296. Best Meeting Point

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 of someone in the group. 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.

Solution 1. Median in Math. Collect coordinates in sorted order.

public class Solution {
    public int minTotalDistance(int[][] grid) {
            int m = grid.length, n = grid[0].length;
        int total = 0, Z[] = new int[m*n];
        for (int dim=0; dim<2; ++dim) {
            int i = 0, j = 0;
            if (dim == 0) {
                for (int x=0; x<n; ++x)
                    for (int y=0; y<m; ++y)
                        if (grid[y][x] == 1)
                            Z[j++] = x;
            } else {
                for (int y=0; y<m; ++y)
                    for (int g : grid[y])
                        if (g == 1)
                            Z[j++] = y;
            }
            while (i < --j)
                total += Z[j] - Z[i++];
        }
        return total;
    }
}

Solution 2. Count how many people live in each row and each column. Only O(m+n) space.

public class Solution {
    public int minTotalDistance(int[][] grid) {
        int m = grid.length, n = grid[0].length;
        int[] I = new int[m], J = new int[n];
        for (int i=0; i<m; ++i)
            for (int j=0; j<n; ++j)
                if (grid[i][j] == 1) {
                    ++I[i];
                    ++J[j];
                }
        
        int total = 0;
        for (int[] K : new int[][]{ I, J }) {
            int i = 0, j = K.length - 1;
            while (i < j) {
                int k = Math.min(K[i], K[j]);
                total += k * (j - i);
                if ((K[i] -= k) == 0) ++i;
                if ((K[j] -= k) == 0) --j;
            }
        }
        return total;
    }
}