<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 {
public:
     vector<int>nextSmaller(vector<int> &arr,int n){
        stack<int>s;
        s.push(-1);
        vector<int>ans(n);

        for(int i=n-1;i>=0;i--){
            int curr = arr[i];
            while(s.top() != -1 && arr[s.top()] >= curr){
                s.pop();
            }
            //ans is stack top
                ans[i] = s.top();
                s.push(i);
        }
         return ans;
    }

   
    vector<int> prevSmaller(vector<int> &arr,int n){
        stack<int>s;
        s.push(-1);
        vector<int>ans(n);

        for(int i=0;i<n;i++){
            int curr = arr[i];
            while(s.top() != -1 && arr[s.top()] >= curr){
                s.pop();
            }
            //ans is stack top
                ans[i] = s.top();
                s.push(i);
        }
         return ans;
    }
    
    
   int largestRectangleArea(vector<int> &heights) {
        int n = heights.size();

        vector<int> next(n);
        next = nextSmaller(heights,n);

        vector<int> prev(n);
        prev = prevSmaller(heights,n);

        int area = INT_MIN;

        for(int i=0;i<n;i++){
            int l = heights[i];

           if(next[i] == -1){
            next[i] = n;
           }
           int b = next[i] - prev[i] - 1;
            int newArea = l*b;
           area = max(area,newArea);
        }
    return area;
    }

    int maximalRectangle(vector<vector<char>>& matrix) {
       vector<int> height(matrix[0].size(),0);

        int maxi=INT_MIN;
        for(int i=0;i<matrix.size();i++)
        {
            //to create height array
            for(int j=0;j<height.size();j++)
            {
                if(matrix[i][j]=='1')
                {
                    height[j]++;
                }
                else height[j]=0;
            }
            maxi=max(maxi,largestRectangleArea(height));
        }

            return maxi;
    }
};
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