<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>

int desperate_optimization(int precision){
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cout.setf(ios::fixed);
    cout.setf(ios::showpoint);
    cout.precision(precision);
    return 0;
}

class Solution {
public:

    vector<int> PSE(vector<int> & nums){
        vector<int> result(nums.size() , -1);
        stack<pair<int,int>> st;
        for(int i = 0; i < nums.size() ; i++){
            while(!st.empty() && st.top().first >= nums[i]) st.pop();
            if(!st.empty()) result[i] = st.top().second;
            st.push({nums[i],i});
        }
        return result;
    }

    vector<int> NSE(vector<int> & nums){
        vector<int> result(nums.size() , nums.size());
        stack<pair<int,int>> st;
        for(int i = nums.size() - 1 ; i >= 0 ; i--){
            while(!st.empty() && st.top().first >= nums[i]){
                 st.pop();
            }
            if(!st.empty()) result[i] = st.top().second;
            st.push({nums[i],i});
        }
        return result;
    }

    int heightOfHistogram(vector<int> & nums){
        vector<int> pse = PSE(nums);
        vector<int> nse = NSE(nums);
        int maxValue = 0;
        for(int i = 0; i < nums.size() ; i++){
            maxValue = max(maxValue,nums[i]*(nse[i] - pse[i] - 1));
        }
        return maxValue;
    }

    Solution(){
        desperate_optimization(10);
    }

    int maximalRectangle(vector<vector<char>>& matrix) {

        if(matrix.empty() || matrix[0].empty()) return 0;
        vector<vector<int>> newMatrix(matrix.size() , vector<int>(matrix[0].size(),0));
        for(int i = 0 ;i < matrix.size() ; i++){
            for(int j = 0 ; j < matrix[0].size() ; j++){
                if(i == 0) newMatrix[i][j] = (matrix[i][j] - '0');
                else{
                    newMatrix[i][j] = (matrix[i][j] == '0') ? 0 : newMatrix[i-1][j] + 1;
                }
            }
        }

        int maxValue = 0;

        for(int i = 0 ; i < matrix.size() ; i++){
            maxValue = max(maxValue, heightOfHistogram(newMatrix[i]));
        }

        return maxValue;
    }
};
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