# 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)

```java
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)

```java
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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