Introduction

In this blog post, we will explore how to solve the LeetCode problem "2463" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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 ith robot. You are also given a 2D integer array factory where factory[j] = [positionj, limitj] indicates that positionj is the position of the jth factory and that the jth factory can repair at most limitj 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], positionj <= 109 0 <= limitj <= robot.length The input will be generated such that it is always possible to repair every robot.

Explanation

Here's a breakdown of the solution, followed by the Python code:

• Key Idea: The problem can be optimally solved using dynamic programming. Sort both robots and factories by their positions. The DP state represents the minimum cost to repair the first i robots using the first j factories.
• Transition: For each robot, we can either assign it to the current factory or skip the current factory. The cost of assigning a robot to a factory is the distance between them. We need to track the number of robots already assigned to a factory and ensure it doesn't exceed the factory's limit.

• Optimization: Using memoization to avoid recomputation of overlapping subproblems.

• Complexity:

  • Runtime: O(R F R), where R is the number of robots and F is the number of factories.
  • Space: O(R * F), for the memoization table.

Code

    def minimumTotalDistance(robot, factory):
    robot.sort()
    factory.sort()
    n = len(robot)
    m = len(factory)

    memo = {}

    def dp(i, j, capacity):
        if i == n:
            return 0
        if j == m:
            return float('inf')

        if (i, j, capacity) in memo:
            return memo[(i, j, capacity)]

        # Option 1: Skip the current factory
        ans = dp(i, j + 1, 0)

        # Option 2: Use the current factory to repair the current robot
        if capacity < factory[j][1]:
            ans = min(ans, abs(robot[i] - factory[j][0]) + dp(i + 1, j, capacity + 1))

        memo[(i, j, capacity)] = ans
        return ans

    return dp(0, 0, 0)