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:
The rectangle inside the matrix must have an area > 0.
What if the number of rows is much larger than the number of columns?
Solution：
2d累计遍历 + Binary Search via hashset (TreeSet)
思路：
Time Complexity: O([min(m,n)^2max(m,n)log(max(m,n))]
) Space Complexity: O()
Solution Code：
class Solution {
public int maxSumSubmatrix(int[][] matrix, int k) {
int m = matrix.length, n = 0;
if (m > 0) n = matrix[0].length;
if (m * n == 0) return 0;
int M = Math.max(m, n);
int N = Math.min(m, n);
int ans = Integer.MIN_VALUE;
for (int x = 0; x < N; x++) {
int sums[] = new int[M];
for (int y = x; y < N; y++) {
TreeSet<Integer> set = new TreeSet<Integer>();
int curSum = 0;
for (int z = 0; z < M; z++) {
sums[z] += m > n ? matrix[z][y] : matrix[y][z];
curSum += sums[z];
if (curSum <= k) ans = Math.max(ans, curSum);
Integer i = set.ceiling(curSum - k);
if (i != null) ans = Math.max(ans, curSum - i);
set.add(curSum);
}
}
}
return ans;
}
}
