/* first consider the situation matrix is 1D
we can save every sum of 0~i(0<=i<len) and binary search previous sum to find
possible result for every index, time complexity is O(NlogN).
so in 2D matrix, we can sum up all values from row i to row j and create a 1D array
to use 1D array solution.
If col number is less than row number, we can sum up all values from col i to col j
then use 1D array solution.
*/
public int maxSumSubmatrix(int[][] matrix, int target) {
int row = matrix.length;
if(row==0)return 0;
int col = matrix[0].length;
int m = Math.min(row,col);
int n = Math.max(row,col);
//indicating sum up in every row or every column
boolean colIsBig = col>row;
int res = Integer.MIN_VALUE;
for(int i = 0;i<m;i++){
int[] array = 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<Integer>();
set.add(0);
//traverse every column/row and sum up
for(int k = 0;k<n;k++){
array[k]=array[k]+(colIsBig?matrix[j][k]:matrix[k][j]);
val = val + array[k];
//use TreeMap to binary search previous sum to get possible result
Integer subres = set.ceiling(valtarget);
if(null!=subres){
res=Math.max(res,valsubres);
}
set.add(val);
}
}
}
return res;
}
Java Binary Search solution time complexity min(m,n)^2*max(m,n)*log(max(m,n))



@wuzixigua said in Java Binary Search solution time complexity min(m,n)^2*max(m,n)*log(max(m,n)):
l = val + array[k];
Thanks for your idea.In a 1D matrix,why the time complex is O(nlogn)? I think the sum array is not sorted because of the negative number,so we can't do a binary search directly.


@futurehau Because the sum is added to a treeset， i think this step finished the sorting process.