What is wrong with implementing a queue with a minimum?

Question

Python

There is a task:

Minimum-aware queue Implement a minimum
-aware queue.

Specifications Input

The first line of the input contains the number n — the number of operations with the queue. Each next line contains the number a (0≤a≤10000). If a>0, then this number must be added to the queue. If a=0, then this is a request to remove an element from the queue.

Output

For each request to remove an element from the queue, it is necessary to display the value of the minimum element of the queue (taking into account the value of the element being removed). If the delete request is called on an empty queue, then print -1.

There is the following implementation of the queue with a minimum of two stacks.

class Stack_min:

    def __init__(self):
        self.stack = []
        self.stack_min = []

    def push(self, e):
        self.stack.append(e)
        if self.stack_min and self.stack_min[-1] < e:
            self.stack_min.append(self.stack_min[-1])
        else:
            self.stack_min.append(e)
        return self

    def pop(self):
        if self.stack != []:
            self.stack_min.pop()
            return self.stack.pop()

    def min(self):
        if self.stack_min:
            return self.stack_min[-1]

    def aslist(self):
        return self.stack

    def back(self):
        if self.stack != []:
            return self.stack[-1]

    def size(self):
        return len(self.stack)

    def clear(self):
        self.stack.clear()
        self.stack_min.clear()
        return self


class Queue_min:

    def __init__(self):
        self.s1 = Stack_min()
        self.s2 = Stack_min()

    def _copy(self):
        if self.s2.size() == 0:
            while self.s1.size():
                self.s2.push(self.s1.pop())

    def push(self, e):
        self.s1.push(e)
        return self

    def pop(self):
        self._copy()
        return self.s2.pop()

    def front(self):
        self._copy()
        return self.s2.back()

    def size(self):
        return self.s1.size() + self.s2.size()

    def min(self):
        self._copy()
        return self.s2.min()

    def clear(self):
        self.s1.clear()
        self.s2.clear()
        return self


queue = Queue_min()

for i in range(int(input())):
    inp = int(input())
    if inp != 0:
        queue.push(inp)
    else:
        if not queue.size():
            print(-1)
        else:
            print(queue.min())
            queue.pop()

When submitting a task, the system returns an error "The program gives an incorrect answer." What could be the problem?

Answer 1

When calculating the minimum, you only consider one of the two stacks, and the minimum can be in either of the two. Consider this test with 7 queries:

7
5
6
7
0
1
2
0

After the first operation of requesting the minimum, you will have two elements in s2. Then you add 2 smaller elements to s1. During the second minimum query, the elements from s1 to s2 will not be transferred and the minimum will be returned only among the elements of s2.