<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> nextsmallerelement(vector<int>& input){

    stack<int> s;
    s.push(-1);

    vector<int> ans(input.size());

    for(int i = input.size()-1 ; i>=0 ; i--){

        int x = input[i];

        while(s.top() !=-1 && input[s.top()] >= x){
            //greater
            s.pop();
        }

        //smaller
        ans[i] = s.top();
        s.push(i);

        }
        return ans;
    }

    vector<int> prevsmallerelement(vector<int>& input){
        stack<int> st;
        st.push(-1);

        vector<int> ans;

        for(int i = 0 ; i<input.size() ; i++){
            int x = input[i];

            while(st.top() != -1 && input[st.top()] >= x){
                //greater
                st.pop();
            }

            //smaller
            ans.push_back(st.top());
            st.push(i);
        }
        return ans;
    }

    int largestRectangleArea(vector<int>& heights) {

        vector<int> prev = prevsmallerelement(heights);

        vector<int> next = nextsmallerelement(heights);

        int maxarea=INT_MIN;

        for(int i=0 ; i<heights.size() ; i++){

            int len = heights[i];

            if(next[i] == -1){
                next[i] = heights.size();
            }
            int width = next[i] - prev[i] - 1;
            int area = len * width;
            maxarea = max(maxarea , area);
        
        }

        return maxarea;
    }

    int maximalRectangle(vector<vector<char>>& matrix) {
        if (matrix.empty() || matrix[0].empty()) {
            return 0;
        }

        int rows = matrix.size();
        int cols = matrix[0].size();

        vector<int> heights(cols, 0);
        int maxRectangleArea = 0;

        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < cols; j++) {
                if (matrix[i][j] == '1') {
                    heights[j]++;
                } else {
                    heights[j] = 0;
                }
            }

            maxRectangleArea = max(maxRectangleArea, largestRectangleArea(heights));
        }

        return maxRectangleArea;
    }
};
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