<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>& v) {
        v.push_back(0); // inserting 0 at end as at the end of loop it
        // will calculate max area for all remaining data in stack.  
        int n = v.size();
        vector<int>st;
        vector<int>dp(n,0); // stores the leftmost point for all height_value where 
        // it can go and till that point height is >= height_value 

        // dp only stores index of monotonically increasing height
        int ans = 0; // max area
        for(int i=0;i<n;i++){
            if(st.empty() || v[st.back()]<v[i]){ // as stack is empty means may be it is 
                    dp[i]=i;                     //start or there was disconinous in height
                st.push_back(i);              // by some height = 0 entry  => any new height can start and its 
                                            // left most point is (i) itself as no height or new height
            }else if (v[st.back()]>v[i]){ // new height is less so have to remove all big heights which is less than it.
                int j = i; // store the possible left most point for new height(new entry) 
                while( !st.empty() && v[st.back()]>v[i]){
                    int h = v[st.back()];
                    int ind = dp[st.back()];
                    j = ind;
                    ans=max(ans,(i-ind)*h); // dynamically calculate area as we already know for all 
                    st.pop_back();   // height its right most point is i and left most is stored and stack is monotono up 
                }
                if(st.empty() || v[st.back()]!=v[i]){ // if new height is same as old height in stack we skipping the 
                    dp[i]=j;                          // insertion as for that heigth a index is already present 
                    st.push_back(i);
                }
            }
            
        }
        return ans; // returning the final answer
    }
};
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