Minimum Total Distance Traveled

Hard

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ArraysDynamic Programming

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.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

Can you clarify the data type and range of values in the input array? Could the array contain negative numbers, zeros, or floating-point numbers?
What should be returned if the input array is empty or null?
Are the locations in the input array one-dimensional or two-dimensional? If two-dimensional, what does each coordinate represent?
If there are multiple optimal solutions (i.e., multiple points that result in the same minimum total distance), is any one of them acceptable, or is there a specific solution that should be prioritized?
What is the real-world context of this problem? Knowing the context might help clarify assumptions about the meaning of 'distance' and how it should be calculated (e.g., Euclidean, Manhattan, or some other distance metric).

Brute Force Solution

Approach

The brute force approach to minimizing total distance traveled involves examining every single possible location where we could meet. It's like trying every possible meeting spot to see which one results in the least amount of travel overall. We compare the total travel distance for each candidate location to find the absolute minimum.

Here's how the algorithm would work step-by-step:

Consider every possible location as a potential meeting point.
For each potential meeting point, calculate the total distance everyone would have to travel to reach it.
Keep track of the smallest total distance found so far.
Compare the total distance for the current meeting point with the smallest total distance we've seen so far.
If the current meeting point's total distance is smaller, update the smallest total distance.
Repeat this process for every single possible location.
After checking every possible location, the smallest total distance you've kept track of is the answer.

Code Implementation

def minimum_total_distance_brute_force(locations):
    minimum_distance = float('inf')

    # Iterate through all possible meeting points
    for meeting_point in locations:

        total_distance_for_current_meeting_point = 0

        # Calculate the total distance for the current meeting point
        for individual_location in locations:
            total_distance_for_current_meeting_point += abs(meeting_point - individual_location)

        # Update the minimum distance if necessary
        if total_distance_for_current_meeting_point < minimum_distance:

            minimum_distance = total_distance_for_current_meeting_point

    return minimum_distance

Big(O) Analysis

Time Complexity

O(n*m) – The brute force approach considers every possible location in a grid or space of size n*m (assuming n and m are dimensions of the space) as a potential meeting point. For each potential meeting point, we calculate the total distance to every person. Therefore, the total number of locations we check is n*m. For each location, we need to iterate through all the people in the grid. So we multiply n*m by the number of people, but since the number of people is constant, we can disregard it and say the complexity is dominated by n*m, leading to O(n*m).

Space Complexity

O(1) – The described brute force approach only requires storing a variable to track the smallest total distance found so far. No additional data structures that scale with the input size are needed to store intermediate results or potential meeting locations. Therefore, the auxiliary space complexity is constant, regardless of the number of possible meeting locations, N.

Optimal Solution

Approach

The best way to solve this problem is to realize the optimal meeting point minimizes the total travel. The key insight is that the meeting point should be the middle location if you line up all the locations.

Here's how the algorithm would work step-by-step:

First, sort all the locations in ascending order.
If there's an odd number of locations, pick the middle location as the meeting point.
If there's an even number of locations, the meeting point can be any point between the two middle locations. For simplicity, choose one of these middle locations.
Calculate the distance each person needs to travel by subtracting the chosen meeting point from their location and taking the absolute value (to make sure the distance is positive).
Sum up all these distances to get the minimum total distance traveled.

Code Implementation

def minimum_total_distance(locations):
    locations.sort()

    number_of_locations = len(locations)

    # Find the meeting point; it's the median.
    if number_of_locations % 2 == 0:
        meeting_point = locations[number_of_locations // 2 - 1]
    else:
        meeting_point = locations[number_of_locations // 2]

    total_distance = 0

    # Calculate the total distance traveled.
    for location in locations:

        distance_to_meeting_point = abs(location - meeting_point)

        total_distance += distance_to_meeting_point

    return total_distance

Big(O) Analysis

Time Complexity

O(n log n) – The dominant operation in this solution is sorting the locations in ascending order. Sorting algorithms like merge sort or quicksort, which are typically used in standard library implementations, have a time complexity of O(n log n), where n is the number of locations. Calculating the distances and summing them up involves iterating through the sorted array once, which takes O(n) time. Since O(n log n) grows faster than O(n), the overall time complexity is determined by the sorting step. Therefore, the algorithm's time complexity is O(n log n).

Space Complexity

O(1) – The algorithm sorts the input array in place (or uses an existing sorting algorithm, which we assume has O(1) auxiliary space as most standard library sorts do for primitive types), and then uses a few integer variables to store the middle index and calculate the total distance. The space required for these index variables and the total distance accumulator does not depend on the input size N (the number of locations). Therefore, the auxiliary space complexity is constant.

Edge Cases

Case	How to Handle
Empty input array	Return 0 as no distance needs to be traveled if there are no people to move.
Array with only one element	Return 0 as the single person doesn't need to move.
All locations are the same	Return 0 as everyone is already at the same location and no distance needs to be traveled.
Integer overflow when calculating distances	Use long or appropriate data type to store distances to prevent overflow.
Large input array causing time limit exceedance	Optimize the algorithm to have a better time complexity, perhaps O(n log n) or O(n) where possible using median selection.
Negative location values	The problem may require shifting all values to be non-negative if the chosen median selection is index-based.
Locations represented as floating point numbers, leading to precision errors in distance calculations	Employ an appropriate tolerance when comparing floating-point numbers to mitigate precision errors; use integer arithmetic if possible after scaling up the inputs.
Optimal meeting point is not an integer (if input is integer locations)	If meeting point must be an integer, round to the nearest integer for minimum distance (can test floor and ceiling of the mean).

Empty input array

How to Handle:

Return 0 as no distance needs to be traveled if there are no people to move.

Array with only one element

How to Handle:

Return 0 as the single person doesn't need to move.

All locations are the same

How to Handle:

Return 0 as everyone is already at the same location and no distance needs to be traveled.

Integer overflow when calculating distances

How to Handle:

Use long or appropriate data type to store distances to prevent overflow.

Large input array causing time limit exceedance

How to Handle:

Optimize the algorithm to have a better time complexity, perhaps O(n log n) or O(n) where possible using median selection.

Negative location values

How to Handle:

The problem may require shifting all values to be non-negative if the chosen median selection is index-based.

Locations represented as floating point numbers, leading to precision errors in distance calculations

How to Handle:

Employ an appropriate tolerance when comparing floating-point numbers to mitigate precision errors; use integer arithmetic if possible after scaling up the inputs.

Optimal meeting point is not an integer (if input is integer locations)

How to Handle:

If meeting point must be an integer, round to the nearest integer for minimum distance (can test floor and ceiling of the mean).