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?

一刷
題解：
video link for the problem "find the max sum rectangle in 2D array": Maximum Sum Rectangular Submatrix in Matrix dynamic programming/2D kadane
根據這個啟發，我們得到l, r列之間的sum array, sum[i]表示第i行，l, r列之間的和。
然后對于每個array創建一個set, 從頭加到尾，并依次加入set中。如果對于當前的sum, 如果在set中存在一個num, 滿足sum-num<k, 表示我們找到了這個數, 為了快速尋找，我們采用treeset。update res

class Solution {
    public int maxSumSubmatrix(int[][] matrix, int k) {
    //2D Kadane's algorithm + 1D maxSum problem with sum limit k
    //2D subarray sum solution
    
    //boundary check
    if(matrix.length == 0) return 0;
    
    int m = matrix.length, n = matrix[0].length;
    int result = Integer.MIN_VALUE;
    
    //outer loop should use smaller axis
    //now we assume we have more rows than cols, therefore outer loop will be based on cols 
    for(int left = 0; left < n; left++){
        //array that accumulate sums for each row from left to right 
        int[] sums = new int[m];
        for(int right = left; right < n; right++){
            //update sums[] to include values in curr right col
            for(int i = 0; i < m; i++){
                sums[i] += matrix[i][right];
            }
            
            //we use TreeSet to help us find the rectangle with maxSum <= k with O(logN) time
            TreeSet<Integer> set = new TreeSet<Integer>();
            //add 0 to cover the single row case
            set.add(0);
            int currSum = 0;
            
            for(int sum : sums){
                currSum += sum;
                //we use sum subtraction (curSum - sum) to get the subarray with sum <= k
                //therefore we need to look for the smallest sum >= currSum - k
                Integer num = set.ceiling(currSum - k);
                if(num != null) result = Math.max( result, currSum - num );
                set.add(currSum);
            }
        }
    }
    
    return result;
}
}