Time Taken to Cross the Door

Hard

Google

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Topics:

Arrays

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.

Time Taken to Cross the Door

Hard

Google

2 views

Topics:

Arrays

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.

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:

Can you please specify the data type of the time taken by each person, and are negative or zero values possible?
What are the possible values for the number of people, and should I expect an empty input?
If it's impossible for everyone to cross the door, what should the function return?
Is the time taken by each person distinct, or can multiple people have the same time?
Are there any constraints on memory usage, or specific limitations on auxiliary data structures I can use?

Brute Force Solution

Approach

The brute force strategy focuses on exploring every possible path to cross the door. We essentially simulate every combination of movements to find the fastest route. This involves trying everything, no matter how inefficient.

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

Consider all possible sequences of movements you can make to cross the door.
Calculate how much time each sequence takes to execute.
Compare the total time taken for each sequence.
Select the sequence with the minimum total time. This is the brute force solution.

Code Implementation

def time_taken_to_cross_the_door_brute_force(movements, door_width):
    # Initialize to infinity to find minimum.
    minimum_time = float('inf')

    def calculate_time(movement_sequence):
        current_position = 0
        total_time = 0
        for movement in movement_sequence:
            current_position += movement[0]
            total_time += movement[1]
        return current_position, total_time

    def find_all_sequences(current_sequence):
        nonlocal minimum_time
        position, time = calculate_time(current_sequence)

        # Once past the door, update the minimum time.
        if position >= door_width:

            minimum_time = min(minimum_time, time)
            return

        # Stop if we've exceeded a reasonable time to avoid infinite recursion.
        if time > 1000:
            return

        for movement in movements:
            find_all_sequences(current_sequence + [movement])

    # Start exploring movement sequences from an empty sequence.
    find_all_sequences([])

    # Returns the minimum time or -1 if no path was found
    if minimum_time == float('inf'):
        return -1
    else:
        return minimum_time

Big(O) Analysis

Time Complexity

O(k^n) – The brute force approach considers all possible sequences of movements to cross the door. If each movement has k possible choices (e.g., move left, move right, jump), and the number of movements needed to cross the door is potentially proportional to n (the width of the door, for example), then we explore k^n possible paths. Calculating the time for each path takes O(n) time in the worst case. Consequently, the overall time complexity of this brute force approach is O(n * k^n). Since exponential terms dominate, this simplifies to O(k^n).

Space Complexity

O(N!) – The brute force approach considers all possible sequences of movements. To generate and store these sequences, we implicitly create a branching structure resembling a tree. In the worst-case scenario, where there are many possible movements, the algorithm might explore a large number of paths, requiring storage for each path being explored at any given time. The number of possible sequences can grow factorially with the number of movements (N), where N represents the maximum number of movements required to cross the door, leading to the storage of up to N! different sequences in the recursion stack and/or auxiliary data structures holding possible paths. Therefore, the auxiliary space complexity is O(N!).

Optimal Solution

Approach

The key is to find the most efficient way to schedule people to cross the door. We want to reduce unnecessary backtracking and utilize pairs to speed things up, focusing on minimizing the total time by strategically pairing the fastest and slowest people.

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

Identify the two fastest and the two slowest people.
Consider two possible strategies: either the two fastest people escort the slower ones, or the fastest person always returns with someone.
Calculate the total time for the first strategy where the two fastest people escort everyone else across the door.
Calculate the total time for the second strategy where the fastest person escorts, returning each time.
Compare the total times of these two strategies and pick the shorter one.
The shorter time will be the minimum time to get everyone across the door.

Code Implementation

def time_to_cross_the_door(people_times):
    number_of_people = len(people_times)
    if number_of_people <= 1:
        return 0
    
    people_times.sort()

    first_person_time = people_times[0]
    second_person_time = people_times[1]
    slowest_person_time = people_times[-1]
    second_slowest_person_time = people_times[-2]

    # Strategy 1: Two fastest escort the slower ones
    strategy_one_time = second_person_time
    strategy_one_time += first_person_time
    strategy_one_time += slowest_person_time
    strategy_one_time += second_person_time

    strategy_one_time += second_slowest_person_time

    # Strategy 2: Fastest person always returns
    strategy_two_time = slowest_person_time
    strategy_two_time += first_person_time
    strategy_two_time += second_slowest_person_time
    strategy_two_time += first_person_time

    strategy_two_time += second_person_time

    # Need to determine the optimal strategy to get everyone across
    return min(strategy_one_time, strategy_two_time)

Big(O) Analysis

Time Complexity

O(n log n) – The dominant operations are identifying the two fastest and two slowest people which involves sorting the array of n people's times. Sorting algorithms like merge sort or heap sort have a time complexity of O(n log n). The rest of the operations like calculating the total time for both strategies involve a few arithmetic calculations which are constant time operations and therefore do not contribute significantly to the overall time complexity. Therefore, the overall time complexity is O(n log n).

Space Complexity

O(1) – The algorithm identifies the two fastest and two slowest people and calculates total times for two strategies. It only uses a few constant extra variables to store the times and intermediate results, such as the minimum time. The number of variables does not depend on N, the number of people. Therefore, the space complexity is constant.

Edge Cases

Case	How to Handle
Null or empty input list of people	Return 0, assuming no one crossing the door means no time taken.
People list contains non-positive integer values representing time	Throw an IllegalArgumentException or return an error code indicating invalid input, as travel time cannot be negative or zero.
List contains a single person with a valid time	Return that person's time, as they cross alone.
Large number of people causing potential integer overflow when summing times	Use a data type with a larger range like long or handle overflow by throwing an exception.
All people have the same crossing time	The optimal strategy may involve the two fastest (identical) people always returning.
List of times is already sorted in ascending or descending order	The algorithm should still correctly determine the optimal strategy regardless of the initial order.
The two fastest people are significantly faster than the rest	Verify the algorithm correctly leverages the fastest pair to minimize return trips.
Input list has extreme values for crossing times (very small and very large numbers)	Ensure calculations avoid precision issues, and the algorithm correctly handles the wide range of values.