<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>& hgt) {
                int mx = 0 ; 
        stack<int>st;
        st.push(-1);
        int ans =0 ; 
        for(int i =0 ; i < hgt.size() ; i++)
        {   
            int num = hgt[i];
           
             while( st.top() != - 1 && hgt[st.top()] >= num )
                {   
                    int ht= hgt[st.top()];

                    st.pop();
                    int value=  ht * ( i- st.top() -1 )  ;


                    ans = max( ans , value ) ;
            }
            
            st.push(i);

        }
        // ans = max( ans , st.top());
         while( st.top() != - 1)
                {   
                    int num =st.top();
                    num = hgt[num];
                    st.pop();
                
                    int value=  num * ( hgt.size()- st.top() -1 )  ;

                    ans = max( ans , value ) ;
            }
        return ans;

 


    }
       

    
};
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