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Best Meeting Point

Hard

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Topics:

ArraysGreedy Algorithms

A group of two or more people wants to meet and minimize the total travel distance. You are given a 2D grid of values 0 or 1, where each 1 marks the home location of one of the people. The distance is calculated using Manhattan Distance, where distance(p1, p2) = |p2.x - p1.x| + |p2.y - p1.y|.

For example, given three people living at (0,0), (0,4), and (2,2):

1 - 0 - 0 - 0 - 1
|
0 - 0 - 0 - 0 - 0
|
0 - 0 - 1 - 0 - 0

The point (0,2) is an ideal meeting point, as the total travel distance of 2 + 2 + 2 = 6 is minimal. So return 6.

Example:

Input: grid = [[1,0,0,0,1],[0,0,0,0,0],[0,0,1,0,0]]
Output: 6
Explanation: Return the min distance 2 + 2 + 2 = 6

Constraints:

m == grid.length
n == grid[i].length
1 <= m, n <= 200
grid[i][j] is either 0 or 1.
There will be at least two people in the grid.

Try coding it in LeetCode

Best Meeting Point

Hard

Asked by:

52 views

Topics:

ArraysGreedy Algorithms

For example, given three people living at (0,0), (0,4), and (2,2):

1 - 0 - 0 - 0 - 1
|
0 - 0 - 0 - 0 - 0
|
0 - 0 - 1 - 0 - 0

The point (0,2) is an ideal meeting point, as the total travel distance of 2 + 2 + 2 = 6 is minimal. So return 6.

Example:

Input: grid = [[1,0,0,0,1],[0,0,0,0,0],[0,0,1,0,0]]
Output: 6
Explanation: Return the min distance 2 + 2 + 2 = 6

Constraints:

m == grid.length
n == grid[i].length
1 <= m, n <= 200
grid[i][j] is either 0 or 1.
There will be at least two people in the grid.

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:

What are the constraints on the dimensions m and n of the grid? What is the maximum value for m and n?
Can the grid be empty (i.e., contain no homes represented by 1s)? If so, what should the return value be?
Are the grid cells guaranteed to only contain 0s and 1s, or could there be other values?
If there are multiple meeting points that result in the same minimal total Manhattan distance, is any one of them acceptable or is there a preferred meeting point (e.g., lexicographically smallest coordinates)?
Is the grid guaranteed to be rectangular (i.e., all rows have the same number of columns)?

Brute Force Solution

Approach

The brute force approach to finding the best meeting point involves checking every possible location. For each location, we calculate the total distance everyone has to travel to meet there, and then we choose the location with the smallest total distance.

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

Consider every single possible location as a potential meeting point.
For each of these potential meeting points, calculate the distance from that point to each person's location.
Add up all the distances to get the total travel distance for that potential meeting point.
Keep track of the location with the smallest total travel distance seen so far.
After considering every possible location, choose the location that resulted in the smallest total travel distance. That is the best meeting point.

Code Implementation

def best_meeting_point_brute_force(person_locations):
    # Initialize the best meeting point to None and the minimum total distance to infinity.
    best_meeting_point = None
    minimum_total_distance = float('inf')

    # Find the range of possible X and Y coordinates.
    x_coordinates = [location[0] for location in person_locations]
    y_coordinates = [location[1] for location in person_locations]
    min_x = min(x_coordinates)
    max_x = max(x_coordinates)
    min_y = min(y_coordinates)
    max_y = max(y_coordinates)

    # Iterate through all possible locations within the grid.
    for potential_x in range(min_x, max_x + 1):
        for potential_y in range(min_y, max_y + 1):
            total_distance = 0

            # Calculate the total distance from the potential meeting point to each person.
            for person_x, person_y in person_locations:
                total_distance += abs(potential_x - person_x) + abs(potential_y - person_y)

            # Update the best meeting point if the current total distance is smaller.
            if total_distance < minimum_total_distance:
                minimum_total_distance = total_distance
                best_meeting_point = (potential_x, potential_y)

    return best_meeting_point

Big(O) Analysis

Time Complexity

O(n*m) – Let n be the number of rows and m be the number of columns in the grid. The brute force approach iterates through every cell in the grid as a potential meeting point. For each potential meeting point (of which there are n*m), we calculate the Manhattan distance to each person's location. Calculating the sum of distances to all people takes constant time since the number of people is not related to the input size. Therefore, we perform a constant time operation n*m times. Thus, the overall time complexity is O(n*m).

Space Complexity

O(1) – The brute force approach described only needs to store a few variables: one to track the current potential meeting point, another to accumulate the total distance for that meeting point, and one to store the minimum total distance found so far along with the corresponding best meeting point. The number of these variables remains constant regardless of the number of possible locations or people involved. Therefore, the auxiliary space used is constant, independent of the input size N, resulting in O(1) space complexity.

Optimal Solution

Approach

The key is to realize the best meeting point will minimize the total distance everyone travels. This can be found by independently finding the optimal location along each axis (horizontal and vertical) and combining them. Finding the optimal location along a single axis involves finding the median of the locations.

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

Separate the locations into their horizontal (x) and vertical (y) coordinates.
Sort the horizontal coordinates.
Sort the vertical coordinates.
Find the middle value of the sorted horizontal coordinates. If there's an even number of coordinates, take the average of the two middle values.
Find the middle value of the sorted vertical coordinates. If there's an even number of coordinates, take the average of the two middle values.
The best meeting point is at the location defined by the horizontal and vertical medians you just found.

Code Implementation

def find_best_meeting_point(locations):
    horizontal_coordinates = []
    vertical_coordinates = []
    for location in locations:
        horizontal_coordinates.append(location[0])
        vertical_coordinates.append(location[1])

    horizontal_coordinates.sort()
    vertical_coordinates.sort()

    # Finding median minimizes total distance along x-axis
    if len(horizontal_coordinates) % 2 == 0:
        median_horizontal = (
            horizontal_coordinates[len(horizontal_coordinates) // 2 - 1]
            + horizontal_coordinates[len(horizontal_coordinates) // 2]
        ) / 2
    else:
        median_horizontal = horizontal_coordinates[len(horizontal_coordinates) // 2]

    # Finding median minimizes total distance along y-axis
    if len(vertical_coordinates) % 2 == 0:
        median_vertical = (
            vertical_coordinates[len(vertical_coordinates) // 2 - 1]
            + vertical_coordinates[len(vertical_coordinates) // 2]
        ) / 2
    else:
        median_vertical = vertical_coordinates[len(vertical_coordinates) // 2]

    best_meeting_point_x = median_horizontal
    best_meeting_point_y = median_vertical

    total_distance = 0
    for location in locations:
        total_distance += abs(location[0] - best_meeting_point_x) + abs(location[1] - best_meeting_point_y)

    return int(total_distance)

Big(O) Analysis

Time Complexity

O(n log n) – The time complexity is dominated by the sorting steps. Separating the coordinates takes O(n) time, where n is the number of locations. Sorting the horizontal coordinates takes O(n log n) time, and sorting the vertical coordinates also takes O(n log n) time using an efficient sorting algorithm like merge sort or quicksort. Finding the median values then takes O(1) time. Therefore, the overall time complexity is O(n) + O(n log n) + O(n log n) + O(1), which simplifies to O(n log n).

Space Complexity

O(N) – The algorithm creates two lists to store the horizontal and vertical coordinates, each potentially containing all N coordinates from the input. Sorting these lists is typically done in-place or with minimal extra space depending on the sorting algorithm, but the lists themselves require O(N) auxiliary space. Therefore, the dominant factor in space complexity is the storage of the coordinate lists.

Edge Cases

Case	How to Handle
Empty grid (m=0 or n=0)	Return 0 immediately as there are no homes to meet.
Grid with no homes (all cells are 0)	Return 0 immediately as there are no homes to meet.
Grid with only one home	Return 0 immediately as the meeting point is the home itself.
Large grid dimensions (m and n are very large)	The median approach scales efficiently, but ensure no integer overflow occurs when calculating Manhattan distances.
Homes clustered in one corner of the grid	The median-finding approach correctly identifies the optimal meeting point in this skewed distribution.
Homes distributed evenly across the grid	The median-finding approach correctly identifies the optimal meeting point in this even distribution.
Integer overflow when calculating Manhattan distance.	Use long data type for intermediate calculations to prevent overflow.
Multiple potential meeting points yield the same minimal distance	The median guarantees only one optimal meeting point but several points may yield the same distance; return the first one that satisfies the condition.

Empty grid (m=0 or n=0)

How to Handle:

Return 0 immediately as there are no homes to meet.

Grid with no homes (all cells are 0)

How to Handle:

Return 0 immediately as there are no homes to meet.

Grid with only one home

How to Handle:

Return 0 immediately as the meeting point is the home itself.

Large grid dimensions (m and n are very large)

How to Handle:

The median approach scales efficiently, but ensure no integer overflow occurs when calculating Manhattan distances.

Homes clustered in one corner of the grid

How to Handle:

The median-finding approach correctly identifies the optimal meeting point in this skewed distribution.

Homes distributed evenly across the grid

How to Handle:

The median-finding approach correctly identifies the optimal meeting point in this even distribution.

Integer overflow when calculating Manhattan distance.

How to Handle:

Use long data type for intermediate calculations to prevent overflow.

Multiple potential meeting points yield the same minimal distance

How to Handle:

The median guarantees only one optimal meeting point but several points may yield the same distance; return the first one that satisfies the condition.

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 are the constraints on the dimensions m and n of the grid? What is the maximum value for m and n?
Can the grid be empty (i.e., contain no homes represented by 1s)? If so, what should the return value be?
Are the grid cells guaranteed to only contain 0s and 1s, or could there be other values?
If there are multiple meeting points that result in the same minimal total Manhattan distance, is any one of them acceptable or is there a preferred meeting point (e.g., lexicographically smallest coordinates)?
Is the grid guaranteed to be rectangular (i.e., all rows have the same number of columns)?

Brute Force Solution

Approach

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

Consider every single possible location as a potential meeting point.
For each of these potential meeting points, calculate the distance from that point to each person's location.
Add up all the distances to get the total travel distance for that potential meeting point.
Keep track of the location with the smallest total travel distance seen so far.
After considering every possible location, choose the location that resulted in the smallest total travel distance. That is the best meeting point.

Code Implementation

def best_meeting_point_brute_force(person_locations):
    # Initialize the best meeting point to None and the minimum total distance to infinity.
    best_meeting_point = None
    minimum_total_distance = float('inf')

    # Find the range of possible X and Y coordinates.
    x_coordinates = [location[0] for location in person_locations]
    y_coordinates = [location[1] for location in person_locations]
    min_x = min(x_coordinates)
    max_x = max(x_coordinates)
    min_y = min(y_coordinates)
    max_y = max(y_coordinates)

    # Iterate through all possible locations within the grid.
    for potential_x in range(min_x, max_x + 1):
        for potential_y in range(min_y, max_y + 1):
            total_distance = 0

            # Calculate the total distance from the potential meeting point to each person.
            for person_x, person_y in person_locations:
                total_distance += abs(potential_x - person_x) + abs(potential_y - person_y)

            # Update the best meeting point if the current total distance is smaller.
            if total_distance < minimum_total_distance:
                minimum_total_distance = total_distance
                best_meeting_point = (potential_x, potential_y)

    return best_meeting_point

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Separate the locations into their horizontal (x) and vertical (y) coordinates.
Sort the horizontal coordinates.
Sort the vertical coordinates.
Find the middle value of the sorted horizontal coordinates. If there's an even number of coordinates, take the average of the two middle values.
Find the middle value of the sorted vertical coordinates. If there's an even number of coordinates, take the average of the two middle values.
The best meeting point is at the location defined by the horizontal and vertical medians you just found.

Code Implementation

def find_best_meeting_point(locations):
    horizontal_coordinates = []
    vertical_coordinates = []
    for location in locations:
        horizontal_coordinates.append(location[0])
        vertical_coordinates.append(location[1])

    horizontal_coordinates.sort()
    vertical_coordinates.sort()

    # Finding median minimizes total distance along x-axis
    if len(horizontal_coordinates) % 2 == 0:
        median_horizontal = (
            horizontal_coordinates[len(horizontal_coordinates) // 2 - 1]
            + horizontal_coordinates[len(horizontal_coordinates) // 2]
        ) / 2
    else:
        median_horizontal = horizontal_coordinates[len(horizontal_coordinates) // 2]

    # Finding median minimizes total distance along y-axis
    if len(vertical_coordinates) % 2 == 0:
        median_vertical = (
            vertical_coordinates[len(vertical_coordinates) // 2 - 1]
            + vertical_coordinates[len(vertical_coordinates) // 2]
        ) / 2
    else:
        median_vertical = vertical_coordinates[len(vertical_coordinates) // 2]

    best_meeting_point_x = median_horizontal
    best_meeting_point_y = median_vertical

    total_distance = 0
    for location in locations:
        total_distance += abs(location[0] - best_meeting_point_x) + abs(location[1] - best_meeting_point_y)

    return int(total_distance)

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty grid (m=0 or n=0)	Return 0 immediately as there are no homes to meet.
Grid with no homes (all cells are 0)	Return 0 immediately as there are no homes to meet.
Grid with only one home	Return 0 immediately as the meeting point is the home itself.
Large grid dimensions (m and n are very large)	The median approach scales efficiently, but ensure no integer overflow occurs when calculating Manhattan distances.
Homes clustered in one corner of the grid	The median-finding approach correctly identifies the optimal meeting point in this skewed distribution.
Homes distributed evenly across the grid	The median-finding approach correctly identifies the optimal meeting point in this even distribution.
Integer overflow when calculating Manhattan distance.	Use long data type for intermediate calculations to prevent overflow.
Multiple potential meeting points yield the same minimal distance	The median guarantees only one optimal meeting point but several points may yield the same distance; return the first one that satisfies the condition.

Empty grid (m=0 or n=0)

How to Handle:

Return 0 immediately as there are no homes to meet.

Grid with no homes (all cells are 0)

How to Handle:

Return 0 immediately as there are no homes to meet.

Grid with only one home

How to Handle:

Return 0 immediately as the meeting point is the home itself.

Large grid dimensions (m and n are very large)

How to Handle:

The median approach scales efficiently, but ensure no integer overflow occurs when calculating Manhattan distances.

Homes clustered in one corner of the grid

How to Handle:

The median-finding approach correctly identifies the optimal meeting point in this skewed distribution.

Homes distributed evenly across the grid

How to Handle:

The median-finding approach correctly identifies the optimal meeting point in this even distribution.

Integer overflow when calculating Manhattan distance.

How to Handle:

Use long data type for intermediate calculations to prevent overflow.

Multiple potential meeting points yield the same minimal distance

How to Handle:

The median guarantees only one optimal meeting point but several points may yield the same distance; return the first one that satisfies the condition.