<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:
    //  vector<int> NSR(vector<int>& heights){
    //     int n = heights.size();
    //     vector<int> nsr(n);
    //     stack<int> st;
    //     for(int i=0;i<n;i++){
    //         while(!st.empty() && heights[i]<heights[st.top()]){
    //             nsr[st.top()] = i;
    //             st.pop();
    //         }
    //         st.push(i);
    //     }
    //     while(!st.empty()){
    //         nsr[st.top()] = n;
    //         st.pop();
    //     }
    //     return nsr;
    //  }

    //  vector<int> NSL(vector<int>& heights){
    //     int n = heights.size();
    //     vector<int> nsl(n);
    //     stack<int> st;
    //     for(int i=n-1;i>=0;i--){
    //         while(!st.empty() && heights[i]<heights[st.top()]){
    //             nsl[st.top()] = i;
    //             st.pop();
    //         }
    //         st.push(i);
    //     }
    //     while(!st.empty()){
    //         nsl[st.top()] = -1;
    //         st.pop();
    //     }
    //     return nsl;
    //  }
    //  In the above case time complexity is O(3n) we are traversing the loop 3 times 
    int largestRectangleArea(vector<int>& heights) {
         int n = heights.size();
         // In the optimized Approach there is no need of Array 
         stack<int> st; // Only need of Stack 

         int maxArea = 0;
         for(int i=0;i<n;i++)  {
            while(!st.empty() && heights[i]<heights[st.top()]){
                int index = st.top();st.pop();
                if(!st.empty()){
                    maxArea = max(maxArea,heights[index]*(i-st.top()-1));
                }
                else
                {
                    maxArea =  max(maxArea,heights[index]*(i));
                }
            }
            st.push(i);
         }
         while(!st.empty()){
            int index = st.top();st.pop();
             if(!st.empty()){
                    maxArea = max(maxArea,heights[index]*(n-st.top()-1));
                }
                else
                {
                    maxArea =  max(maxArea,heights[index]*(n));
                }
         }


         return maxArea;
    }
};
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