Find the Minimum Area to Cover All Ones I
<p>You are given a 2D <strong>binary</strong> array <code>grid</code>. Find a rectangle with horizontal and vertical sides with the<strong> smallest</strong> area, such that all the 1&#39;s in <code>grid</code> lie inside this rectangle.</p>

<p>Return the <strong>minimum</strong> possible area of the rectangle.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">grid = [[0,1,0],[1,0,1]]</span></p>

<p><strong>Output:</strong> <span class="example-io">6</span></p>

<p><strong>Explanation:</strong></p>

<p><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/3100-3199/3195.Find%20the%20Minimum%20Area%20to%20Cover%20All%20Ones%20I/images/examplerect0.png" style="padding: 10px; background: rgb(255, 255, 255); border-radius: 0.5rem; width: 279px; height: 198px;" /></p>

<p>The smallest rectangle has a height of 2 and a width of 3, so it has an area of <code>2 * 3 = 6</code>.</p>
</div>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">grid = [[1,0],[0,0]]</span></p>

<p><strong>Output:</strong> <span class="example-io">1</span></p>

<p><strong>Explanation:</strong></p>

<p><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/3100-3199/3195.Find%20the%20Minimum%20Area%20to%20Cover%20All%20Ones%20I/images/examplerect1.png" style="padding: 10px; background: rgb(255, 255, 255); border-radius: 0.5rem; width: 204px; height: 201px;" /></p>

<p>The smallest rectangle has both height and width 1, so its area is <code>1 * 1 = 1</code>.</p>
</div>

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

<ul>
	<li><code>1 &lt;= grid.length, grid[i].length &lt;= 1000</code></li>
	<li><code>grid[i][j]</code> is either 0 or 1.</li>
	<li>The input is generated such that there is at least one 1 in <code>grid</code>.</li>
</ul>
Array; Matrix
function minimumArea(grid: number[][]): number {
    const [m, n] = [grid.length, grid[0].length];
    let [x1, y1] = [m, n];
    let [x2, y2] = [0, 0];
    for (let i = 0; i < m; ++i) {
        for (let j = 0; j < n; ++j) {
            if (grid[i][j] === 1) {
                x1 = Math.min(x1, i);
                y1 = Math.min(y1, j);
                x2 = Math.max(x2, i);
                y2 = Math.max(y2, j);
            }
        }
    }
    return (x2 - x1 + 1) * (y2 - y1 + 1);
}
