Time to Cross a Bridge

Hard

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ArraysGreedy AlgorithmsDynamic Programming

There are k workers who want to move n boxes from the right (old) warehouse to the left (new) warehouse. You are given the two integers n and k, and a 2D integer array time of size k x 4 where time[i] = [right_i, pick_i, left_i, put_i].

The warehouses are separated by a river and connected by a bridge. Initially, all k workers are waiting on the left side of the bridge. To move the boxes, the i^th worker can do the following:

Cross the bridge to the right side in right_i minutes.
Pick a box from the right warehouse in pick_i minutes.
Cross the bridge to the left side in left_i minutes.
Put the box into the left warehouse in put_i minutes.

The i^th worker is less efficient than the j^th worker if either condition is met:

left_i + right_i > left_j + right_j
left_i + right_i == left_j + right_j and i > j

The following rules regulate the movement of the workers through the bridge:

Only one worker can use the bridge at a time.
When the bridge is unused prioritize the least efficient worker (who have picked up the box) on the right side to cross. If not, prioritize the least efficient worker on the left side to cross.
If enough workers have already been dispatched from the left side to pick up all the remaining boxes, no more workers will be sent from the left side.

Return the elapsed minutes at which the last box reaches the left side of the bridge.

Example 1:

Input: n = 1, k = 3, time = [[1,1,2,1],[1,1,3,1],[1,1,4,1]]

Output: 6

Explanation:

From 0 to 1 minutes: worker 2 crosses the bridge to the right.
From 1 to 2 minutes: worker 2 picks up a box from the right warehouse.
From 2 to 6 minutes: worker 2 crosses the bridge to the left.
From 6 to 7 minutes: worker 2 puts a box at the left warehouse.
The whole process ends after 7 minutes. We return 6 because the problem asks for the instance of time at which the last worker reaches the left side of the bridge.

Example 2:

Input: n = 3, k = 2, time = [[1,5,1,8],[10,10,10,10]]

Output: 37

Explanation:

The last box reaches the left side at 37 seconds. Notice, how we do not put the last boxes down, as that would take more time, and they are already on the left with the workers.

Constraints:

1 <= n, k <= 10⁴
time.length == k
time[i].length == 4
1 <= left_i, pick_i, right_i, put_i <= 1000

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:

What are the constraints on the number of workers and the length of the bridge? Also, what are the maximum values for the time it takes each worker to cross alone or with a colleague?
Can the time it takes for a worker to cross alone or with a colleague be zero?
Is there any relationship between the time it takes for a worker to cross alone versus with a colleague? For example, is it always the case that crossing with a colleague takes longer or is there no predictable pattern?
If it's impossible for all workers to cross the bridge, what should the function return (e.g., -1, null, throw an exception)?
Is there an inherent order to how workers cross? Can I optimize by strategically selecting which workers cross together or alone at different stages?

Brute Force Solution

Approach

The brute force method for figuring out the fastest way to get everyone across the bridge involves checking absolutely every possible order in which people could cross. We consider all combinations, even the really slow ones, until we find the quickest.

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

First, list out all the different orders in which people can cross the bridge.
For each of those orders, figure out who goes across together each time, keeping in mind only two people can cross at once.
Calculate how long each pair takes to cross based on the slower person's speed in the pair.
Add up all the crossing times and return trip times for each order to get a total crossing time.
Compare all of the total crossing times and choose the shortest one.

Code Implementation

def time_to_cross_a_bridge_brute_force(people_crossing_times):

    fastest_time = float('inf')

    def find_min_time(left_side, right_side, flashlight_on_left, current_time):
        nonlocal fastest_time

        if not left_side:
            # Base case: everyone is on the right side
            fastest_time = min(fastest_time, current_time)
            return

        if flashlight_on_left:
            # Explore all possible pairs crossing to the right
            for i in range(len(left_side)): 
                for j in range(i + 1, len(left_side)): 
                    
                    # This ensures that the same person never crosses with themselves
                    crossing_pair = [left_side[i], left_side[j]]
                    remaining_left = left_side[:i] + left_side[i+1:j] + left_side[j+1:]
                    
                    crossing_time = max(crossing_pair)

                    # Explore all possible people returning with the flashlight
                    for returning_person in crossing_pair:
                        new_right_side = right_side + crossing_pair
                        new_right_side.sort()

                        return_time = returning_person

                        find_min_time(sorted(remaining_left + [returning_person]), sorted([person for person in new_right_side if person not in crossing_pair or person == returning_person and new_right_side.count(person) == 1]), False, current_time + crossing_time + return_time)
        else:
            # Explore all possible people crossing to the left
            for person_returning in right_side:

                #Only one person returns with the flashlight.
                remaining_right = right_side[:right_side.index(person_returning)] + right_side[right_side.index(person_returning)+1:]
                return_time = person_returning

                # This moves one person from right to left.
                find_min_time(sorted(left_side + [person_returning]), sorted(remaining_right), True, current_time + return_time)

    # Begin with everyone on the left side
    find_min_time(sorted(people_crossing_times), [], True, 0)

    return fastest_time

Big(O) Analysis

Time Complexity

O(n! * 2^n) – The brute force approach involves generating all possible permutations of the n people to determine the crossing order. There are n! (n factorial) such permutations. For each permutation, we need to consider all possible pairings of people for each crossing, where at most two people can cross together. In the worst case, each person may need to return with the flashlight, leading to roughly 2^n possible groupings for each permutation. Thus, the time complexity is proportional to n! multiplied by 2^n, giving O(n! * 2^n).

Space Complexity

O(N!) – The brute force approach requires generating all possible permutations of the people crossing the bridge. Storing all these permutations requires creating a list of lists, where each inner list represents a different order of people. Since there are N! (N factorial) possible permutations for N people, the space required to store these permutations grows factorially with the input size N. This results in an auxiliary space complexity of O(N!).

Optimal Solution

Approach

The key idea is to recognize that the fastest two people should be helping the slower ones cross. We want to minimize the time the two fastest people spend going back and forth.

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

Consider the two fastest people as the 'helpers' and the two slowest as the 'burden'.
There are two main strategies: either the two fastest people repeatedly take one person across at a time, or the two fastest people take each other back and forth.
The first strategy involves the fastest person returning each time with someone to bring the slowest person over.
The second strategy involves the two fastest going across, the fastest returns, the two slowest go across, and the second fastest returns.
Calculate the total time for both strategies.
Choose the strategy that results in the shortest total time.

Code Implementation

def time_to_cross_a_bridge(people_crossing_times):
    people_crossing_times.sort()
    number_of_people = len(people_crossing_times)
    total_time = 0

    while number_of_people > 0:
        if number_of_people == 1:
            total_time += people_crossing_times[0]
            number_of_people -= 1
        elif number_of_people == 2:
            total_time += people_crossing_times[1]
            number_of_people -= 2
        else:
            # Determine the fastest two and slowest two
            fastest_person = people_crossing_times[0]
            second_fastest_person = people_crossing_times[1]
            slowest_person = people_crossing_times[-1]
            second_slowest_person = people_crossing_times[-2]

            # Calculate the two strategies.
            strategy_one_time = second_fastest_person + fastest_person + \
                                 slowest_person + second_fastest_person
            strategy_two_time = fastest_person + slowest_person + \
                                 fastest_person + second_slowest_person

            # This picks the optimal strategy for crossing.
            if strategy_one_time < strategy_two_time:
                total_time += strategy_one_time
            else:
                total_time += strategy_two_time

            # Remove the slowest two people from the list.
            people_crossing_times = people_crossing_times[:-2]
            number_of_people -= 2

    return total_time

Big(O) Analysis

Time Complexity

O(1) – The algorithm operates on a fixed-size group of four people (or fewer when n < 4). Regardless of the number of people initially (n), the core logic considers only the two fastest and two slowest, performing a constant number of calculations to compare two strategies. Therefore, the time complexity is independent of n and can be considered O(1).

Space Complexity

O(1) – The described algorithm primarily uses a fixed number of variables to store the times of the fastest and slowest people and to calculate the total time for each strategy. It does not create any auxiliary data structures like lists or hash maps that scale with the input size N (number of people). Therefore, the extra space required remains constant regardless of the input, resulting in O(1) space complexity.

Edge Cases

Case	How to Handle
workers is zero or negative	Return 0 if the number of workers is zero and throw an exception if negative, as no crossing is possible.
bridge length is zero or negative	Return 0 if the bridge length is zero and throw an exception if negative, as no crossing is possible.
Individual times or pair times are zero	If individual time is zero, worker instantly crosses; if pair time is zero, handle appropriately during pair selection, ensuring not dividing by zero.
times array is null or empty	Throw an IllegalArgumentException if the arrays are null or empty to prevent NullPointerExceptions or incorrect processing.
Number of workers is greater than the length of the times array.	Handle only available workers, adjusting based on the number of workers provided vs. the length of the `times` array.
Workers times contains large numbers	Use `long` data type to prevent integer overflow when calculating the total time and intermediate sums.
All workers have the same crossing time	Optimize the pairing strategy, possibly always sending the two slowest, since individual differences won't matter.
Only one worker	Return the single worker's individual crossing time as the result, avoiding pairing logic.

workers is zero or negative

How to Handle:

Return 0 if the number of workers is zero and throw an exception if negative, as no crossing is possible.

bridge length is zero or negative

How to Handle:

Return 0 if the bridge length is zero and throw an exception if negative, as no crossing is possible.

Individual times or pair times are zero

How to Handle:

If individual time is zero, worker instantly crosses; if pair time is zero, handle appropriately during pair selection, ensuring not dividing by zero.

times array is null or empty

How to Handle:

Throw an IllegalArgumentException if the arrays are null or empty to prevent NullPointerExceptions or incorrect processing.

Number of workers is greater than the length of the times array.

How to Handle:

Handle only available workers, adjusting based on the number of workers provided vs. the length of the `times` array.

Workers times contains large numbers

How to Handle:

Use `long` data type to prevent integer overflow when calculating the total time and intermediate sums.

All workers have the same crossing time

How to Handle:

Optimize the pairing strategy, possibly always sending the two slowest, since individual differences won't matter.

Only one worker

How to Handle:

Return the single worker's individual crossing time as the result, avoiding pairing logic.