Daily LeetCode - by Vítor Belim

Description

There are some robots and factories on the X-axis. You are given an integer array robot where robot[i] is the position of the i^th robot. You are also given a 2D integer array factory where factory[j] = [position_j, limit_j] indicates that position_j is the position of the j^th factory and that the j^th factory can repair at most limit_j robots.

The positions of each robot are unique. The positions of each factory are also unique. Note that a robot can be in the same position as a factory initially.

All the robots are initially broken; they keep moving in one direction. The direction could be the negative or the positive direction of the X-axis. When a robot reaches a factory that did not reach its limit, the factory repairs the robot, and it stops moving.

At any moment, you can set the initial direction of moving for some robot. Your target is to minimize the total distance traveled by all the robots.

Return the minimum total distance traveled by all the robots. The test cases are generated such that all the robots can be repaired.

Note that

All robots move at the same speed.
If two robots move in the same direction, they will never collide.
If two robots move in opposite directions and they meet at some point, they do not collide. They cross each other.
If a robot passes by a factory that reached its limits, it crosses it as if it does not exist.
If the robot moved from a position x to a position y, the distance it moved is |y - x|.

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], position_j <= 10⁹
0 <= limit_j <= robot.length
The input will be generated such that it is always possible to repair every robot.

Solutions

This code solves the robot-to-factory assignment problem using dynamic programming and is optimized for space efficiency. It first sorts both arrays and expands factories into individual positions (if a factory can repair multiple robots, it's added that many times to an array). Then it uses bottom-up DP with two arrays (current and next) representing minimum costs: current[j] is the minimum total distance to assign all remaining robots to factories from index j onward. For each robot (processed from last to first), it decides whether to assign the robot to the current factory (paying its distance cost and recursing) or skip to the next factory. The recurrence is current[j] = min(skip factory, assign to factory), where skipping means current[j+1] and assigning means abs(robot_position - factory_position) + next[j+1]. After processing each robot, it swaps the arrays so the previous state becomes the current state for the next iteration. The final answer is current[0], representing the minimum cost to assign all robots starting from the first factory.

Language: javascript(2026-04-14 08:56)DONE

CPU Performance14.29%

Memory Performance14.29%

/**
 * @param {number[]} robots
 * @param {number[][]} factories
 * @return {number}
 */
var minimumTotalDistance = function(robots, factories) {
    robots.sort((a, b) => a - b);
    factories.sort((a, b) => a[0] - b[0]);

    const factoryPositions = [];
    for (const [pos, rep] of factories) {
        for (let i = 0; i < rep; i++) {
            factoryPositions.push(pos);
        }
    }

    let robotCount = robots.length;
    let factoryCount = factoryPositions.length;
    let next = new Array(factoryCount + 1).fill(0);
    let current = new Array(factoryCount + 1).fill(0);

    current[factoryCount] = Infinity;

    for (let i = robotCount - 1; i >= 0; i--) {
        for (let j = factoryCount - 1; j >= 0; j--) {
            current[j] = Math.min(
                current[j + 1],
                Math.abs(robots[i] - factoryPositions[j]) + next[j + 1],
            );
        }

        next = current.slice(0, factoryCount + 1);
    }

    return current[0];
};

Description

Solutions

AI Explanation