LeetCode-Best Meeting Point

296. Best Meeting Point

 

 

 

• Total Accepted: 6496
• Total Submissions: 13313
• Difficulty: Hard

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.

Analysis:

https://discuss.leetcode.com/topic/51109/8ms-c-dp-solution-with-o-mn-time-o-m-n-space

Solution:

public class Solution {
    public int minTotalDistance(int[][] grid) {
        if (grid.length==0 || grid[0].length==0) return 0;
        
        int rows = grid.length, cols = grid[0].length;
        int[] xNums = new int[cols];
        int[] yNums = new int[rows];
        int initDis = 0; // Distance to (0,0)

        for (int i=0;i<rows;i++){
            int sum = 0;
            for (int j=0;j<cols;j++){
                sum += grid[i][j];
                xNums[j] +=sum;
                // Update init distance
                if (grid[i][j] == 1){
                    initDis += (i+j);
                }
            }
            yNums[i] = xNums[cols-1];
        }

        int totalNum = xNums[cols-1];
        // find out the optimal point on xNums.
        int minDis = initDis;
        for (int i=1;i<cols;i++){
            int curDis = minDis + xNums[i-1] - (totalNum - xNums[i-1]);
            if (curDis < minDis){
                minDis = curDis;
            } else break;
        }

        // find out the optimal point on yNums.
        for (int i=1;i<rows;i++){
            int curDis = minDis + yNums[i-1] - (totalNum -yNums[i-1]);
            if (curDis < minDis) {
                minDis = curDis;
            } else break;
        }
        return minDis; 
    }
}