# 363. Max Sum of Rectangle No Larger Than K

### Problem:

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?

### Solutions:

``````public class Solution {
public int maxSumSubmatrix(int[][] matrix, int k) {
if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
return 0;
}
int[][] maxs = new int[matrix.length][matrix[0].length];

init(maxs, matrix, k);
for (int i = 1; i < matrix.length; i ++) {
for (int j = 1; j < matrix[0].length; j ++) {
maxs[i][j] = maxs[i - 1][j] + maxs[i][j - 1] - maxs[i - 1][j - 1] + matrix[i][j];
}
}
int max = Integer.MIN_VALUE;
for (int i = 0; i < maxs.length; i ++) {
for (int j = 0; j < maxs[0].length; j ++) {
for (int m = i; m < maxs.length; m ++) {
for (int n = j; n < maxs[0].length; n ++) {
int tmp = maxs[m][n];
if (i - 1 >= 0) {
tmp -= maxs[i - 1][n];
}
if (j - 1 >= 0) {
tmp -= maxs[m][j - 1];
}
if (i - 1 >= 0 && j - 1 >= 0) {
tmp += maxs[i - 1][j - 1];
}
if (tmp <= k) {
max = Math.max(max, tmp);
}
}
}
}
}
return max;
}
private void init(int[][] maxs, int[][] matrix, int k) {
for (int i = 0; i < matrix.length; i ++) {
if (i == 0) {
maxs[i][0] = matrix[i][0];
}
else {
maxs[i][0] = maxs[i - 1][0] + matrix[i][0];
}
}
for (int j = 0; j < matrix[0].length; j ++) {
if (j == 0) {
maxs[0][j] = matrix[0][j];
}
else {
maxs[0][j] = maxs[0][j - 1] + matrix[0][j];
}
}
}
}
``````

### Solutions:

``````public class Solution {
public int maxSumSubmatrix(int[][] matrix, int k) {
if(matrix==null||matrix.length==0||matrix[0].length==0)
return 0;

int m=matrix.length;
int n=matrix[0].length;

int result = Integer.MIN_VALUE;

for(int c1=0; c1<n; c1++){
int[] each = new int[m];
for(int c2=c1; c2>=0; c2--){

for(int r=0; r<m; r++){
each[r]+=matrix[r][c2];
}

result = Math.max(result, getLargestSumCloseToK(each, k));
}
}

return result;
}

public int getLargestSumCloseToK(int[] arr, int k){
int sum=0;
TreeSet<Integer> set = new TreeSet<Integer>();
int result=Integer.MIN_VALUE;

for(int i=0; i<arr.length; i++){
sum=sum+arr[i];

Integer ceiling = set.ceiling(sum-k);
if(ceiling!=null){
result = Math.max(result, sum-ceiling);
}