<p>Given a <code>matrix</code>&nbsp;and a <code>target</code>, return the number of non-empty submatrices that sum to <font face="monospace">target</font>.</p>

<p>A submatrix <code>x1, y1, x2, y2</code> is the set of all cells <code>matrix[x][y]</code> with <code>x1 &lt;= x &lt;= x2</code> and <code>y1 &lt;= y &lt;= y2</code>.</p>

<p>Two submatrices <code>(x1, y1, x2, y2)</code> and <code>(x1&#39;, y1&#39;, x2&#39;, y2&#39;)</code> are different if they have some coordinate&nbsp;that is different: for example, if <code>x1 != x1&#39;</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/09/02/mate1.jpg" style="width: 242px; height: 242px;" />
<pre>
<strong>Input:</strong> matrix = [[0,1,0],[1,1,1],[0,1,0]], target = 0
<strong>Output:</strong> 4
<strong>Explanation:</strong> The four 1x1 submatrices that only contain 0.
</pre>

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

<pre>
<strong>Input:</strong> matrix = [[1,-1],[-1,1]], target = 0
<strong>Output:</strong> 5
<strong>Explanation:</strong> The two 1x2 submatrices, plus the two 2x1 submatrices, plus the 2x2 submatrix.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> matrix = [[904]], target = 0
<strong>Output:</strong> 0
</pre>

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

<ul>
	<li><code>1 &lt;= matrix.length &lt;= 100</code></li>
	<li><code>1 &lt;= matrix[0].length &lt;= 100</code></li>
	<li><code>-1000 &lt;= matrix[i][j] &lt;= 1000</code></li>
	<li><code>-10^8 &lt;= target &lt;= 10^8</code></li>
</ul>

class Solution {
public:
    int numSubmatrixSumTarget(vector<vector<int>>& matrix, int target) {
        int n = matrix.size();
        int m = matrix[0].size();

        // Prefix sum along the rows
        vector<vector<int>> prefix(n, vector<int>(m + 1, 0));
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                prefix[i][j + 1] = prefix[i][j] + matrix[i][j];
            }
        }

        int cnt = 0;

        // Iterate over all possible column pairs
        for (int col1 = 1; col1 <= m; col1++) {
            for (int col2 = col1; col2 <= m; col2++) {

                // Build the array of sums for the current column range
                vector<int> arr(n);
                for (int i = 0; i < n; i++) {
                    arr[i] = prefix[i][col2] - prefix[i][col1 - 1];
                }

                // Find the maximum sum of subarrays with sum <= k using a set
                unordered_map<int,int> mp;
                int sum = 0;
                int ans = 0;
                mp[sum] = 1;

                for (int i = 0; i < n; i++) {
                    sum += arr[i];
                    cnt += mp[sum-target];
                    mp[sum]++;
                }
            }
        }

        return cnt;
    }
};
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