<p>Given an array of integers <code>heights</code> representing the histogram&#39;s bar height where the width of each bar is <code>1</code>, return <em>the area of the largest rectangle in the histogram</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/01/04/histogram.jpg" style="width: 522px; height: 242px;" />
<pre>
<strong>Input:</strong> heights = [2,1,5,6,2,3]
<strong>Output:</strong> 10
<strong>Explanation:</strong> The above is a histogram where width of each bar is 1.
The largest rectangle is shown in the red area, which has an area = 10 units.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/01/04/histogram-1.jpg" style="width: 202px; height: 362px;" />
<pre>
<strong>Input:</strong> heights = [2,4]
<strong>Output:</strong> 4
</pre>

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

<ul>
	<li><code>1 &lt;= heights.length &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= heights[i] &lt;= 10<sup>4</sup></code></li>
</ul>

class Solution {
public:
    int largestRectangleArea(vector<int>& heights) {
        int n = heights.size();
      stack < int > st;
      int leftsmall[n], rightsmall[n];
      for (int i = 0; i < n; i++) {
        while (!st.empty() && heights[st.top()] >= heights[i]) {
          st.pop();
        }
        if (st.empty())
          leftsmall[i] = 0;
        else
          leftsmall[i] = st.top() + 1;
        st.push(i);
      }
      // clear the stack to be re-used
      while (!st.empty())
        st.pop();

      for (int i = n - 1; i >= 0; i--) {
        while (!st.empty() && heights[st.top()] >= heights[i])
          st.pop();

        if (st.empty())
          rightsmall[i] = n - 1;
        else
          rightsmall[i] = st.top() - 1;

        st.push(i);
      }
      int maxA = 0;
      for (int i = 0; i < n; i++) {
        maxA = max(maxA, heights[i] * (rightsmall[i] - leftsmall[i] + 1));
      }
      return maxA;
    }
};
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