Minimize Manhattan Distances

Hard

Amazon

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Topics:

ArraysGreedy Algorithms

You are given an array points representing integer coordinates of some points on a 2D plane, where points[i] = [x_i, y_i].

The distance between two points is defined as their Manhattan distance.

Return the minimum possible value for maximum distance between any two points by removing exactly one point.

Example 1:

Input: points = [[3,10],[5,15],[10,2],[4,4]]

Output: 12

Explanation:

The maximum distance after removing each point is the following:

After removing the 0^th point the maximum distance is between points (5, 15) and (10, 2), which is |5 - 10| + |15 - 2| = 18.
After removing the 1^st point the maximum distance is between points (3, 10) and (10, 2), which is |3 - 10| + |10 - 2| = 15.
After removing the 2^nd point the maximum distance is between points (5, 15) and (4, 4), which is |5 - 4| + |15 - 4| = 12.
After removing the 3^rd point the maximum distance is between points (5, 15) and (10, 2), which is |5 - 10| + |15 - 2| = 18.

12 is the minimum possible maximum distance between any two points after removing exactly one point.

Example 2:

Input: points = [[1,1],[1,1],[1,1]]

Output: 0

Explanation:

Removing any of the points results in the maximum distance between any two points of 0.

Constraints:

3 <= points.length <= 10⁵
points[i].length == 2
1 <= points[i][0], points[i][1] <= 10⁸

Minimize Manhattan Distances

Hard

Amazon

2 views

Topics:

ArraysGreedy Algorithms

You are given an array points representing integer coordinates of some points on a 2D plane, where points[i] = [x_i, y_i].

The distance between two points is defined as their Manhattan distance.

Return the minimum possible value for maximum distance between any two points by removing exactly one point.

Example 1:

Input: points = [[3,10],[5,15],[10,2],[4,4]]

Output: 12

Explanation:

The maximum distance after removing each point is the following:

After removing the 0^th point the maximum distance is between points (5, 15) and (10, 2), which is |5 - 10| + |15 - 2| = 18.
After removing the 1^st point the maximum distance is between points (3, 10) and (10, 2), which is |3 - 10| + |10 - 2| = 15.
After removing the 2^nd point the maximum distance is between points (5, 15) and (4, 4), which is |5 - 4| + |15 - 4| = 12.
After removing the 3^rd point the maximum distance is between points (5, 15) and (10, 2), which is |5 - 10| + |15 - 2| = 18.

12 is the minimum possible maximum distance between any two points after removing exactly one point.

Example 2:

Input: points = [[1,1],[1,1],[1,1]]

Output: 0

Explanation:

Removing any of the points results in the maximum distance between any two points of 0.

Constraints:

3 <= points.length <= 10⁵
points[i].length == 2
1 <= points[i][0], points[i][1] <= 10⁸

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:

What is the data type of the coordinates? Are they integers or floats, and what is the expected range for x and y?
Can the input arrays be empty or null? What should I return in such cases?
Are there any constraints on the number of locations provided in the input arrays?
If multiple locations minimize the total Manhattan distance, should I return one of them, and if so, how should I choose which one?
Can locations have the same coordinates (i.e., are duplicates allowed)? If so, how should I handle them?

Brute Force Solution

Approach

The goal is to find the best single location to minimize the total travel distance to several destinations. The brute force method involves checking every possible location within a defined area to calculate the total travel distance to all destinations from that location.

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

First, define the area where we will consider possible locations. This could be a grid or a specific range of coordinates.
Pick a location within the defined area. This location will be a potential meeting point.
Calculate the total travel distance from that location to all the given destinations. We add up the distances to each one.
Store this total travel distance, and remember which location it corresponds to.
Repeat steps 2 and 3 for every possible location within the defined area.
Once we have checked all the locations, compare the total travel distances we stored.
The location with the smallest total travel distance is our answer; that's the location that minimizes the total distance everyone needs to travel.

Code Implementation

def minimize_manhattan_distances_brute_force(destinations):
    min_total_distance = float('inf')
    best_location = None

    # Define the area to search based on the destinations
    min_x = min(x for x, y in destinations)
    max_x = max(x for x, y in destinations)
    min_y = min(y for x, y in destinations)
    max_y = max(y for x, y in destinations)

    for potential_x in range(min_x, max_x + 1):
        for potential_y in range(min_y, max_y + 1):
            current_location = (potential_x, potential_y)

            # Calculate the total Manhattan distance to all destinations
            total_distance = 0

            for destination_x, destination_y in destinations:
                total_distance += abs(potential_x - destination_x) + abs(potential_y - destination_y)

            # Update the best location if this one is better
            if total_distance < min_total_distance:

                # We have a new minimum, so keep track of it
                min_total_distance = total_distance
                best_location = current_location

    return best_location

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Big(O) Analysis

Time Complexity

O(m*n) – The brute force solution iterates through every possible location within a defined area. Assume 'm' represents the number of possible locations considered within this area. For each of these 'm' locations, we calculate the Manhattan distance to each of the 'n' destination points. Therefore, the total number of distance calculations is proportional to m*n. This results in a time complexity of O(m*n).

Space Complexity

O(1) – The brute force approach, as described, stores only the current minimum total travel distance and the corresponding location. These are constant-size variables. The amount of space required doesn't grow with the number of destinations; it's fixed, regardless of how many locations we check. Therefore, the auxiliary space complexity is O(1).

Optimal Solution

Approach

The best location to minimize the sum of Manhattan distances to all given points is the median point. To find this, we consider the x and y coordinates separately, finding the median of each, and combining them to get the optimal location.

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

Consider the x-coordinates of all the points. Find the middle x-coordinate value. This is the median.
Now, do the same for the y-coordinates. Find the middle y-coordinate value.
The point formed by the median x-coordinate and the median y-coordinate is the location that minimizes the total Manhattan distance.

Code Implementation

def minimize_manhattan_distance(points):
    x_coordinates = [point[0] for point in points]
    y_coordinates = [point[1] for point in points]

    x_coordinates.sort()
    y_coordinates.sort()

    # Need the median x to minimize the total distance on the x-axis
    if len(x_coordinates) % 2 == 0:
        median_x = (x_coordinates[len(x_coordinates) // 2 - 1] + x_coordinates[len(x_coordinates) // 2]) / 2
    else:
        median_x = x_coordinates[len(x_coordinates) // 2]

    # Need the median y to minimize the total distance on the y-axis
    if len(y_coordinates) % 2 == 0:
        median_y = (y_coordinates[len(y_coordinates) // 2 - 1] + y_coordinates[len(y_coordinates) // 2]) / 2
    else:
        median_y = y_coordinates[len(y_coordinates) // 2]

    optimal_location = (median_x, median_y)
    return optimal_location

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Big(O) Analysis

Time Complexity

O(n log n) – The dominant operations are finding the median x and y coordinates. To find the median, we need to sort the x-coordinates and y-coordinates separately. Sorting n elements typically takes O(n log n) time using efficient algorithms like merge sort or quicksort. Since we perform sorting twice (once for x and once for y), the total time complexity remains O(n log n) + O(n log n), which simplifies to O(n log n).

Space Complexity

O(1) – The solution involves finding the median of x-coordinates and y-coordinates separately. Although not explicitly stated, to find the median, sorting the x and y coordinates is required. If in-place sorting algorithms like heapsort are utilized, then no additional space is required to hold the sorted x and y coordinates. Only a few variables are needed to store the median x and y values regardless of the number of input points N, hence the space complexity is constant.

Edge Cases

Case	How to Handle
Empty input list of coordinates	Return 0, as no distance needs to be minimized when there are no points.
List contains only one coordinate	Return 0, as no pair exists for calculating distance.
All coordinates are identical	The median will be any of these coordinates, resulting in minimal (zero) distance.
Coordinates with large integer values causing potential overflow during distance calculation	Use appropriate data types (e.g., long) to prevent integer overflow when calculating distances or consider modular arithmetic if appropriate.
Coordinates with negative values	The Manhattan distance calculation handles negative values correctly as it uses absolute differences.
Skewed distribution: one coordinate far from all others	Median calculation ensures the selected coordinate minimizes the total distance despite the outlier.
A very large number of coordinates (scalability)	Ensure sorting algorithms used for median calculation are efficient (e.g., O(n log n)) or use a linear-time median finding algorithm to maintain scalability.
Duplicate coordinates present in the input list	The median calculation will still correctly identify a minimizing point, irrespective of coordinate duplicates.