<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:
int largestRectangleArea(vector<int>& heights) {
        int n=heights.size();
        stack<int>st;
        vector<int>right(n);
       int ans=0;
       int idx;
       for(int i=0;i<n;i++){
        while(!st.empty()&&heights[st.top()]>heights[i]){
            idx=st.top();
            st.pop();
            if(!st.empty()){
                ans=max(ans,heights[idx]*(i-st.top()-1));
            }
            else ans=max(ans,heights[idx]*(i));
        }
        st.push(i);
       }
       while(!st.empty()){
            idx=st.top();
            st.pop();
            if(!st.empty()){
                ans=max(ans,heights[idx]*(n-st.top()-1));
            }
            else ans=max(ans,heights[idx]*(n));
        }
       
        return ans;
    }
    int maximalRectangle(vector<vector<char>>& matrix) {
        int ans=0;
        int row=matrix.size();
        int col=matrix[0].size();
        vector<int>heights(col,0);
        for(int i=0;i<row;i++){
            for(int j=0;j<col;j++){
                if(matrix[i][j]=='0') heights[j]=0;
                else heights[j]+=1;
            }
            ans=max(ans,largestRectangleArea(heights));
        }
        return ans;
    }
};
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