<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:
        robot.sort()
        factory.sort()
        n, m = len(robot), len(factory)
        costs = [inf] * n
        for pos, capacity in factory:
            distf =[(0, -1)]
            cumcost = 0
            for idx, rpos in enumerate(robot): 
                cumcost += abs(rpos - pos)
                while distf and idx - distf[0][1] > capacity:
                    heappop(distf)
                heappush(distf, (costs[idx] - cumcost, idx))
                costs[idx] = cumcost + distf[0][0]
        return costs[-1]
64