CreatingQueueWithTwoStacks:
def __init__(self):
self.stack_1 = Stack()
self.stack_2 = Stack()
def enqueue(self, item):
self.stack_1.push(item)
def dequeue(self):
if not self.stack_1.is_empty():
while self.stack_1.size() > 0:
self.stack_2.push(self.stack_1.pop())
res = self.stack_2.pop()
while self.stack_2.size() > 0:
self.stack_1.push(self.stack_2.pop())
return res
if __name__ == '__main__':
q = CreatingQueueWithTwoStacks()
q.enqueue(1)
q.enqueue(2)
q.enqueue(3)
a = q.dequeue()
print(a)
b = q.dequeue()
print(b)
c = q.dequeue()
print(c)
d = q.dequeue()
print(d)
q.enqueue(5)
q.enqueue(6)
print(q.dequeue())
First, create a stack object. Then create a queue out of 2 stacks. Since a Stack = FIFO (first in first out), and Queue = LIFO (last in first out), add all the items to the "in stack" and then pop them into the output.
class Stack:
def __init__(self):
self.items = []
def push(self, item):
self.items.append(item)
def pop(self):
return self.items.pop()
def size(self):
return len(self.items)
def is_empty(self):
return self.items == []
class Queue2Stacks(object):
def __init__(self):
# Two Stacks
self.in_stack = Stack()
self.out_stack = Stack()
def enqueue(self, item):
self.in_stack.push(item)
def dequeue(self):
if self.out_stack.is_empty:
while self.in_stack.size()>0:
self.out_stack.push(self.in_stack.pop())
return self.out_stack.items.pop()
#driver code
q = Queue2Stacks()
for i in range(5):
q.enqueue(i)
for i in range(5):
print(q.dequeue(i))
Gives you 0,1,2,3,4
Stack1, Stack2.
Enqueue:
Push el into stack1.
Dequeue:
While (!empty(Stack1))
el = Pop from stack1
Push el into stack2
returnEl = Pop from Stack2
While (!empty(Stack2))
el = Pop from stack2
Push el into stack1
return returnEl
That is a way of implementing the algorithm in pseudocode, it shouldn`t be difficult to implement it in python knowing the basic syntax.
I found this solution that works for implementing a queue using two stacks. I use set instead of queue. We can use the following implementation. for the time cost of m function calls on your queue. This optimization can be any mix of enqueue and dequeue calls.
#
#
class Stack():
def __init__(self):
self.stk = []
def pop(self):
"""raises IndexError if you pop when it's empty"""
return self.stk.pop()
def push(self, elt):
self.stk.append(elt)
def is_empty(self):
return len(self.stk) == 0
def peek(self):
if not self.stk.is_empty():
return self.stk[-1]
class Queue():
def __init__(self):
self.q = Stack() # the primary queue
self.b = Stack() # the reverse, opposite q (a joke: q vs b)
self.front = None
def is_empty(self):
return self.q.is_empty()
def peek(self):
if self.q.is_empty():
return None
else:
return self.front
def enqueue(self, elt):
self.front = elt
self.q.push(elt)
def dequeue(self):
"""raises IndexError if you dequeue from an empty queue"""
while not self.q.is_empty() > 0:
elt = self.q.pop()
self.b.push(elt)
val = self.b.pop()
elt = None
while not self.b.is_empty() > 0:
elt = self.b.pop()
self.q.push(elt)
self.front = elt
return val
# Now let's test
class TestQueueTwoStacks(unittest.TestCase):
def setUp(self):
self.q = Queue()
def test_queuedequue(self):
"""queue up 5 integers, check they are in there, dequeue them, check for emptiness, perform other blackbox and whitebox tests"""
self.assertTrue(self.q.is_empty())
self.assertTrue(self.q.q.is_empty())
self.assertTrue(self.q.b.is_empty())
l = range(5)
for i in l:
self.q.enqueue(i)
self.assertEqual(4, self.q.peek())
self.assertEqual(l, self.q.q.stk)
s = []
l.reverse()
for i in l:
elt = self.q.dequeue()
s.append(elt)
self.assertTrue(self.q.is_empty())
self.assertTrue(self.q.q.is_empty())
self.assertTrue(self.q.b.is_empty())
l.reverse()
self.assertEqual(s, l)
self.assertEqual([], self.q.b.stk)
self.assertEqual([], self.q.q.stk)
if __name__ == "__main__":
# unittest.main()
suite = unittest.TestLoader().loadTestsFromTestCase(TestQueueTwoStacks)
unittest.TextTestRunner(verbosity=2).run(suite)
I have 2 issues with the code below:
push(o) throws an exception TypeError: can only assign an iterable.
Should I throw an exception if pop() is invoked on an empty stack ?
class Stack(object):
def __init__(self):
self.storage = []
def isEmpty(self):
return len(self.storage) == 0
def push(self,p):
self.storage[:0] = p
def pop(self):
"""issue: throw exception?"""
return None
No need to jump through these loops, See 5.1.1 Using Lists as Stacks
If you insist on having methods isEmpty() and push() you can do:
class stack(list):
def push(self, item):
self.append(item)
def isEmpty(self):
return not self
You are right to use composition instead of inheritance, because inheritance brings methods in that you don't want to expose.
class Stack:
def __init__(self):
self.__storage = []
def isEmpty(self):
return len(self.__storage) == 0
def push(self,p):
self.__storage.append(p)
def pop(self):
return self.__storage.pop()
This way your interface works pretty much like list (same behavior on pop for example), except that you've locked it to ensure nobody messes with the internals.
I won't talk about the list structure as that's already been covered in this question. Instead I'll mention my preferred method for dealing with stacks:
I always use the Queue module. It supports FIFO and LIFO data structures and is thread safe.
See the docs for more info. It doesn't implement a isEmpty() function, it instead raises a Full or Empty exception if a push or pop can't be done.
Stack follows LIFO mechanism.You can create a list and do a normal append() to append the element to list and do pop() to retrieve the element out of the list which you just inserted.
Here is an example for stack class
class Stack(object):
def __init__(self):
self.items = []
def push(self, item):
self.items.append(item)
def pop(self):
return self.items.pop()
def peek(self):
return self.items[-1]
def isEmpty(self):
return len(self.items) == 0
class Stack:
def __init__(self):
self.items=[]
def isEmpty(self):
return self.items==[]
def push(self , item):
self.items.append(item)
def pop(self):
return self.items.pop()
def size(self):
return len(self.items)
def peek(self):
return self.items[-1]
Create a stack
To create a new stack we can simply use Stack()
for example:
s=Stack()
"s" is the name of new stack
isEmpty
By using isEmpty() we can check our stack is empty or not
for example:
we have two stacks name s1=(0,1,4,5,6) and s2=()
if we use print(s1.isEmpty()) it will return False
if we use print(s2.isEmpty()) it will return True
push
By using push operation we can add items to top of the stack
we can add "6" to the stack name "s" using
s.push(6)
pop
we can use pop operation to remove and return the top item of a stack
if there is a stack name "s" with n amount items (n>0)
we can remove it's top most item by using
s.pop()
size
This operation will return how many items are in the stack
if there is a stack name "s" s=(1,2,3,4,5,3)
print(s.size())
will return "6"
peek
This operation returns the top item without removing it
print(s.peek())
"we can print items of the stack using print(s.items)"
class Stack:
def __init__(self):
self.stack = []
def pop(self):
if self.is_empty():
return None
else:
return self.stack.pop()
def push(self, d):
return self.stack.append(d)
def peek(self):
if self.is_empty():
return None
else:
return self.stack[-1]
def size(self):
return len(self.stack)
def is_empty(self):
return self.size() == 0
class stack:
def __init__(self,n):##constructor
self.no = n ##size of stack
self.Stack = [] ##list for store stack items
self.top = -1
def push(self):##push method
if self.top == self.no - 1 :##check full condition
print("Stack Overflow.....")
else:
n = int(input("enter an element :: "))
self.Stack.append(n) ## in list add stack items use of append method
self.top += 1##increment top by 1
def pop(self):## pop method
if self.top == -1: #check empty condition
print("Stack Underflow....")
else:
self.Stack.pop()## delete item from top of stack using pop method
self.top -= 1 ## decrement top by 1
def peep(self): ##peep method
print(self.top,"\t",self.Stack[-1]) ##display top item
def disp (self): #display method
if self.top == -1:# check empty condition
print("Stack Underflow....")
else:
print("TOP \tELEMENT")
for i in range(self.top,-1,-1): ## print items and top
print(i," \t",self.Stack[i])
n = int(input("Enter Size :: ")) # size of stack
stk = stack(n) ## object and pass n as size
while(True): ## loop for choice as a switch case
print(" 1: PUSH ")
print(" 2: POP ")
print(" 3: PEEP ")
print(" 4: PRINT ")
print(" 5: EXIT ")
option = int(input("enter your choice :: "))
if option == 1:
stk.push()
elif option == 2:
stk.pop()
elif option == 3:
stk.peep()
elif option == 4:
stk.disp()
elif option == 5:
print("you are exit!!!!!")
break
else:
print("Incorrect option")