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
function numSubmatrixSumTarget(matrix: number[][], target: number): number {
    const m = matrix.length;
    const n = matrix[0].length;
    let ans = 0;
    for (let i = 0; i < m; ++i) {
        const col: number[] = new Array(n).fill(0);
        for (let j = i; j < m; ++j) {
            for (let k = 0; k < n; ++k) {
                col[k] += matrix[j][k];
            }
            ans += f(col, target);
        }
    }
    return ans;
}

function f(nums: number[], target: number): number {
    const d: Map<number, number> = new Map();
    d.set(0, 1);
    let cnt = 0;
    let s = 0;
    for (const x of nums) {
        s += x;
        if (d.has(s - target)) {
            cnt += d.get(s - target)!;
        }
        d.set(s, (d.get(s) || 0) + 1);
    }
    return cnt;
}
