<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 solve(vector<int> arr)
    {
        int maxi = 0;
        stack<int> st;

        for (int i = 0; i < arr.size(); i++)
        {
            while (!st.empty() && arr[st.top()] > arr[i])
            {
                int num = arr[st.top()];
                st.pop();

                int nse = i;  // Next Smaller Element index
                int pse = st.empty() ? -1 : st.top();  // Previous Smaller Element index

                maxi = max(maxi, num * (nse - pse - 1));
            }

            st.push(i);
        }

        while (!st.empty())
        {
            int num = arr[st.top()];
            st.pop();

            int nse = arr.size();  // Use the size of the array as the next smaller element
            int pse = st.empty() ? -1 : st.top();  // Handle empty stack condition

            maxi = max(maxi, num * (nse - pse - 1));
        }

        return maxi;
    }

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

        vector<vector<int>> matrix1(matrix.size(), vector<int>(matrix[0].size(), 0));

        // Build up the histogram row by row
        for (int i = 0; i < matrix.size(); i++)
        {
            for (int j = 0; j < matrix[0].size(); j++)
            {
                if (matrix[i][j] == '1')
                {
                    matrix1[i][j] = (i == 0 ? 1 : matrix1[i-1][j] + 1);
                }
                else
                {
                    matrix1[i][j] = 0;
                }
            }
        }

        int maxi = 0;

        // Solve for the maximum area in each row's histogram
        for (int i = 0; i < matrix1.size(); i++)
        {
            maxi = max(maxi, solve(matrix1[i]));
        }

        return maxi;
    }
};

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