<p>Given an <code>m x n</code> matrix <code>matrix</code> and an integer <code>k</code>, return <em>the max sum of a rectangle in the matrix such that its sum is no larger than</em> <code>k</code>.</p>

<p>It is <strong>guaranteed</strong> that there will be a rectangle with a sum no larger than <code>k</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/03/18/sum-grid.jpg" style="width: 255px; height: 176px;" />
<pre>
<strong>Input:</strong> matrix = [[1,0,1],[0,-2,3]], k = 2
<strong>Output:</strong> 2
<strong>Explanation:</strong> Because the sum of the blue rectangle [[0, 1], [-2, 3]] is 2, and 2 is the max number no larger than k (k = 2).
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> matrix = [[2,2,-1]], k = 3
<strong>Output:</strong> 3
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == matrix.length</code></li>
	<li><code>n == matrix[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 100</code></li>
	<li><code>-100 &lt;= matrix[i][j] &lt;= 100</code></li>
	<li><code>-10<sup>5</sup> &lt;= k &lt;= 10<sup>5</sup></code></li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow up:</strong> What if the number of rows is much larger than the number of columns?</p>

class Solution {
    int diff_ = INT_MAX;

    void FindOptimal(const vector<int>& col_sum, int k) {
        int n = col_sum.size();
        vector<int> prefix(n + 1);
        set<int> se{0};
        for (int i = 0; i < n && diff_ != 0; ++i) {
            int num = prefix[i+1] = prefix[i] + col_sum[i];
            if (auto it = se.lower_bound(num - k); it != se.end()) {
                diff_ = min(diff_, k - (num - *it));
            }
            se.insert(num);
        }
    }
public:
    int maxSumSubmatrix(vector<vector<int>>& matrix, int k) {
        int m = matrix.size(), n = matrix[0].size();
        for (int i = 0; i < m; ++i) {
            vector<int> col_sum(n);
            for (int r = i; r < m; ++r) {
                for (int c = 0; c < n; ++c) {
                    col_sum[c] += matrix[r][c];
                }
                FindOptimal(col_sum, k);
                printf("i:%d, r:%d, diff:%d\n", i, r, diff_);
                if (diff_ == 0) {
                    return k;
                }
            }
        }
        return k - diff_;
    }
};
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