<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 {
    int largestRectangleArea(vector<int>heights) {
        stack<int> st;
        int n = heights.size();
        int maxi = 0;
        for (int i = 0; i < n; i++) {
            while (!st.empty() && heights[st.top()] > heights[i]) {
                int curr = st.top();
                st.pop();
                int area =
                    heights[curr] * (st.empty() ? i : (i - st.top() - 1));
                maxi = max(maxi, area);
            }
            st.push(i);
        }

        while (!st.empty()) {
            int curr = st.top();
            st.pop();
            int area = heights[curr] * (st.empty() ? n : (n - st.top() - 1));
            maxi = max(maxi, area);
        }
        return maxi;
    }
public:
    int maximalRectangle(vector<vector<char>>& matrix) {
        int m=matrix[0].size();
        int n=matrix.size();
        vector<int>heights(m,0);
        int ans=0;
        for(int i=0; i<n; i++){
            for(int j=0; j<m; j++){
                if(matrix[i][j]=='1')heights[j]++;
                else heights[j]=0;
            }
            int area=largestRectangleArea(heights);
            ans=max(ans,area);
        }
        return ans;
    }
};
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