Time Taken to Cross the Door - Video Solutions
Time Taken to Cross the Door - LeetCode Premium
Time Taken to Cross the Door - Video Solution
Watch 3 video solutions for Time Taken to Cross the Door, a hard level problem involving Array, Queue, Simulation. This walkthrough by LeetCode Україна has 1,837 views views. Want to try solving it yourself? Practice on FleetCode or read the detailed text solution.
Problem Statement
There are n persons numbered from 0 to n - 1 and a door. Each person can enter or exit through the door once, taking one second.
You are given a non-decreasing integer array arrival of size n, where arrival[i] is the arrival time of the ith person at the door. You are also given an array state of size n, where state[i] is 0 if person i wants to enter through the door or 1 if they want to exit through the door.
If two or more persons want to use the door at the same time, they follow the following rules:
- If the door was not used in the previous second, then the person who wants to exit goes first.
- If the door was used in the previous second for entering, the person who wants to enter goes first.
- If the door was used in the previous second for exiting, the person who wants to exit goes first.
- If multiple persons want to go in the same direction, the person with the smallest index goes first.
Return an array answer of size n where answer[i] is the second at which the ith person crosses the door.
Note that:
- Only one person can cross the door at each second.
- A person may arrive at the door and wait without entering or exiting to follow the mentioned rules.
Example 1:
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.
Example 2:
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.
Constraints:
n == arrival.length == state.length1 <= n <= 1050 <= arrival[i] <= narrivalis sorted in non-decreasing order.state[i]is either0or1.
Approach Overview
Problem Overview: You are given two arrays arrival and state. Each person reaches a single door at a specific second and either wants to enter (0) or exit (1). Only one person can use the door per second, and the direction priority depends on how the door was used in the previous second. The task is to compute the exact time each person crosses the door.
Approach 1: Direct Timeline Simulation (Brute Force) (Time: O(T + n), Space: O(n))
A straightforward method simulates every second from time 0 until all people cross the door. At each second, scan the list of people who have already arrived and determine who is waiting to enter or exit. Apply the priority rule based on the previous second’s door usage. While this approach models the problem clearly, repeatedly scanning waiting people is inefficient when many seconds pass without events. The complexity depends on the total simulated time T, which may exceed n significantly.
Approach 2: Queue + Simulation (Optimal) (Time: O(n), Space: O(n))
Maintain two queues: one for people waiting to enter and another for people waiting to exit. Iterate through time while pushing newly arrived people into their respective queues. At each second, decide which queue gets access to the door using the rules: if the door was unused last second, exiting people get priority; otherwise the previous direction keeps priority. If the preferred queue is empty, serve the other queue. When both queues are empty, jump the simulation time directly to the next arrival to avoid idle iteration.
This design avoids repeated scans by storing waiting people in FIFO order. Each person enters and leaves a queue exactly once, so the simulation runs in linear time. The logic closely mirrors how operating systems simulate resource scheduling and is a common pattern in queue-based scheduling problems.
Recommended for interviews: The queue-based simulation is the expected solution. It processes each person once, runs in O(n) time, and cleanly models direction priority using a single variable tracking the last door usage. Interviewers often look for this pattern when testing understanding of array-driven events and simulation logic.
Complexity Analysis
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Direct Timeline Simulation | O(T + n) | O(n) | Useful for understanding the door priority rules before optimizing |
| Queue + Simulation | O(n) | O(n) | Best general solution; processes arrivals once and models the door with two queues |