Minimum Total Distance Traveled

Problem Statement

On an X-axis, there are various robots and factories positioned uniquely. For every robot, its initial position is provided in an array robot, where each entry robot[i] denotes the location of the ith robot. Similarly, factories are given in a 2D array factory where each entry factory[j] = [positionj, limitj] specifies the position of the jth factory and the maximum number of robots it can repair.

Given these robots start broken, they need repairs to stop moving and their motion is linear along the X-axis either leftward or rightward. The goal is to strategically direct or allow robots to move such that the cumulative distance they travel is minimized. Once a robot encounters a factory with remaining repair capacity (limitj has not been reached), it gets repaired and its movement ceases. Robots are noted to have unique starting points and similarly, factories have unique positions, which may coincide with a robot's start such that a robot may need no movement for repair.

Key points to consider:

• Regardless of their numbers, robots in motion along the same path or converging paths do not interact or impact each other's travel.
• A factory that has attended to its maximum (limitj) will be treated as non-existent by any subsequent robot requiring repair.
• The robots can begin their initial movement toward any direction, which can be adjusted to minimize overall travel distance.

The challenge then becomes calculating the minimal possible sum of distances all robots need to traverse to get repaired, ensuring every robot is eventually repaired, given the constraints that are guaranteed by the problem's conditions.

Examples

Example 1

Input:

robot = [0,4,6], factory = [[2,2],[6,2]]

Output:

4

Explanation:

As shown in the figure:
- The first robot at position 0 moves in the positive direction. It will be repaired at the first factory.
- The second robot at position 4 moves in the negative direction. It will be repaired at the first factory.
- The third robot at position 6 will be repaired at the second factory. It does not need to move.
The limit of the first factory is 2, and it fixed 2 robots.
The limit of the second factory is 2, and it fixed 1 robot.
The total distance is |2 - 0| + |2 - 4| + |6 - 6| = 4. It can be shown that we cannot achieve a better total distance than 4.

Example 2

Input:

robot = [1,-1], factory = [[-2,1],[2,1]]

Output:

2

Explanation:

As shown in the figure:
- The first robot at position 1 moves in the positive direction. It will be repaired at the second factory.
- The second robot at position -1 moves in the negative direction. It will be repaired at the first factory.
The limit of the first factory is 1, and it fixed 1 robot.
The limit of the second factory is 1, and it fixed 1 robot.
The total distance is |2 - 1| + |(-2) - (-1)| = 2. It can be shown that we cannot achieve a better total distance than 2.

Constraints

• 1 <= robot.length, factory.length <= 100
• factory[j].length == 2
• -109 <= robot[i], positionj <= 109
• 0 <= limitj <= robot.length
• The input will be generated such that it is always possible to repair every robot.

Approach and Intuition

To address this problem, the understanding boils down to:

1. Parsing through each robot's starting position and determining the optimal direction (either toward the nearest factory on the left or right) requiring the least movement. This hinges on whether the nearby factory has enough repair capacity remaining.

2. The problem can potentially be approached by sorting robots and factories based on their position on the X-axis which helps in efficiently determining the nearest factories.

   • For a robot at position x, check both directions for the closest available factory.
   • Deciding the direction:
     • Calculate the distance to the nearest right-side factory and its available capacity.
     • Likewise, calculate for the left side.
     • Choose the direction where the robot covers the lesser distance and the factory can still accommodate a repair.

3. Loop through each robot and, based on their positions and the available factories, adjust their directions and calculate the distances moved.

   • This involves checking if moving to the closest factory (in either direction) is possible without exceeding its limit.

4. Sum these minimal distances for all robots to provide the answer.

These steps respect each constraint provided by the problem, notably handling factories' capacities, unique positions, and non-collision nature of the robots. While simple on the surface, the implementation must be meticulously designed to always opt for the optimal repair point, thereby ensuring minimal total travel. Moreover, the example scenarios describe how different configurations might lead to the same or different outcomes in terms of total repair distances, reinforcing the need for a smart, dynamic approach to solve the problem efficiently.

Solutions

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];
    }
};