<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 {
 public:
  int maxSumSubmatrix(vector<vector<int>>& matrix, int k) {
    const int m = matrix.size();
    const int n = matrix[0].size();
    int ans = INT_MIN;

    for (int baseCol = 0; baseCol < n; ++baseCol) {
      // sums[i] := sum(matrix[i][baseCol..j])
      vector<int> sums(m, 0);
      for (int j = baseCol; j < n; ++j) {
        for (int i = 0; i < m; ++i)
          sums[i] += matrix[i][j];
        // Find the maximum sum <= k of all the subarrays.
        set<int> accumulate{0};
        int prefix = 0;
        for (const int sum : sums) {
          prefix += sum;
          if (const auto it = accumulate.lower_bound(prefix - k);
              it != accumulate.cend())
            ans = max(ans, prefix - *it);
          accumulate.insert(prefix);
        }
      }
    }

    return ans;
  }
};
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