<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://assets.leetcode.com/uploads/2024/05/08/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://assets.leetcode.com/uploads/2024/05/08/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>

class Solution:
    def minimumArea(self, grid: List[List[int]]) -> int:
        R = len(grid)
        C = len(grid[0])
        ll,rr,uu,bb = C+1,-1,R+1,-1
        for r in range(R):
            for c in range(C):
                if grid[r][c] == 1:
                    ll = min(ll,c)
                    rr = max(rr,c)
                    uu = min(uu,r)
                    bb = max(bb,r)
                
        return (rr-ll+1) * (bb-uu+1)
        
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