Campus Bikes II

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Dynamic ProgrammingBit Manipulation

On a campus represented as a 2D grid, there are n workers and m bikes, with n <= m. You are given a list of the workers' locations: workers[i] = (xi, yi) and a list of bike locations: bikes[j] = (xj, yj). All the workers and bikes are distinct.

Assign a bike to each worker such that the sum of the Manhattan distances between each worker and their assigned bike is minimized.

The Manhattan distance between two points (a, b) and (c, d) is |a - c| + |b - d|.

Return the minimum possible sum of Manhattan distances between each worker and their assigned bike.

Example 1:

Input: workers = [[0,0],[2,1]], bikes = [[1,2],[3,3]]
Output: 6
Explanation: 
Assign worker[0] to bike[0], worker[1] to bike[1]. The Manhattan distance of worker[0] to bike[0] is |0-1| + |0-2| = 3.
The Manhattan distance of worker[1] to bike[1] is |2-3| + |1-3| = 3.
The total Manhattan distance is 3 + 3 = 6.

Example 2:

Input: workers = [[0,0],[1,1],[2,0]], bikes = [[1,0],[2,2],[2,1]]
Output: 4
Explanation: 
Assign worker[0] to bike[0], worker[1] to bike[1], worker[2] to bike[2]. The Manhattan distance of worker[0] to bike[0] is |0-1| + |0-0| = 1.
The Manhattan distance of worker[1] to bike[1] is |1-2| + |1-2| = 2.
The Manhattan distance of worker[2] to bike[2] is |2-2| + |0-1| = 1.
The total Manhattan distance is 1 + 2 + 1 = 4.

Constraints:

n == workers.length
m == bikes.length
1 <= n <= m <= 10
workers[i].length == bikes[j].length == 2
0 <= xi, yi < 1000
0 <= xj, yj < 1000
All the workers and bikes are distinct.

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:

Can a worker or bike be unassigned after being assigned?
What is the expected return format? Should I return a single integer representing the minimum total Manhattan distance, or should I return a list/map of worker-bike pairings?
What is the maximum number of workers and bikes? Are there any constraints on the coordinate values (e.g., are they integers, and what's their range)?
Is it guaranteed that the number of workers and bikes are equal?
If there are multiple possible assignments with the same minimum total Manhattan distance, is any one of them acceptable?

Brute Force Solution

Approach

The problem wants us to find the best way to assign bikes to workers. The brute force way is to just try out every possible matching between workers and bikes and see which assignment is the best.

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

Consider the first worker and try assigning them each bike, one at a time.
For each of those bike assignments, consider the second worker and assign them each of the remaining bikes, one at a time.
Keep doing this for each worker, each time assigning them one of the bikes that hasn't already been assigned.
Eventually, all workers will have a bike.
For each complete assignment of workers to bikes, calculate the total distance traveled.
Keep track of the assignment that has the smallest total distance.
After trying every single possible way of assigning workers to bikes, the assignment with the smallest total distance is the answer.

Code Implementation

def campus_bikes_brute_force(workers, bikes):
    minimum_distance = float('inf')

    def calculate_distance(worker_index, assigned_bikes, current_distance):
        nonlocal minimum_distance

        # Base case: all workers have been assigned a bike
        if worker_index == len(workers):
            minimum_distance = min(minimum_distance, current_distance)
            return

        # Try assigning each unassigned bike to the current worker
        for bike_index in range(len(bikes)):
            if bike_index not in assigned_bikes:

                # Calculate distance between the current worker and bike
                distance = abs(workers[worker_index][0] - bikes[bike_index][0]) + \
                           abs(workers[worker_index][1] - bikes[bike_index][1])

                # Recursively call function for next worker
                calculate_distance(
                    worker_index + 1,
                    assigned_bikes | {bike_index},
                    current_distance + distance
                )
   # Initialize the recursive calls with initial values
    calculate_distance(0, set(), 0)
   # After trying every possible combination, return the minimum distance
    return minimum_distance

Big(O) Analysis

Time Complexity

O(n!) – The brute force solution explores every possible assignment of bikes to workers. If there are n workers and n bikes, the first worker has n choices, the second has (n-1) choices, the third has (n-2) choices, and so on. This results in n * (n-1) * (n-2) * ... * 1, which is n! (n factorial) possible assignments. Calculating the total distance for each assignment takes O(n) time, but the dominant factor is the number of possible assignments. Therefore, the overall time complexity is O(n!).

Space Complexity

O(N) – The provided brute-force approach uses recursion to explore all possible worker-bike assignments. The maximum depth of the recursion is proportional to the number of workers, which we can denote as N. Each recursive call stores the current assignment state (which bikes are assigned) on the call stack. Therefore, the auxiliary space used by the recursion stack is O(N), where N is the number of workers. No other significant auxiliary data structures are used.

Optimal Solution

Approach

The core idea is to explore possible assignments of workers to bikes and to do so in a way that avoids redundant calculations. We will use a technique to only investigate the most promising assignments and discard less promising ones early on to dramatically reduce the overall computation.

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

Think of this problem as assigning each worker to a bike, and we want the total distance to be as small as possible.
Start by figuring out all the possible pairings of workers and bikes.
Organize these pairings by how far apart the worker and bike are. Start with the closest pairings.
Try assigning workers to bikes starting with the closest pairings. Keep track of the total distance as you go.
If you find a way to assign all workers to bikes that has a lower total distance than anything you've seen before, remember that distance.
As you explore different pairings, if the total distance you've already calculated is higher than the best total distance you've already found, stop exploring that particular set of pairings. It can't possibly be the best.
Keep going until you've considered all possibilities that could lead to a better solution. This is much faster than checking every single combination.

Code Implementation

def campus_bikes_two(workers, bikes):

def calculate_distance(worker, bike):
	return abs(worker[0] - bike[0]) + abs(worker[1] - bike[1])

	number_of_workers = len(workers)
	number_of_bikes = len(bikes)

	# Initialize the best total distance found so far to infinity.
	minimum_total_distance = float('inf')

	def find_minimum_distance(worker_index, assigned_bikes, current_total_distance):
		nonlocal minimum_total_distance

		# If we've assigned all workers to bikes,
		if worker_index == number_of_workers:
			# update the minimum total distance if necessary.
			minimum_total_distance = min(minimum_total_distance, current_total_distance)
			return

		# If the current total distance is already greater than the minimum,
		# it's impossible to find a better solution from this point forward.
		if current_total_distance > minimum_total_distance:
			return

		for bike_index in range(number_of_bikes):
			# Only consider bikes that haven't been assigned yet.
			if bike_index not in assigned_bikes:
				distance = calculate_distance(workers[worker_index], bikes[bike_index])
				# Recursively explore assigning the current worker to the current bike.
				find_minimum_distance(
					worker_index + 1,
					assigned_bikes | {bike_index},
					current_total_distance + distance,
				)

	# Start the process by assigning the first worker to bikes.
	find_minimum_distance(0, set(), 0)

	return minimum_total_distance

Big(O) Analysis

Time Complexity

O(M * N * 2^(M+N)) – The algorithm explores possible assignments of workers to bikes using a technique similar to backtracking with pruning. In the worst-case scenario, the algorithm may have to consider a significant portion of all possible worker-bike assignments. Calculating distances between each worker and bike takes O(M * N) where M is the number of workers and N is the number of bikes. The pruning strategy helps, but the maximum number of branches explored can still grow exponentially. Consequently, it approaches O(M * N * 2^(M+N)) reflecting the cost of distance calculation multiplied by a factor related to the exponential nature of searching possible assignments even with pruning.

Space Complexity

O(P * 2^min(M, N)) – The space complexity stems primarily from the implicit exploration of worker-bike pairings, which can be visualized as a decision tree. Each level of the tree corresponds to a worker, and each branch at that level corresponds to assigning that worker to a bike. The algorithm keeps track of the 'best total distance' found so far, a single variable. However, the space is dominated by the potential number of active branches representing the explored pairings (partial assignments). In the worst-case, where no pruning occurs, this could approach exploring all possible subsets of bike assignments for each worker, up to 2^min(M, N), where M is the number of workers, N is the number of bikes, and P is the space needed to store information about each partial assignment which would include the distance so far and which workers/bikes are assigned. Thus, the overall space complexity is O(P * 2^min(M, N)).

Edge Cases

Case	How to Handle
Empty workers or bikes array	Return 0 since no assignments are possible if either workers or bikes are empty.
Workers and bikes arrays have significantly different lengths (e.g., one worker and many bikes)	The cost calculation and minimization should still function correctly, exploring all possible bike assignments to the single worker.
Large input size approaching the memory limit for recursive solutions	Consider using dynamic programming with memoization instead of recursion to avoid stack overflow errors and improve memory usage.
All workers are at the same location or all bikes are at the same location	The distance calculation should still function correctly, resulting in either 0 or the distance to the nearest other location.
Integer overflow when calculating Manhattan distance with large coordinate values	Use a data type like long to store the Manhattan distance to prevent integer overflow.
Workers or bikes have negative coordinate values	The Manhattan distance calculation should handle negative coordinates correctly as the absolute difference is taken.
Coordinate values are zero	The distance calculation should handle zero coordinates correctly as the absolute difference is taken.
Duplicate worker or bike locations	The algorithm should correctly handle duplicate locations by exploring all possible assignments, not treating them as a single entity.

Empty workers or bikes array

How to Handle:

Return 0 since no assignments are possible if either workers or bikes are empty.

Workers and bikes arrays have significantly different lengths (e.g., one worker and many bikes)

How to Handle:

The cost calculation and minimization should still function correctly, exploring all possible bike assignments to the single worker.

Large input size approaching the memory limit for recursive solutions

How to Handle:

Consider using dynamic programming with memoization instead of recursion to avoid stack overflow errors and improve memory usage.

All workers are at the same location or all bikes are at the same location

How to Handle:

The distance calculation should still function correctly, resulting in either 0 or the distance to the nearest other location.

Integer overflow when calculating Manhattan distance with large coordinate values

How to Handle:

Use a data type like long to store the Manhattan distance to prevent integer overflow.

Workers or bikes have negative coordinate values

How to Handle:

The Manhattan distance calculation should handle negative coordinates correctly as the absolute difference is taken.

Coordinate values are zero

How to Handle:

The distance calculation should handle zero coordinates correctly as the absolute difference is taken.

Duplicate worker or bike locations

How to Handle:

The algorithm should correctly handle duplicate locations by exploring all possible assignments, not treating them as a single entity.