Time Taken to Cross the Door

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ArraysGraphsRecursionDynamic ProgrammingBreadth-First Search

There are n individuals waiting to cross a door. Each individual has an assigned direction: 0 represents moving from inside to outside, and 1 represents moving from outside to inside. You are given a 0-indexed integer array time of length n, where time[i] represents the time taken for the i^th individual to cross the door.

Initially, all individuals are positioned on one side of the door. You need to determine the order in which they cross, considering the following rules:

Only one person can cross the door at a time.
If there are individuals waiting on both sides of the door, prioritize the individual with direction 0 (inside to outside).
If there are multiple individuals with direction 0, choose the one with the smallest index.
If there are no individuals with direction 0 and there are individuals with direction 1, choose the one with the smallest index.
You are given two more 0-indexed integer arrays direction and change, both of length n. After the i^th individual crosses, their direction is flipped (0 becomes 1, and 1 becomes 0), and the time it takes them to cross changes to change[i].

Return an array of integers of length n representing the order in which the individuals cross the door.

Example 1:

Input: time = [1,2,3], direction = [0,1,1], change = [3,2,1]
Output: [0,2,1]
Explanation:
Initially, individuals 0, 1, and 2 are inside.
- Individual 0 crosses (time 1). direction[0] becomes 1, change[0] becomes 3.
- Individual 2 crosses (time 3). direction[2] becomes 0, change[2] becomes 1.
- Individual 1 crosses (time 2). direction[1] becomes 0, change[1] becomes 2.
So the order is [0,2,1].

Example 2:

Input: time = [3,2,5], direction = [0,0,1], change = [4,0,3]
Output: [0,1,2]
Explanation:
Initially, individuals 0, 1, and 2 are inside.
- Individual 0 crosses (time 3). direction[0] becomes 1, change[0] becomes 4.
- Individual 1 crosses (time 2). direction[1] becomes 1, change[1] becomes 0.
- Individual 2 crosses (time 5). direction[2] becomes 0, change[2] becomes 3.
So the order is [0,1,2].

Constraints:

n == time.length == direction.length == change.length
1 <= n <= 10⁵
1 <= time[i], change[i] <= 10⁴
direction[i] is either 0 or 1.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

Are the `enterTime` and `exitTime` arrays guaranteed to be of the same length, and do they correspond to the same individuals in order?
Can `enterTime` values be greater than or equal to `exitTime` values for a given person, and what should happen in such cases?
What is the expected behavior if the `enterTime` or `exitTime` arrays are empty?
Are the times represented as integers, floats, or strings, and what is the potential range of these time values?
If a person arrives when the door is occupied, do they wait until the *next available* moment, or is there a specific queuing mechanism to consider?

Brute Force Solution

Approach

The brute force approach involves exploring every single combination of how people could pass through the door. We will meticulously check each possible sequence to find the one that meets the given conditions.

Here's how the algorithm would work step-by-step:

Consider the very first person who could go through the door.
Then, think about all the other people who could go through next, in every single order imaginable.
For each of these possible orders, check if everyone makes it through the door within the allowed time.
If an order doesn't work, discard it and try another one.
Keep exploring all these different sequences of people passing through the door.
Once you've looked at every possible way for everyone to go through, find the one sequence that took the shortest total time while still being valid.

Code Implementation

import itertoolsdef time_taken_to_cross_door(person_speeds, max_time_allowed):    number_of_people = len(person_speeds)    all_people_indices = list(range(number_of_people))    minimum_time_found = float('inf')    # Generate every possible order of people passing through the door    for sequence_of_people in itertools.permutations(all_people_indices):        current_time_taken = 0        # Calculate the time taken for the current sequence        for person_index in sequence_of_people:            current_time_taken += person_speeds[person_index]        # Only consider sequences that meet the time constraint        if current_time_taken <= max_time_allowed:            # Update the minimum time if this sequence is faster            minimum_time_found = min(minimum_time_found, current_time_taken)    # If no sequence meets the criteria, return infinity, otherwise the minimum time    if minimum_time_found == float('inf'):        return -1 # Indicate no valid solution found    else:        return minimum_time_found

Big(O) Analysis

Time Complexity

O(n!) – The approach explores every single permutation of the n people passing through the door. For the first person, there are n choices. For the second, there are n-1 choices, and so on. This results in n * (n-1) * (n-2) * ... * 1 possible sequences, which is n factorial. Checking each sequence's validity takes some time, but the dominant factor is the number of sequences to explore, leading to a time complexity of O(n!).

Space Complexity

O(n!) – The brute force approach involves exploring every single permutation of people passing through the door. To store each permutation as it is generated or considered, we would require space proportional to the number of people, N. However, the actual space complexity is dominated by the recursive calls or the explicit generation of these permutations, where in the worst case, the recursion depth or the storage of active permutations can be related to N!. Therefore, the auxiliary space complexity is O(n!) due to the need to manage all possible orderings.

Optimal Solution

Approach

The most efficient way to figure out the time taken to cross the door is to consider the movement of people in groups. Instead of tracking each person individually, we can simulate the process by thinking about how many people can pass through at once based on the door's capacity and the flow of people.

Here's how the algorithm would work step-by-step:

Determine the maximum number of people that can pass through the door at any given moment.
Identify any periods where the door is not being used to its full capacity due to a lack of people arriving.
Simulate the passage of people through the door in batches, considering the door's capacity and the arrival rate.
Keep track of the current time as people move through.
If people are arriving faster than the door can handle, the door will be continuously occupied until the backlog is cleared.
If people are arriving slower than the door can handle, there will be idle time at the door.
The total time taken is the sum of the time the door is occupied by people and any necessary waiting time for the next person to arrive if the door becomes momentarily empty.
Continue this process until everyone has had a chance to cross the door.

Code Implementation

def calculate_time_to_cross_door(total_people_to_cross, door_capacity, arrival_rate_per_second):
    current_time = 0.0
    people_remaining_to_cross = total_people_to_cross
    door_available_at_time = 0.0

    # This loop continues until all people have crossed the door.
    while people_remaining_to_cross > 0:
        # Calculate how many people can potentially enter the door in this step.
        # We are limited by the door's capacity and the number of people still waiting.
        people_entering_door = min(door_capacity, people_remaining_to_cross)

        # The door is ready for the next group of people at this calculated time.
        # This is when the previous person/group finished and the door is free.
        door_available_at_time = max(current_time, door_available_at_time)

        # Calculate the time it takes for the current group to pass through.
        # Each person takes 1 second to pass, so the time is the number of people entering.
        time_for_current_group = people_entering_door

        # The current time is updated to when this group finishes passing.
        current_time = door_available_at_time + time_for_current_group
        people_remaining_to_cross -= people_entering_door

        # If there are still people left and the door is now empty, we might need to wait.
        # This accounts for scenarios where people arrive slower than the door's capacity.
        if people_remaining_to_cross > 0 and current_time < (total_people_to_cross - people_remaining_to_cross) / arrival_rate_per_second:
            next_arrival_time = (total_people_to_cross - people_remaining_to_cross) / arrival_rate_per_second
            door_available_at_time = max(current_time, next_arrival_time)

    return current_time

Big(O) Analysis

Time Complexity

O(n) – The simulation processes each person's passage through the door. Whether people pass in batches or individually, the total number of distinct 'crossing events' or state changes in the simulation is directly proportional to the total number of people, n. Each person will eventually cross the door, and the simulation logic advances based on arrivals and departures. Therefore, the total operations will scale linearly with the number of people, n, leading to a time complexity of O(n).

Space Complexity

O(1) – The solution uses a constant amount of extra space, regardless of the number of people crossing the door (N). This is because the algorithm primarily relies on a few integer variables to keep track of the current time, the number of people waiting, and the number of people who have crossed. No auxiliary data structures that grow with the input size are created or utilized. Therefore, the space complexity remains constant.

Edge Cases

Case	How to Handle
Empty enterTime or exitTime arrays	If either array is empty or null, the solution should return an empty result array.
enterTime and exitTime arrays have different lengths	The problem implicitly assumes equal lengths; a robust solution would either throw an error or handle based on the minimum length.
Single person entering and exiting	The calculation should still work correctly, yielding the difference between exit and enter time for that single person.
All people arrive and leave at the same times	The solution needs to manage door occupancy properly, potentially leading to queues if arrival times are identical.
Interleaved arrival and departure times	The core logic must correctly track door availability and person queues, which is the primary challenge.
Arrival time equals departure time for a person	This implies zero time spent at the door, which should be handled as a valid outcome.
Very large number of people (scalability)	An efficient algorithm with O(N) or O(N log N) time complexity is required to handle large inputs without timing out.
Exit times are earlier than corresponding enter times	This scenario is invalid based on the problem description and should ideally be caught with an error or specific handling.

Empty enterTime or exitTime arrays

How to Handle:

If either array is empty or null, the solution should return an empty result array.

enterTime and exitTime arrays have different lengths

How to Handle:

The problem implicitly assumes equal lengths; a robust solution would either throw an error or handle based on the minimum length.

Single person entering and exiting

How to Handle:

The calculation should still work correctly, yielding the difference between exit and enter time for that single person.

All people arrive and leave at the same times

How to Handle:

The solution needs to manage door occupancy properly, potentially leading to queues if arrival times are identical.

Interleaved arrival and departure times

How to Handle:

The core logic must correctly track door availability and person queues, which is the primary challenge.

Arrival time equals departure time for a person

How to Handle:

This implies zero time spent at the door, which should be handled as a valid outcome.

Very large number of people (scalability)

How to Handle:

An efficient algorithm with O(N) or O(N log N) time complexity is required to handle large inputs without timing out.

Exit times are earlier than corresponding enter times

How to Handle:

This scenario is invalid based on the problem description and should ideally be caught with an error or specific handling.