As per the problem statement, you are only supposed to use valid queue operations i.e. push to back, peek/pop from front, size and is empty. So by that logic top() method is incorrect. If you write the for loop in push() then you can use peek() method in top().
Leetcode says: "You must use only standard operations of a queue, which means that only `push to back`, `peek/pop from front`,`size` and `is empty` operations are valid." And here you used self.q[-1] which is accessing the last element which is not allowed
I'm so glad you point out at the beginning that it would take O(n) to implement the pop operation. I was thinking to myself "that doesn't even make sense to use a queue, I'll have to iterate through the whole thing to pop the next stack item. (I literally was wracking my brain before watching the solution video about who I could possibly do that in O(1)).
The reason leetcode is showing poor performance is that there are lot of people who have implemented it like a stack and not queue that behaves like stack
Here, we are using deque funtionality in case of getting top element But we should use it as queue So we cannot directly get element from rear So it is wrong i think For that we should first perform operation like pop then we can get top element without popping
I came across the same question with you.There are some difference between queue and deque, so if we solve the problem by queue,the solution will be different.
"You must use only standard operations of a queue, which means that only `push to back`, `peek/pop from front`,`size` and `is empty` operations are valid." And here you used self.q[-1] which is essentially peeking from the back which is not allowed Must get accepted because it's an easy but double queue solution is better
Hi Mr.Neet, you have about covered most graph topic but one that seems to be missing is strongly connected components. Would really appreciate it if you could teach that when you have time.
javascript solution on neetcode pops with n^2 complexity as iterates through the array and removes element from the start causing other elements to shift left
For any one who is saying that the top() function is implemented properly, the idea behind using the index of [-1] as he did was that there is already a function for queues called "peek" that you could use that gives you the last element directly, so it is safe to say that it is implemented, and we always have an idea of what the top element would be. However, if anyone faces an issue in interview where the interviewer still wants an implementation for the top() function, you could make sure to keep O(1) for it by using a value that always tracks the top value, which would be always the last one pushed, or the 2nd to last one after popping Code Example: from collections import deque class MyStack: def __init__(self): self.q1 = deque() self.peek = None def push(self, x: int) -> None: self.q1.appendleft(x) self.peek = x def pop(self) -> int: n = len(self.q1) for i in range(n - 1): poped = self.q1.pop() self.q1.appendleft(poped) if i == n-2: self.peek = poped return self.q1.pop() def top(self) -> int: return self.peek def empty(self) -> bool: return len(self.q1) == 0
for queues, peek gets the first element, not the last. top is not implemented correctly here because you can't grab the last element like that w/ the constraints given in the problem
Better to use queue, from queue import Queue. This makes more sense bcoz u r using deque and only implementing queue feature, so y not use queue. By the way nice explanation.
Queue is a synchronized queue class for multithreaded programming. You can get away with it on LeetCode but it's not recommended for use as a general data structure because there's overhead for the synchronization.
Then you're using stack methods, so you're not implementing it using a queue. You're implementing a stack using a stack... NeetCode shouldn't be doing q[-1] because that's not a queue method. It's a mistake in the solution.
The top() could be implemented much simpler by reusing pop() and push() def push(self, x: int) -> None: self.q.append(x) def pop(self) -> int: for i in range(len(self.q) - 1): self.push(self.q.popleft()) return self.q.popleft() def top(self) -> int: res = self.pop() self.push(res) return res
If you use Python it only compares to other Python solutions. Mostly it's just randomness in the timing. When everyone does the same solution the timings get tightly clustered. I have almost the same solution and it's 40ms which is faster than 76% of solutions. Also some people are implementing it using a stack which is faster but not a correct solution. That messes up the timing for correct solutions.
Hello can anyone please clarify if this a valid solution the time complexity is good and it beats 92% but just don't know if it is right way to do. class MyQueue: def __init__(self): self.num = []
As per the problem statement, you are only supposed to use valid queue operations i.e. push to back, peek/pop from front, size and is empty.
So by that logic top() method is incorrect. If you write the for loop in push() then you can use peek() method in top().
You can also initialize a stacktop class variable to keep track of the top and just return that
Came here to say this also. Thanks for mentioning this.
Yeah this is a poor solution. Should be done using the basic FIFO queue module which doesn't allow half of what he's used.
front and back are valid operations for a queue. So getting the last element i.e. back is correct.
Leetcode says:
"You must use only standard operations of a queue, which means that only `push to back`, `peek/pop from front`,`size` and `is empty` operations are valid."
And here you used self.q[-1] which is accessing the last element which is not allowed
Looks like you are starting to make these every day which is great news! Thank you
I'm so glad you point out at the beginning that it would take O(n) to implement the pop operation. I was thinking to myself "that doesn't even make sense to use a queue, I'll have to iterate through the whole thing to pop the next stack item. (I literally was wracking my brain before watching the solution video about who I could possibly do that in O(1)).
The reason leetcode is showing poor performance is that there are lot of people who have implemented it like a stack and not queue that behaves like stack
Here, we are using deque funtionality in case of getting top element
But we should use it as queue
So we cannot directly get element from rear
So it is wrong i think
For that we should first perform operation like pop then we can get top element without popping
I came across the same question with you.There are some difference between queue and deque, so if we solve the problem by queue,the solution will be different.
performance is relative, and many people have implemented it using stacks which pushes your solution towards the less efficient ones.
This explanation is top tier. Thank you Neet
how are you allowed to use q[-1] if you need to use only queue operations
"You must use only standard operations of a queue, which means that only `push to back`, `peek/pop from front`,`size` and `is empty` operations are valid."
And here you used self.q[-1] which is essentially peeking from the back which is not allowed
Must get accepted because it's an easy but double queue solution is better
Hi Mr.Neet, you have about covered most graph topic but one that seems to be missing is strongly connected components. Would really appreciate it if you could teach that when you have time.
javascript solution on neetcode pops with n^2 complexity as iterates through the array and removes element from the start causing other elements to shift left
For any one who is saying that the top() function is implemented properly, the idea behind using the index of [-1] as he did was that there is already a function for queues called "peek" that you could use that gives you the last element directly, so it is safe to say that it is implemented, and we always have an idea of what the top element would be.
However, if anyone faces an issue in interview where the interviewer still wants an implementation for the top() function, you could make sure to keep O(1) for it by using a value that always tracks the top value, which would be always the last one pushed, or the 2nd to last one after popping
Code Example:
from collections import deque
class MyStack:
def __init__(self):
self.q1 = deque()
self.peek = None
def push(self, x: int) -> None:
self.q1.appendleft(x)
self.peek = x
def pop(self) -> int:
n = len(self.q1)
for i in range(n - 1):
poped = self.q1.pop()
self.q1.appendleft(poped)
if i == n-2:
self.peek = poped
return self.q1.pop()
def top(self) -> int:
return self.peek
def empty(self) -> bool:
return len(self.q1) == 0
for queues, peek gets the first element, not the last. top is not implemented correctly here because you can't grab the last element like that w/ the constraints given in the problem
Excellent Solution, Easly Understood
Dear Neet, can you teach us about 232. Implement Queue using Stacks as well? 232 is on the Grind 75 list
th-cam.com/video/nPvjcQBplH4/w-d-xo.html
Better to use queue, from queue import Queue. This makes more sense bcoz u r using deque and only implementing queue feature, so y not use queue. By the way nice explanation.
Exactly. If you are going to use deque , you can remove from either side ( .popleft() and pop())
Queue is a synchronized queue class for multithreaded programming. You can get away with it on LeetCode but it's not recommended for use as a general data structure because there's overhead for the synchronization.
Push is O(n), pop/top is O(1)
Amazing explanation
If you know its length and are accessing it using -1 anyways then why not just use pop or return the -1 element and then do del q[-1]?
Then you're using stack methods, so you're not implementing it using a queue. You're implementing a stack using a stack...
NeetCode shouldn't be doing q[-1] because that's not a queue method. It's a mistake in the solution.
The top() could be implemented much simpler by reusing pop() and push()
def push(self, x: int) -> None:
self.q.append(x)
def pop(self) -> int:
for i in range(len(self.q) - 1):
self.push(self.q.popleft())
return self.q.popleft()
def top(self) -> int:
res = self.pop()
self.push(res)
return res
smart solution
Great explanation ! Tx !
May i ask what app do you use to annotate in the video?
Sure, I just use Paint3D
When using two queues the runtime and space is over 70% ...strange
Might be me, but I find this problem to be unnecessarily complex. It is not hard, but it is just annoying.
My Solution:
class MyStack:
def __init__(self):
self.qStack = []
self.qEmpty = []
def push(self, x: int) -> None:
self.qEmpty.append(x)
for elem in self.qStack:
self.qEmpty.append(elem)
self.qStack = self.qEmpty
self.qEmpty = []
def pop(self) -> int:
if len(self.qStack) > 0:
return self.qStack.pop(0)
def top(self) -> int:
if len(self.qStack) > 0:
return self.qStack[0]
def empty(self) -> bool:
return len(self.qStack) == 0
But, it's not using two stack.
if using dequeue we can simply use pop() right it will pop from rear abd behave as a stack why to use popleft
this problem is silly
Maybe it's not efficient because you're using python and python in itself is... not efficient.
If you use Python it only compares to other Python solutions. Mostly it's just randomness in the timing. When everyone does the same solution the timings get tightly clustered. I have almost the same solution and it's 40ms which is faster than 76% of solutions.
Also some people are implementing it using a stack which is faster but not a correct solution. That messes up the timing for correct solutions.
Hello can anyone please clarify if this a valid solution the time complexity is good and it beats 92% but just don't know if it is right way to do.
class MyQueue:
def __init__(self):
self.num = []
def push(self, x: int) -> None:
self.num.append(x)
def pop(self) -> int:
return self.num.pop(0)
def peek(self) -> int:
return self.num[0]
def empty(self) -> bool:
return len(self.num) == 0
top operation is not legal.
There is no solution for Leetcode 332 Reconstruct Itinerary. Please provide one.
the solution is giving wrong answer @neet
For pop when cant we just use self.q.popleft() thats it
that's also my doubt