<p>There are some robots and factories on the X-axis. You are given an integer array <code>robot</code> where <code>robot[i]</code> is the position of the <code>i<sup>th</sup></code> robot. You are also given a 2D integer array <code>factory</code> where <code>factory[j] = [position<sub>j</sub>, limit<sub>j</sub>]</code> indicates that <code>position<sub>j</sub></code> is the position of the <code>j<sup>th</sup></code> factory and that the <code>j<sup>th</sup></code> factory can repair at most <code>limit<sub>j</sub></code> robots.</p>

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

<p>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.</p>

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

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

<p><strong>Note that</strong></p>

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

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/09/15/example1.jpg" style="width: 500px; height: 320px;" />
<pre>
<strong>Input:</strong> robot = [0,4,6], factory = [[2,2],[6,2]]
<strong>Output:</strong> 4
<strong>Explanation:</strong> 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.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/09/15/example-2.jpg" style="width: 500px; height: 329px;" />
<pre>
<strong>Input:</strong> robot = [1,-1], factory = [[-2,1],[2,1]]
<strong>Output:</strong> 2
<strong>Explanation:</strong> 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.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= robot.length, factory.length &lt;= 100</code></li>
	<li><code>factory[j].length == 2</code></li>
	<li><code>-10<sup>9</sup> &lt;= robot[i], position<sub>j</sub> &lt;= 10<sup>9</sup></code></li>
	<li><code>0 &lt;= limit<sub>j</sub> &lt;= robot.length</code></li>
	<li>The input will be generated such that it is always possible to repair every robot.</li>
</ul>

class Solution:
    def minimumTotalDistance(self, robot: List[int], factory: List[List[int]]) -> int:
        """
        Calculates the minimum total distance traveled by all robots to reach factories.

        Args:
        robot (List[int]): Positions of the robots on the X-axis.
        factory (List[List[int]]): Each sublist contains [position, limit] of a factory.

        Returns:
        int: The minimum total distance all robots have to travel.
        """
        # Sort robots and factories based on their positions
        robot.sort()
        factory.sort()

        n = len(robot)
        m = len(factory)

        # Initialize the DP table with None
        dp = [[None] * (m + 1) for _ in range(n + 1)]

        # Precompute the distances between robots and factories
        # distances[j][i]: distance from robot i to factory j
        distances = [[abs(robot[i] - factory[j][0]) for i in range(n)] for j in range(m)]

        # Precompute cumulative distances for each factory
        # cumulative_distances[j][i]: sum of distances from robot[0] to robot[i] to factory j
        cumulative_distances = []
        for j in range(m):
            cum_dist = [0] * n
            cum_dist[0] = distances[j][0]
            for i in range(1, n):
                cum_dist[i] = cum_dist[i - 1] + distances[j][i]
            cumulative_distances.append(cum_dist)

        def dfs(i, j):
            # Base case: all robots are assigned
            if i == n:
                return 0
            # Base case: no factories left
            if j == m:
                return float('inf')  # Cannot assign remaining robots
            # Check if result is already computed
            if dp[i][j] is not None:
                return dp[i][j]

            res = dfs(i, j + 1)  # Option 1: Skip this factory

            max_assign = min(factory[j][1], n - i)  # Max robots we can assign to this factory
            for k in range(1, max_assign + 1):
                # Calculate cost of assigning k robots starting from robot i to factory j
                if i == 0:
                    cost = cumulative_distances[j][i + k - 1]
                else:
                    cost = cumulative_distances[j][i + k - 1] - cumulative_distances[j][i - 1]
                # Recursive call for remaining robots and factories
                next_res = dfs(i + k, j + 1)
                if next_res != float('inf'):
                    total_cost = cost + next_res
                    res = min(res, total_cost)

            dp[i][j] = res
            return res

        result = dfs(0, 0)
        return result
        
1226