<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:
    void get_prevSmall(vector<int>& matrix, vector<int>& prevSmall){
        stack<int> st;
        st.push(-1);
        for(int i = 0; i < matrix.size(); i++){
            while(st.top() != -1 && matrix[st.top()] >= matrix[i]) st.pop();
            prevSmall[i] = st.top();
            st.push(i); 
        }
    }

    void get_nextSmall(vector<int>& matrix, vector<int>& nextSmall){
        stack<int> st;
        st.push(matrix.size());
        for(int i = matrix.size() - 1; i >= 0; i--){
            while(st.top() != matrix.size() && matrix[st.top()] >= matrix[i]) st.pop();
            nextSmall[i] = st.top();
            st.push(i); 
        }
    }

    int maximalRectangle(vector<vector<char>>& matrix) {
        vector<vector<int>> matrix2(matrix.size(), vector<int>(matrix[0].size()));
        for(int i = 0; i < matrix.size(); i++){
            for(int j = 0; j < matrix[0].size(); j++){
                if(i > 0 && matrix[i-1][j] != '0' && matrix[i][j] != '0') matrix2[i][j] = ((matrix[i][j]) - '0') + matrix2[i-1][j];
                else matrix2[i][j] = matrix[i][j] - '0';
            }
        }
        vector<int> prevSmall(matrix2[0].size());
        vector<int> nextSmall(matrix2[0].size());
        int ans = 0;
        for(int i = 0; i < matrix2.size(); i++){
            get_prevSmall(matrix2[i], prevSmall);
            get_nextSmall(matrix2[i], nextSmall);
            for(int j = 0; j < matrix2[0].size(); j++){
                ans = max(matrix2[i][j] * (nextSmall[j] - prevSmall[j] - 1), ans);
            }
        }
        return ans;
    }
};
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