Problem

Given an integer matrix, find a submatrix where the sum of numbers is zero. Your code should return the coordinate of the left-up and right-down number.

Example

Given matrix

[
  [1 ,5 ,7],
  [3 ,7 ,-8],
  [4 ,-8 ,9],
]

return [(1,1), (2,2)]

Challenge

O(n3) time.

Note

原理是这样的：
先对矩阵matrix的每个点到左顶点之间的子矩阵求和，存在新矩阵sum上。注意，sum[i+1][j+1]代表的是matrix[0][0]到matrix[i][j]的子矩阵求和。如下所示：

Given matrix[m][n]
------------------
[
  [1 ,5 ,7],
  [3 ,7 ,-8],
  [4 ,-8 ,9],
]
Get sum[m+1][n+1]
-----------------
0  0  0  0
0  1  6  13
0  4  16 15
0  8  12 20

然后，我们进行这样的操作，从0开始，确定两行i1、i2，再将第k列的sum[i1][k]和sum[i2][k]相减，就得到matrix[i1][0]到matrix[i2][k-1]的子矩阵（i2-i1行，k列）求和diff，存入map。还是在这两行，假设计算到第j列，得到了相等的和diff。说明从i1到i2行，从k到j列的子矩阵求和为0，即相当于两个平行放置的矩形，若左边的值为diff，左边与右边之和也是diff，那么右边的值一定为0。下面是map中存放操作的例子：

diff-j chart
------------
diff    j

1       1       i1 = 0, i2 = 1
6       2
13      3
4       1       i1 = 0, i2 = 2
16      2
15      3
8       1       i1 = 0, i2 = 3
12      2
20      3
3       1       i1 = 1, i2 = 2
10      2
2       3
------------------------------
(above res has no same pair in same region)
7       1       i1 = 1, i2 = 3
6       2       
7       3       diff-j pair exists in map

res[0][0] = i1 = 1, res[0][1] = map.get(diff) = 1,
res[1][0] = i2 - 1 = 3 - 1 = 2, res[1][1] = j = 2,

so res = [(1, 1), (2, 2)]

Solution

public class Solution {
    public int[][] submatrixSum(int[][] matrix) {
        int m = matrix.length, n = matrix[0].length;
        int[][] sum = new int[m+1][n+1];
        int[][] res = new int[2][2];
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                sum[i+1][j+1] = matrix[i][j] + sum[i][j+1] + sum[i+1][j] - sum[i][j];
            }
        }
        for (int i1 = 0; i1 < m; i1++) {
            for (int i2 = i1+1; i2 <= m; i2++) {
                Map<Integer, Integer> map = new HashMap<Integer, Integer>();
                for (int j = 0; j <= n; j++) {
                    int diff = sum[i2][j] - sum[i1][j];
                    if (map.containsKey(diff)) {
                        res[0][0] = i1; res[0][1] = map.get(diff);
                        res[1][0] = i2-1; res[1][1] = j-1;
                        return res;
                    }
                    else map.put(diff, j);
                }
            }
        }
        return res;
    }
}