296. Best Meeting Point

题目：

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.

Hint:

1. Try to solve it in one dimension first. How can this solution apply to the two dimension case?

链接： http://leetcode.com/problems/best-meeting-point/

题解：

很有意思的一道题目，假设二维数组中一个点到其他给定点的Manhattan Distance最小，求distance和。 因为在一维数组中这个distance最小的点就是给定所有点的median，题目又给定使用曼哈顿距离，我们就可以把二维计算分解成为两个一维的计算。应该还可以用DP的方法解决，判断用哪一种方法其实非常复杂，依赖于mn和排序的比较。我们使用一个getMin方法来计算x方向或者y方向到他们中点的距离和。

Time Complexity - Math.max(O(mn), O(nlogn))， Space Complexity - O(mn)。   

public class Solution {
    public int minTotalDistance(int[][] grid) {
        List<Integer> xAxis = new ArrayList<>();
        List<Integer> yAxis = new ArrayList<>();
        
        for(int i = 0; i < grid.length; i++) {
            for(int j = 0; j < grid[0].length; j++) {
                if(grid[i][j] == 1) {
                    xAxis.add(i);
                    yAxis.add(j);        
                }
            }
        }
        
        return getMin(xAxis) + getMin(yAxis);
    }
    
    private int getMin(List<Integer> list) {
        Collections.sort(list);
        int res = 0;
        int lo = 0, hi = list.size() - 1;
        while(lo < hi) {
            res += list.get(hi--) - list.get(lo++);   // hi - mid +  mid - lo = hi - lo
        } 
        return res;
    }
}

二刷:

还是用了简单地先遍历一遍数组，收集行坐标和列坐标，然后对两个list分别求一维Manhattan距离的方法。这里对列坐标list进行了排序。

Java:

Time Complexity - Math.max(O(mn), O(nlogn))， Space Complexity - O(mn)。   

public class Solution {
    public int minTotalDistance(int[][] grid) {
        List<Integer> rows = new ArrayList<>(), cols = new ArrayList<>();
        for (int i = 0; i < grid.length; i++) {
            for (int j = 0; j < grid[0].length; j++) {
                if (grid[i][j] == 1) {
                    rows.add(i);
                    cols.add(j);
                }
            }
        }
        Collections.sort(cols);
        return getMinDist(rows) + getMinDist(cols);
    }
    
    private int getMinDist(List<Integer> list) {
        if (list == null || list.size() == 0) return Integer.MAX_VALUE;
        int median = list.get(list.size() / 2);
        int minDist = 0;
        for (int idx : list) {
            if (idx < median) minDist += median - idx;
            else minDist += idx - median;
        }
        return minDist;
    }
}

Update:

遍历两次数组，分别对行列坐标进行收集，速度反而比较快。应该是不少test case中m < logn的缘故。

Time Complexity - O(mn), O(nlogn)， Space Complexity - O(mn)。   

public class Solution {
    public int minTotalDistance(int[][] grid) {
        List<Integer> rows = new ArrayList<>(), cols = new ArrayList<>();
        for (int i = 0; i < grid.length; i++) {
            for (int j = 0; j < grid[0].length; j++) {
                if (grid[i][j] == 1) rows.add(i);
            }
        }
        for (int j = 0; j < grid[0].length; j++) {
            for (int i = 0; i < grid.length; i++) {
                if (grid[i][j] == 1) cols.add(j);
            }
        }
        return getMinDist(rows) + getMinDist(cols);
    }
    
    private int getMinDist(List<Integer> list) {
        if (list == null || list.size() == 0) return Integer.MAX_VALUE;
        int median = list.get(list.size() / 2);
        int minDist = 0;
        for (int idx : list) {
            if (idx < median) minDist += median - idx;
            else minDist += idx - median;
        }
        return minDist;
    }
}

Update:

不计算median，利用median - lo + hi - median = hi - lo，同时计算lo和hi到median的距离。来自大神larrywang2014的写法。

public class Solution {
    public int minTotalDistance(int[][] grid) {
        List<Integer> rows = new ArrayList<>(), cols = new ArrayList<>();
        for (int i = 0; i < grid.length; i++) {
            for (int j = 0; j < grid[0].length; j++) {
                if (grid[i][j] == 1) rows.add(i);
            }
        }
        for (int j = 0; j < grid[0].length; j++) {
            for (int i = 0; i < grid.length; i++) {
                if (grid[i][j] == 1) cols.add(j);
            }
        }
        return getMinDist(rows) + getMinDist(cols);
    }
    
    private int getMinDist(List<Integer> list) {
        if (list == null || list.size() == 0) return Integer.MAX_VALUE;
        int minDist = 0;
        int lo = 0, hi = list.size() - 1;
        while (lo < hi) {
            minDist += list.get(hi--) - list.get(lo++);       //  median - lo + hi - median = hi - lo
        }
        return minDist;
    }
}

Reference:

https://leetcode.com/discuss/65336/14ms-java-solution

https://leetcode.com/discuss/65366/o-mn-java-2ms

https://leetcode.com/discuss/65464/java-python-40ms-pointers-solution-median-sort-explanation

https://leetcode.com/discuss/66401/the-only-person-dont-know-median-could-give-shortest-distance

http://math.stackexchange.com/questions/113270/the-median-minimizes-the-sum-of-absolute-deviations

https://leetcode.com/discuss/65510/simple-java-code-without-sorting

http://www.jiuzhang.com/problem/30/