LeetCode – Best Meeting Point (Java)

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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.

Java Solution

This problem is converted to find the median value on x-axis and y-axis.

public int minTotalDistance(int[][] grid) {
    int m=grid.length;
    int n=grid[0].length;
 
    ArrayList<Integer> cols = new ArrayList<Integer>();
    ArrayList<Integer> rows = new ArrayList<Integer>();
    for(int i=0; i<m; i++){
        for(int j=0; j<n; j++){
           if(grid[i][j]==1){
               cols.add(j);
               rows.add(i);
           }
        }
    }
 
    int sum=0;
 
    for(Integer i: rows){
        sum += Math.abs(i - rows.get(rows.size()/2));    
    }
 
    Collections.sort(cols);
 
    for(Integer i: cols){
        sum+= Math.abs(i-cols.get(cols.size()/2));
    }
 
    return sum;
}

3 thoughts on “LeetCode – Best Meeting Point (Java)”

  2. weighted average – (start with index 1 to avoid multiplying by zero)
    x = 1/3(1) + 1/3(1) + 1/3(3) = 5/3 = 1.6666 = 2
    y = 1/3(1) + 1/3(3) + 1/3(5) = 3

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