<p>Given a <code>rows x cols</code>&nbsp;binary <code>matrix</code> filled with <code>0</code>&#39;s and <code>1</code>&#39;s, find the largest rectangle containing only <code>1</code>&#39;s and return <em>its area</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/09/14/maximal.jpg" style="width: 402px; height: 322px;" />
<pre>
<strong>Input:</strong> matrix = [[&quot;1&quot;,&quot;0&quot;,&quot;1&quot;,&quot;0&quot;,&quot;0&quot;],[&quot;1&quot;,&quot;0&quot;,&quot;1&quot;,&quot;1&quot;,&quot;1&quot;],[&quot;1&quot;,&quot;1&quot;,&quot;1&quot;,&quot;1&quot;,&quot;1&quot;],[&quot;1&quot;,&quot;0&quot;,&quot;0&quot;,&quot;1&quot;,&quot;0&quot;]]
<strong>Output:</strong> 6
<strong>Explanation:</strong> The maximal rectangle is shown in the above picture.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> matrix = [[&quot;0&quot;]]
<strong>Output:</strong> 0
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> matrix = [[&quot;1&quot;]]
<strong>Output:</strong> 1
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>rows == matrix.length</code></li>
	<li><code>cols == matrix[i].length</code></li>
	<li><code>1 &lt;= row, cols &lt;= 200</code></li>
	<li><code>matrix[i][j]</code> is <code>&#39;0&#39;</code> or <code>&#39;1&#39;</code>.</li>
</ul>

class Solution {
  

    vector<int>rightSmallerElement(vector<int>&arr)
    {
        
        int n=arr.size();
        vector<int>ans(n);
        stack<int>stk;
        for(int i=n-1;i>=0;i--)
        {
            while(!stk.empty())
            {
                if(arr[stk.top()]<arr[i])
                {
                    ans[i]=stk.top();
                    break;
                }
                else
                {
                    stk.pop();
                }
                
            }
            if(stk.empty())
            {
                ans[i]=n;
            }
            stk.push(i);
        }
        return ans;
    }
      vector<int>leftSmallerElement(vector<int>&arr)
    {
        //for this we would be making a monotonic increasing stack
        //lets not include the element in the left smallest
        vector<int>ans;
        int n=arr.size();
        stack<int>stk;
        for(int i=0;i<arr.size();i++)
        {
            while(!stk.empty())
            {
                if(arr[i]>arr[stk.top()])
                {
                    ans.push_back(stk.top());
                    break;
                }
                else
                {
                    stk.pop();
                }
            }
            if(stk.empty())
            {
                ans.push_back(-1);
            }
            stk.push(i);
        }
        return ans;
    }

    int findLargestArea(vector<int>&arr)
    {
        vector<int>prevSmaller=leftSmallerElement(arr);
        vector<int>nextSmaller=rightSmallerElement(arr);

        int ans=0;
        for(int i=0;i<arr.size();i++)
        {
            int left=prevSmaller[i];
            int right=nextSmaller[i];
            int area=(right-left-1)*arr[i];
            ans=max(ans,area);
        }
        return ans;
    }
public:
    int maximalRectangle(vector<vector<char>>& matrix) {
        int n=matrix.size();
        int m=matrix[0].size();
        int ans=INT_MIN;
        vector<int>height(m,0);
        for(auto arr:matrix)
        {
            for(int i=0;i<m;i++)
            {
                 height[i] = (arr[i] == '1') ? height[i] + 1 : 0;
            }
            ans=max(ans,findLargestArea(height));

        }
        return ans;


    }
};
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