363.Max-Sum-of-Rectangle-No-Larger-Than-K
363. Max Sum of Rectangle No Larger Than K
题目地址
https://leetcode.com/problems/max-sum-of-rectangle-no-larger-than-k/
题目描述
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:
Input: matrix = [[1,0,1],[0,-2,3]], k = 2
Output: 2
Explanation: 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?
代码
Approach #1 Brute Force
Time: O(n^4) && Space: O(n)
class Solution {
public int maxSumSubmatrix(int[][] matrix, int k) {
if (matrix == null || matrix.length == 0 || matrix[0] == 0) return 0;
int rows = matrix.length;
int cols = matrix[0].length;
int[][] areas = new int[rows][cols];
for (int r = 0; r < rows; r++) {
for (int c = 0; c < cols; c++) {
int area = matrix[r][c];
if (r - 1 >= 0) area += areas[r-1][c];
if (c - 1 >= 0) area += areas[r][c - 1];
if (r-1 >= 0 && c - 1 >= 0) area -= areas[r-1][c-1];
areas[r][c] = area;
}
}
int max = Integer.MIN_VALUE;
for (int r1 = 0; r1 < rows; r1++) {
for (int c1 = 0; c1 < cols; c1++) {
for (int r2 = r1; r2 < rows; r2++) {
for (int c2 = c1; c2 < cols; c2++) {
int area = areas[r2][c2];
if (r1 - 1 >= 0) area -= areas[r1-1][c2];
if (c1 - 1 >= 0) area -= areas[r2][c1-1];
if (r1 - 1 >= 0 && c1 - 1 >=0) area += areas[r1-1][c1-1];
if (area <= k) max = Math.max(max, area);
}
}
}
}
return max;
}
}
Approach #2
O(N^3 logn)
class Solution {
public int maxSumSubmatrix(int[][] matrix, int k) {
if (matrix == null || matrix.length == 0 || matrix[0] == 0) return 0;
int rows = matrix.length;
int cols = matrix[0].length;
int[][] areas = new int[rows][cols];
for (int r = 0; r < rows; r++) {
for (int c = 0; c < cols; c++) {
int area = matrix[r][c];
if (r-1 >= 0) area += areas[r-1][c];
if (c-1 >= 0) area += areas[r][c-1];
if (r-1 >= 0 && c-1 >= 0) area -= areas[r-1][c-1];
areas[r][c] = area;
}
}
int max = Integer.MIN_VALUE;
for (int r1 = 0; r1 < rows; r1++) {
for (int r2 = r1; r2 < rows; r2++) {
TreeSet<Integer> tree = new TreeSet<>();
tree.add(0);
for (int c = 0; c < cols; c++) {
int area = areas[r2][c];
if (r1-1 >= 0) area -= areas[r1-1][c];
Integer ceiling = tree.ceiling(area - k);
if (ceiling != null) {
max = Math.max(max, area - ceiling);
}
tree.add(area);
}
}
}
return max;
}
}
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