Special Positions in a Binary Matrix
<p>Given an <code>m x n</code> binary matrix <code>mat</code>, return <em>the number of special positions in </em><code>mat</code><em>.</em></p>

<p>A position <code>(i, j)</code> is called <strong>special</strong> if <code>mat[i][j] == 1</code> and all other elements in row <code>i</code> and column <code>j</code> are <code>0</code> (rows and columns are <strong>0-indexed</strong>).</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1500-1599/1582.Special%20Positions%20in%20a%20Binary%20Matrix/images/special1.jpg" style="width: 244px; height: 245px;" />
<pre>
<strong>Input:</strong> mat = [[1,0,0],[0,0,1],[1,0,0]]
<strong>Output:</strong> 1
<strong>Explanation:</strong> (1, 2) is a special position because mat[1][2] == 1 and all other elements in row 1 and column 2 are 0.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1500-1599/1582.Special%20Positions%20in%20a%20Binary%20Matrix/images/special-grid.jpg" style="width: 244px; height: 245px;" />
<pre>
<strong>Input:</strong> mat = [[1,0,0],[0,1,0],[0,0,1]]
<strong>Output:</strong> 3
<strong>Explanation:</strong> (0, 0), (1, 1) and (2, 2) are special positions.
</pre>

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

<ul>
	<li><code>m == mat.length</code></li>
	<li><code>n == mat[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 100</code></li>
	<li><code>mat[i][j]</code> is either <code>0</code> or <code>1</code>.</li>
</ul>
Array; Matrix
function numSpecial(mat: number[][]): number {
    const m = mat.length;
    const n = mat[0].length;
    const rows: number[] = Array(m).fill(0);
    const cols: number[] = Array(n).fill(0);
    for (let i = 0; i < m; ++i) {
        for (let j = 0; j < n; ++j) {
            rows[i] += mat[i][j];
            cols[j] += mat[i][j];
        }
    }
    let ans = 0;
    for (let i = 0; i < m; ++i) {
        for (let j = 0; j < n; ++j) {
            if (mat[i][j] === 1 && rows[i] === 1 && cols[j] === 1) {
                ++ans;
            }
        }
    }
    return ans;
}
