Minimum Total Distance Traveled

class Solution {
public:
    long long minimumTotalDistance(vector<int>& robots,
                                   vector<vector<int>>& factories) {
        // Order the robots and factories based on position
        sort(begin(robots), end(robots));
        sort(begin(factories), end(factories));

        // Spread out factory positions based on capacities
        vector<int> positions;
        for (const auto& f : factories) {
            for (int i = 0; i < f[1]; i++) {
                positions.push_back(f[0]);
            }
        }

        int robotQty = robots.size(), positionQty = positions.size();
        vector<long long> next(positionQty + 1, 0),
            current(positionQty + 1, 0);

        current[positionQty] = 1e12;

        // Use dynamic programming with two rows
        for (int i = robotQty - 1; i >= 0; i--) {
            for (int j = positionQty - 1; j >= 0; j--) {
                // Consider current matching between robot and factory
                long long include = abs(robots[i] - positions[j]) + next[j + 1];
                // Consider skipping the current factory
                long long exclude = current[j + 1];
                // Select optimal cost
                current[j] = min(include, exclude);
            }
            // Update row for next iteration
            next = current;
        }
        // Output the optimal distribution cost
        return current[0];
    }
};

To tackle the problem of finding the minimum total distance that robots need to travel to factories, the provided C++ solution employs sorting, dynamic programming, and careful management of robot and factory distribution. Here's the breakdown of the solution's approach:

• Sorting Inputs: Initially, the robots and factories are sorted based on their positions. This simplification allows for an easier calculation of distances between each robot and their matched factory.

• Factory Position Management: The solution expands factory positions based on their capacities. Each factory might need to be assigned one or more robots depending on its capacity, which is handled by iterating through each factory and its capacity.

• Dynamic Programming Initialization: Two vectors, next and current, are initialized to store the dynamic programming states. The vector current starts with a large number to ensure any real cost comparison would be chosen over this initial value.

• Robot-Factory Matching: Using dynamic programming, the solution iterates backwards through the list of robots and each possible factory position. For each pairing of robot and factory, the cost of assigning the robot to the current factory (inclusive cost) and the cost of not assigning (exclusive cost) are calculated. The minimum of these two costs determines the optimal strategy for that iteration.

• Minimizing Costs: By updating the current array iteratively from the end state where no robots are left to assign, the solution builds up the minimum possible cost of assigning all robots to factories.

• Result: Ultimately, the current[0] holds the minimum distance cost after all possibilities have been considered, ensuring an optimal solution to the robot-factory assignment problem.

This approach takes advantage of efficient data structures (vectors) and algorithms (sorting, dynamic programming) to manage and solve the challenge efficiently, adhering to C++ programming practices.

class Solution {

    public long calculateMinDistance(List<Integer> robotList, int[][] plantLocations) {
        // Order robot and plant positions
        Collections.sort(robotList);
        Arrays.sort(plantLocations, Comparator.comparingInt(a -> a[0]));

        // Expand plant positions based on capacity
        List<Integer> expandedPlants = new ArrayList<>();
        for (int[] plant : plantLocations) {
            for (int i = 0; i < plant[1]; i++) {
                expandedPlants.add(plant[0]);
            }
        }

        int numOfRobots = robotList.size(), expandedPlantCount = expandedPlants.size();
        long[] previous = new long[expandedPlantCount + 1];
        long[] now = new long[expandedPlantCount + 1];

        previous[expandedPlantCount] = (long) 1e12;

        // Using two rows for memory optimization in DP solution
        for (int i = numOfRobots - 1; i >= 0; i--) {
            for (int j = expandedPlantCount - 1; j >= 0; j--) {
                // Calculate distance when assigning current robot to current plant
                long assign =
                    Math.abs((long) robotList.get(i) - expandedPlants.get(j)) +
                    previous[j + 1];
                // Calculate distance when not assigning current robot to current plant
                long skip = now[j + 1];
                // Choose the lesser distance
                now[j] = Math.min(assign, skip);
            }
            // Prepare for next iteration
            System.arraycopy(now, 0, previous, 0, expandedPlantCount + 1);
        }

        // Optimal distance for assigning robots starting from index 0
        return now[0];
    }
}

The Java program provided is designed to find the minimum total distance traveled by robots to fixed plant locations, given the count of units each plant demands. The algorithm processes input lists of robot positions and plant positions with their capacities, where the robot positions (robotList) and plant positions, along with capacities (plantLocations), are first sorted. During the process:

• Expands the plantLocations array into the expandedPlants list based on the capacity of each plant, thus transforming a 2D array into a 1D list, which simplifies the distance calculation.
• Implements a dynamic programming approach using a sliding window technique (previous and now arrays) to optimize the memory usage.
• Iterates over both robots and plant locations in reverse order, calculating two possible scenarios at each step:
  • Assigning a robot to a plant and then accumulating the total distance.
  • Skipping a robot assignment to keep options open for subsequent robots.
• Chooses the option with the lower accumulated distance at each iteration.
• The calculation process includes magnitudes of positions and ensures the least distance covered by maintaining a minimal cumulative distance from robots to plants.

Key Points of the Algorithm Implementation:

• Sort both robots and plants to align closest distances and minimize traversal.
• Use two arrays to minimize memory overhead while ensuring the calculation of minimal distances through a two-row approach in dynamic programming.
• Every loop iteration calculates and compares two possible distances—assigning or skipping—and updates the results accordingly.
• Efficient handling of the multi-capacity plants is achieved by expanding plant locations in proportion to their capacities, which allows handling them as individual units.

The final distance gets updated iteratively and the result after processing all robots starting with the first robot offers the minimal total distance needed for all robots to satisfy all plant demands successfully. This efficient and memory-optimized dynamic programming solution ensures minimal computation time and resource utilization while solving the problem effectively.

class Operations:
    def computeMinDistance(
        self, bots: List[int], plants: List[List[int]]
    ) -> int:
        # Sort lists
        bots.sort()
        plants.sort()

        # Generate positions from plant capacities
        plants_positions = []
        for plant in plants:
            for count in range(plant[1]):
                plants_positions.append(plant[0])

        bots_len = len(bots)
        plants_len = len(plants_positions)
        forward_dist = [0] * (plants_len + 1)
        interim = [0] * (plants_len + 1)

        interim[plants_len] = float('inf')

        # Update distances iteratively using DP
        for bot_idx in range(bots_len - 1, -1, -1):
            for plant_idx in range(plants_len - 1, -1, -1):
                assign_now = (
                    abs(bots[bot_idx] - plants_positions[plant_idx]) + forward_dist[plant_idx + 1]
                )
                skip_now = interim[plant_idx + 1]
                interim[plant_idx] = min(assign_now, skip_now)

            forward_dist = interim[:]

        return interim[0]

The provided Python script computes the minimum total distance a set of robots must travel to reach specified plant positions. Each plant has a specific number of positions represented through its capacity.

Here is a description of the solution logic:

1. Sort the bot positions and the plant positions for sequential processing.
2. Create a list plants_positions that explicitly enumerates each position for the plants based on their capacity. Each plant's position may be repeated in this list according to its capacity.
3. Initialize arrays forward_dist and interim used for dynamic programming to store the cumulative minimum distances from bots to plant positions as well as provisional calculations, respectively.
4. Iterate over each bot starting from the last to the first, and for each bot, compute two potential distances for each plant position:
   • The distance if the current bot moves to the current plant position combined with the already computed distance for moving to subsequent plant positions.
   • The distance if the current bot does not move to the current plant position but instead uses previous calculations.
5. For each plant position, save the minimum of these two distances.
6. At the end of iterations, interim[0] contains the minimum distance that the bots must travel to reach all positions.

This algorithm relies on dynamic programming to efficiently calculate the solution by breaking the problem down and solving it for smaller sub-problems incrementally while storing intermediate results to avoid redundant calculations..md