Time Taken to Cross the Door Python Solution

lcpy gen -n 2534   # by problem number
lcpy gen -s time_taken_to_cross_the_door   # by problem name

Input: arrival = [0,1,1,2,4], state = [0,1,0,0,1]
Output: [0,3,1,2,4]
Explanation: At each second we have the following:
- At t = 0: Person 0 is the only one who wants to enter, so they just enter through the door.
- At t = 1: Person 1 wants to exit, and person 2 wants to enter. Since the door was used the previous second for entering, person 2 enters.
- At t = 2: Person 1 still wants to exit, and person 3 wants to enter. Since the door was used the previous second for entering, person 3 enters.
- At t = 3: Person 1 is the only one who wants to exit, so they just exit through the door.
- At t = 4: Person 4 is the only one who wants to exit, so they just exit through the door.

Input: arrival = [0,0,0], state = [1,0,1]
Output: [0,2,1]
Explanation: At each second we have the following:
- At t = 0: Person 1 wants to enter while persons 0 and 2 want to exit. Since the door was not used in the previous second, the persons who want to exit get to go first. Since person 0 has a smaller index, they exit first.
- At t = 1: Person 1 wants to enter, and person 2 wants to exit. Since the door was used in the previous second for exiting, person 2 exits.
- At t = 2: Person 1 is the only one who wants to enter, so they just enter through the door.

from collections import deque


class Solution:
    # Time: O(n)
    # Space: O(n)
    def time_taken(self, arrival: list[int], state: list[int]) -> list[int]:
        q: list[deque[int]] = [deque(), deque()]
        n = len(arrival)
        t = i = 0
        st = 1
        ans = [0] * n
        while i < n or q[0] or q[1]:
            while i < n and arrival[i] <= t:
                q[state[i]].append(i)
                i += 1
            if q[0] and q[1]:
                ans[q[st].popleft()] = t
            elif q[0] or q[1]:
                st = 0 if q[0] else 1
                ans[q[st].popleft()] = t
            else:
                st = 1
            t += 1
        return ans