Submatrix Sum

Description

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.

If there are multiple answers, you can return any of them.

Example

Example 1:

Input:
[
  [1, 5, 7],
  [3, 7, -8],
  [4, -8 ,9]
]
Output: [[1, 1], [2, 2]]

Example 2:

Input:
[
  [0, 1],
  [1, 0]
]
Output: [[0, 0], [0, 0]]

Challenge

O(n3) time.

思路：

用前缀和优化, 令 sum[i][j] = sum[0][j] + sum[1][j] + ... + sum[i][j]

然后枚举上下边界, 这样就相当于在一行内, 求一个数组连续子串和为0的问题了.

public class Solution {
    /**
     * @param matrix an integer matrix
     * @return the coordinate of the left-up and right-down number
     */
    public int[][] submatrixSum(int[][] matrix) {
        int[][] result = new int[2][2];
        int M = matrix.length;
        if (M == 0) return result;
        int N = matrix[0].length;
        if (N == 0) return result;
        // pre-compute: sum[i][j] = sum of submatrix [(0, 0), (i, j)]
        int[][] sum = new int[M+1][N+1];
        for (int j = 0; j <= N; ++j) 
            sum[0][j] = 0;
        for (int i = 1; i <= M; ++i) 
            sum[i][0] = 0;
        for (int i = 0; i < M; ++i)
            for (int j = 0; j < N; ++j)
                sum[i+1][j+1] = matrix[i][j] + sum[i+1][j] + sum[i][j+1] - sum[i][j];
                
        for (int l = 0; l < M; ++l) {
            for (int h = l+1; h <= M; ++h) {
                Map<Integer, Integer> map = new HashMap<Integer, Integer>();
                for (int j = 0; j <= N; ++j) {
                    int diff = sum[h][j] - sum[l][j];
                    if (map.containsKey(diff)) {
                        int k = map.get(diff);
                        result[0][0] = l;   result[0][1] = k;
                        result[1][0] = h-1; result[1][1] = j-1;
                        return result;
                    } else {
                        map.put(diff, j);
                    }
                }
            }
        }
        return result;
    }
}