Maximal Rectangle
<p>Given a <code>rows x cols</code>&nbsp;binary <code>matrix</code> filled with <code>0</code>&#39;s and <code>1</code>&#39;s, find the largest rectangle containing only <code>1</code>&#39;s and return <em>its area</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0000-0099/0085.Maximal%20Rectangle/images/maximal.jpg" style="width: 402px; height: 322px;" />
<pre>
<strong>Input:</strong> matrix = [[&quot;1&quot;,&quot;0&quot;,&quot;1&quot;,&quot;0&quot;,&quot;0&quot;],[&quot;1&quot;,&quot;0&quot;,&quot;1&quot;,&quot;1&quot;,&quot;1&quot;],[&quot;1&quot;,&quot;1&quot;,&quot;1&quot;,&quot;1&quot;,&quot;1&quot;],[&quot;1&quot;,&quot;0&quot;,&quot;0&quot;,&quot;1&quot;,&quot;0&quot;]]
<strong>Output:</strong> 6
<strong>Explanation:</strong> The maximal rectangle is shown in the above picture.
</pre>

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

<pre>
<strong>Input:</strong> matrix = [[&quot;0&quot;]]
<strong>Output:</strong> 0
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> matrix = [[&quot;1&quot;]]
<strong>Output:</strong> 1
</pre>

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

<ul>
	<li><code>rows == matrix.length</code></li>
	<li><code>cols == matrix[i].length</code></li>
	<li><code>1 &lt;= row, cols &lt;= 200</code></li>
	<li><code>matrix[i][j]</code> is <code>&#39;0&#39;</code> or <code>&#39;1&#39;</code>.</li>
</ul>
Stack; Array; Dynamic Programming; Matrix; Monotonic Stack
func maximalRectangle(matrix [][]byte) int {
	n := len(matrix[0])
	heights := make([]int, n)
	ans := 0
	for _, row := range matrix {
		for j, v := range row {
			if v == '1' {
				heights[j]++
			} else {
				heights[j] = 0
			}
		}
		ans = max(ans, largestRectangleArea(heights))
	}
	return ans
}

func largestRectangleArea(heights []int) int {
	res, n := 0, len(heights)
	var stk []int
	left, right := make([]int, n), make([]int, n)
	for i := range right {
		right[i] = n
	}
	for i, h := range heights {
		for len(stk) > 0 && heights[stk[len(stk)-1]] >= h {
			right[stk[len(stk)-1]] = i
			stk = stk[:len(stk)-1]
		}
		if len(stk) > 0 {
			left[i] = stk[len(stk)-1]
		} else {
			left[i] = -1
		}
		stk = append(stk, i)
	}
	for i, h := range heights {
		res = max(res, h*(right[i]-left[i]-1))
	}
	return res
}