问一道G家面试题 - 未名空间(mitbbs.com)


问一道G家面试题 - 未名空间(mitbbs.com)
一个matrix,每个matrix[i][j]有一个价格,给你一个budget,问如何求出最大面积的
子矩阵,使得子矩阵的价格之和不超过budget.

Same as max sum rectangle -
One change:get max value that is less than or equal budget.
http://likemyblogger.blogspot.com/2015/08/mj-26-maximum-sum-rectangle-in-2d-matrix.html
    int kadane(vector<int> &col){
        int max_ending_here = INT_MIN, max_ending_sofar = INT_MIN;
        for(auto i:col){
            if(max_ending_here<0) max_ending_here = 0;
            max_ending_here += i;
            max_ending_sofar = max(max_ending_sofar, max_ending_here);
        }
        return max_ending_sofar;
    }
 
    int maxSum2D(vector<vector<int>> &matrix){
        int m = matrix.size();
        if(m==0) return 0;
        int n = matrix[0].size();
        if(n==0) return 0;
 
        int ret = INT_MIN;
        for(int l=0; l<n; ++l){
            vector<int> col(m, 0);
            for(int r=l; r<n; ++r){
                for(int i=0; i<m; ++i){
                    col[i] += matrix[i][r];
                }
                ret = max(ret, kadane(col));
            }
        }
        return ret;
    }

这题看起来和leetcode的Maximal Rectangle 很像,但似乎不可能找到类似的复杂度的
解法。

能想到的solution
1.preprocess:建立一个matrix[][]存[0][0]到[i][j]的sum,这个可以
用dp解决,复杂度O(n^2)
2.check 以每一点为右下点的最大可能的矩阵,这个每个点都是O(i*lg(j))的复
杂度了吧。除了可以用个binary search来减少一点复杂度,我也想不到别的办法了。

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