<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 res = 0;
        vector<int> lefts = smallerLeft(heights);
        vector<int> rights = smallerRight(heights);

        for (int i = 0; i < heights.size(); i++) {
            int l = lefts[i] + 1;
            int r = rights[i] - 1;
            res = max(res, heights[i] * (r - l + 1));
        }
        return res;
    }

private:
    vector<int> smallerLeft(vector<int>& heights) {
        vector<int> res(heights.size());
        stack<int> stack;
        for (int i = 0; i < heights.size(); i++) {
            int h = heights[i];
            while (stack.size() && heights[stack.top()] >= h) {
                stack.pop();
            }
            if (stack.size()) {
                res[i] = stack.top();
            } else {
                res[i] = -1;
            }
            stack.push(i);
        }
        return res;
    }
    vector<int> smallerRight(vector<int>& heights) {
        vector<int> res(heights.size());
        stack<int> stack;
        for (int i = heights.size() - 1; i >= 0; i--) {
            int h = heights[i];
            while (stack.size() && heights[stack.top()] >= h) {
                stack.pop();
            }
            if (stack.size()) {
                res[i] = stack.top();
            } else {
                res[i] = heights.size();
            }
            stack.push(i);
        }
        return res;
    }
};
/*
    for each bar b, find the first bar the left and right that is shorter
    this space in between is the largest rectangle centered at b

    do this for all bars and we can get the max rectagnel possible
 */
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