<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:
    void call1(const vector<int>& nums, vector<int>& right) {
        stack<int> st;
        for (int i = 0; i < nums.size(); ++i) {
            while (!st.empty() && nums[st.top()] > nums[i]) {
                right[st.top()] = i;
                st.pop();
            }
            st.push(i);
        }
    }
    void call2(vector<int> nums,vector<int>&left){
        int n=nums.size();
        stack<int> st;
        for(int i=n-1;i>=0;i--){
            while(!st.empty() && nums[st.top()]>nums[i]){
                left[st.top()]=i;
                st.pop();
            }
            st.push(i);
        }
    }
    int largestRectangleArea(vector<int>& nums) {
        int n = nums.size();
        vector <int> left(n,-1);
        vector<int> right(n,n);
        call1(nums,right);
        call2(nums,left);
        vector <int> w(n,0);
        int mx=INT_MIN;
        for(int i=0;i<n;i++){
            int w=right[i]-left[i]-1;
            mx=max(mx,w*nums[i]);
        }
        return mx;
    }

};
94096