Best Meeting Point

Problem Statement

The problem presents a binary grid where each cell contains either a 0 or a 1. Each '1' represents the home of a friend. The task is to determine the minimal total travel distance for all friends to converge at a single meeting point. This distance should be calculated using the Manhattan Distance formula. Specifically, to find the optimal meeting point that minimizes the sum of individual distances each friend must travel from their home to this point.

Examples

Example 1

Input:

grid = [[1,0,0,0,1],[0,0,0,0,0],[0,0,1,0,0]]

Output:

6

Explanation:

Given three friends living at (0,0), (0,4), and (2,2).
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 2

Input:

grid = [[1,1]]

Output:

1

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 friends in the grid.

Approach and Intuition

The solution revolves around the concept of the Manhattan Distance and finding a meeting point that minimizes this distance sum for all friends. The intuition behind this solution can be best understood by breaking it down step by step:

1. Identify Presence: First, identify all the positions (homes) of friends in the grid marked by '1'.

2. Calculate Median: The Manhattan Distance is minimized when the meeting point is at the median of all the friends' positions. This is due to the nature of absolute value functions, which reach their minimum sum at the median. Hence, separate the positions into their respective x and y coordinates.

   • Sort these x-coordinates and y-coordinates individually.
   • If the count of friends (number of '1's) is odd, the median is the middle value after sorting. If even, any position between the two central points can be considered.

3. Compute Total Distance: Use the median x and y coordinates to determine the optimal meeting point. Sum the Manhattan distances from each friend's home to this point. This can be computed using:

   [ \text{Distance} = |x_{\text{friend}} - x_{\text{median}}| + |y_{\text{friend}} - y_{\text{median}}| ]

   where (x_{\text{friend}}) and (y_{\text{friend}}) are the x and y coordinates of a friend's home respectively, and (x_{\text{median}}) and (y_{\text{median}}) are the median coordinates of all homes.

The use of median reduces the problem significantly by not requiring a check against all possible meeting points, but focusing on the statistically optimal location. This method takes advantage of properties of distances in metric spaces, specifically Manhattan distances applied to a grid system.

With the constraints in place:

• The approach needs to handle grids as large as 200x200, making direct computation feasible.
• Only two entries are required at a minimum, making the problem always solvable.

This systematic approach leverages statistical properties for a computational solution, substantiated by the nature of Manhattan distance and its optimization using median values.

Solutions

public int calculateMinimumDistance(int[][] matrix) {
    List<Integer> rowIndices = getRowIndices(matrix);
    List<Integer> colIndices = getColumnIndices(matrix);
    return calculate1DDistance(rowIndices) + calculate1DDistance(colIndices);
}
    
private int calculate1DDistance(List<Integer> coordinates) {
    int totalDistance = 0;
    int start = 0;
    int end = coordinates.size() - 1;
    while (start < end) {
        totalDistance += coordinates.get(end) - coordinates.get(start);
        start++;
        end--;
    }
    return totalDistance;
}