Number of Submatrices That Sum to Target
<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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1000-1099/1074.Number%20of%20Submatrices%20That%20Sum%20to%20Target/images/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>
Array; Hash Table; Matrix; Prefix Sum
class Solution {
public:
    int numSubmatrixSumTarget(vector<vector<int>>& matrix, int target) {
        int m = matrix.size(), n = matrix[0].size();
        int ans = 0;
        for (int i = 0; i < m; ++i) {
            vector<int> col(n);
            for (int j = i; j < m; ++j) {
                for (int k = 0; k < n; ++k) {
                    col[k] += matrix[j][k];
                }
                ans += f(col, target);
            }
        }
        return ans;
    }

    int f(vector<int>& nums, int target) {
        unordered_map<int, int> d{{0, 1}};
        int cnt = 0, s = 0;
        for (int& x : nums) {
            s += x;
            if (d.count(s - target)) {
                cnt += d[s - target];
            }
            ++d[s];
        }
        return cnt;
    }
};