363. Max Sum of Rectangle No Larger Than K - jiejackyzhang/leetcode-note GitHub Wiki

Given a non-empty 2D matrix matrix and an integer k, find the max sum of a rectangle in the matrix such that its sum is no larger than k.

Example:

Given matrix = [
  [1,  0, 1],
  [0, -2, 3]
]
k = 2

The answer is 2. Because the sum of rectangle [[0, 1], [-2, 3]] is 2 and 2 is the max number no larger than k (k = 2).

Note:

  1. The rectangle inside the matrix must have an area > 0.
  2. What if the number of rows is much larger than the number of columns?

解题思路为：

  • 考虑1D array的情形，我们可以将0~i(0<=i<len)的和保存起来，然后对于每个index，可用binary search找前面的index,两者之差不大于k。
  • 转换到2D matrix的情形，我们可以sum up all values from row i to row j，用一个1D array保存这些和，然后可以用1D array的方法来解。

外部两个循环为O(row^2)，内部1D解法采用binary search，为O((col)log(col))，因此总的时间复杂度为O(row^2*(col)log(col))。

若row比col大，则我们可以取min(row, col)为外部循环。 令m = min(row, col)，n = max(row, col)，则总的时间复杂度为O(m^2*nlogn)。

public class Solution {
    public int maxSumSubmatrix(int[][] matrix, int k) {
        if(matrix == null || matrix.length == 0 || matrix[0] == null || matrix[0].length == 0) return 0;
        int row = matrix.length;
        int col = matrix[0].length;
        int m = Math.min(row, col);
        int n = Math.max(row, col);
        boolean colIsLarger = col > row;
        int res = Integer.MIN_VALUE;
        for(int i = 0; i < m; i++) {
            int[] arr = new int[n];
            // sum from row j to row i
            for(int j = i; j >= 0; j--) {
                int val = 0;
                TreeSet<Integer> set = new TreeSet<>();
                set.add(0); // res can start from index 0
                for(int t = 0; t < n; t++) {
                    arr[t] = arr[t] + (colIsLarger ? matrix[j][t] : matrix[t][j]);
                    // sum from 0 to t
                    val = val + arr[t];
                    Integer subres = set.ceiling(val - k);
                    if(subres != null) res = Math.max(res, val - subres);
                    set.add(val);
                }
            }
        }
        return res;
    }
}
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