LeetCode 0225. Implement Stack using Queues Solution in Java, C++, Python & More | Explanation + Code

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0225. Implement Stack using Queues

Description

Implement a last-in-first-out (LIFO) stack using only two queues. The implemented stack should support all the functions of a normal stack (push, top, pop, and empty).

Implement the MyStack class:

  • void push(int x) Pushes element x to the top of the stack.
  • int pop() Removes the element on the top of the stack and returns it.
  • int top() Returns the element on the top of the stack.
  • boolean empty() Returns true if the stack is empty, false otherwise.

Notes:

  • 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.
  • Depending on your language, the queue may not be supported natively. You may simulate a queue using a list or deque (double-ended queue) as long as you use only a queue's standard operations.

 

Example 1:

Input
["MyStack", "push", "push", "top", "pop", "empty"]
[[], [1], [2], [], [], []]
Output
[null, null, null, 2, 2, false]

Explanation
MyStack myStack = new MyStack();
myStack.push(1);
myStack.push(2);
myStack.top(); // return 2
myStack.pop(); // return 2
myStack.empty(); // return False

 

Constraints:

  • 1 <= x <= 9
  • At most 100 calls will be made to push, pop, top, and empty.
  • All the calls to pop and top are valid.

 

Follow-up: Can you implement the stack using only one queue?

Solutions

Solution 1: Two Queues

We use two queues q_1 and q_2, where q_1 is used to store the elements in the stack, and q_2 is used to assist in implementing the stack operations.

  • push operation: Push the element into q_2, then pop the elements in q_1 one by one and push them into q_2, finally swap the references of q_1 and q_2. The time complexity is O(n).
  • pop operation: Directly pop the front element of q_1. The time complexity is O(1).
  • top operation: Directly return the front element of q_1. The time complexity is O(1).
  • empty operation: Check whether q_1 is empty. The time complexity is O(1).

The space complexity is O(n), where n is the number of elements in the stack.

PythonJavaC++GoTypeScriptRust
class MyStack: def __init__(self): self.q1 = deque() self.q2 = deque() def push(self, x: int) -> None: self.q2.append(x) while self.q1: self.q2.append(self.q1.popleft()) self.q1, self.q2 = self.q2, self.q1 def pop(self) -> int: return self.q1.popleft() def top(self) -> int: return self.q1[0] def empty(self) -> bool: return len(self.q1) == 0 # Your MyStack object will be instantiated and called as such: # obj = MyStack() # obj.push(x) # param_2 = obj.pop() # param_3 = obj.top() # param_4 = obj.empty()(code-box)

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