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
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 <= 100factory[j].length == 2-109 <= robot[i], positionj <= 1090 <= limitj <= robot.lengthThe Greedy Approach involves sorting both the robots and factories based on their positions. We try to assign each robot to the closest possible factory without exceeding the factory's limit, thus minimizing the travel distance one step at a time.
The C solution uses sorting to minimize the distance. First, both arrays are sorted by position. Then, robots are assigned to factories in a greedy manner based on proximity and factory limits until all robots are assigned.
C++
Java
Python
C#
JavaScript
Time Complexity: O(n log n + m log m), where n is the number of robots and m is the number of factories due to the sorting step. The assignment step is O(n + m).
Space Complexity: O(1), not counting input space as we sort in-place.
A Dynamic Programming approach could involve maintaining a DP table where each entry dp[i][j] represents the minimum distance needed to repair the first i robots using the first j factories. This approach could find optimal assignments by considering various combinations and memoizing the results, though it might be more computationally intensive for larger datasets than the greedy approach.
In this Python solution, a DP table is constructed that tracks the minimum distance for repairing a certain number of robots using a given number of factories. By incrementally building up solutions for subproblems, it finds the optimal total travel distance.
Time Complexity: O(n * m * l), where l is the maximum capacity of a factory, due to nested loops.
Space Complexity: O(n * m), for storing the DP table.
| Approach | Complexity |
|---|---|
| Greedy Approach | Time Complexity: O(n log n + m log m), where n is the number of robots and m is the number of factories due to the sorting step. The assignment step is O(n + m). |
| Dynamic Programming Approach | Time Complexity: O(n * m * l), where l is the maximum capacity of a factory, due to nested loops. |
Minimum Total Distance Traveled | Recursion | DP | Leetcode 2463 • Techdose • 6,984 views views
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