Max Sum of Rectangle No Larger Than K

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?

思路：

　　使用l和r划定长方形的左右边界范围，然后在这个范围内，依次记录长方形的上界固定为第一行，下界从第一行到最后一行对应的长方形的和到数组sum。现在问题转换为寻找最合适的sum[j]-sum[i](j和i对应长方形的上下界)，使得该值不大于k，但是最接近k。这个问题可以从Quora上找到解答：

　　You can do this in $\mathrm{O(nlog(n))}$

$\mathrm{First\; thing\; to\; note\; is\; that\; sum\; of\; subarray (i,j] is\; just\; the\; sum\; of\; the\; first j elements\; less\; the\; sum\; of\; the\; first i elements.\; Store\; these\; cumulative\; sums\; in\; the\; array\; cum.\; Then\; the\; problem\; reduces\; to\; finding i,j such\; that i<j and cum[j]-cum[i] is\; as\; close\; to k but\; lower\; than\; it.}$

$\mathrm{To\; solve\; this,\; scan\; from\; left\; to\; right.\; Put\; the cum[i] values\; that\; you\; have\; encountered\; till\; now\; into\; a set.\; When\; you\; are\; processing cum[j] what\; you\; need\; to\; retrieve\; from\; the\; set\; is\; the\; smallest\; number\; in\; the\; set\; such\; which\; is\; not\; smaller\; than cum[j]-k.\; This\; lookup\; can\; be\; done\; in O(log(n)) using\; lower\_bound.\; Hence\; the\; overall\; complexity\; is O(nlog(n)).}$

$\mathrm{Here\; is\; a\; c++\; function\; that\; does\; the\; job,\; assuming\; that\; K>0\; and\; that\; the\; empty\; interval\; with\; sum\; zero\; is\; a\; valid\; answer.\; The\; code\; can\; be\; tweaked\; easily\; to\; take\; care\; of\; more\; general\; cases\; and\; to\; return\; the\; interval\; itself.}$

$对应代码：$

int best_cumulative_sum(int ar[],int N,int K)
{
    set<int> cumset;
    cumset.insert(0);
    int best=0,cum=0;
    for(int i=0;i<N;i++)
    {
        cum+=ar[i];
        set<int>::iterator sit=cumset.lower_bound(cum-K);
        if(sit!=cumset.end())best=max(best,cum-*sit);
        cumset.insert(cum);
    }
    return best;
}

　　在上述基础之上，我们稍加改变，就能够写出下述代码完成此题了。

class Solution {
public:
    int maxSumSubmatrix(vector<vector<int>> &matrix, int k) {
        int row = matrix.size();
        if (row == 0)
            return 0;
        int col = matrix[0].size();
        int ret = INT_MIN;
        for (int l = 0; l < col; l++) {
            vector<int> sums(row, 0);
            for (int r = l; r < col; r++) {
                for (int i = 0; i < row; i++)
                    sums[i] += matrix[i][r];
                // Find the max subarray no more than K 
                set<int> sumSet;
                sumSet.insert(0);
                int curSum = 0;
                int curMax = INT_MIN;
                for (auto sum:sums) {
                    curSum += sum;
                    auto it = sumSet.lower_bound(curSum - k);
                    if (it != sumSet.end())
                        curMax = max(curMax, curSum - *it);
                    sumSet.insert(curSum);
                }
                ret = max(ret, curMax);
            }
        }
        return ret;
    }
};