<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 findLargestAreaInHistogram(int n, vector<int>& rowHistogram) {
        int area = 0;
        for (int i = 0; i < n; i++) {
            int breadth = 1;
            // first find only this area
            area = max(area, rowHistogram[i] * breadth);
            // now start checking for other lengths as well
            // as we might need other heights for more area

            // first check for heights behind it
            for (int j = i - 1; j >= 0; j--) {
                if (rowHistogram[j] >= rowHistogram[i]) {
                    breadth++;
                } else {
                    break;
                }
            }

            // now check for heights after it
            for (int j = i + 1; j < n; j++) {
                if (rowHistogram[j] >= rowHistogram[i]) {
                    breadth++;
                } else {
                    break;
                }
            }

            // now compute the area
            area = max(area, rowHistogram[i] * breadth);
        }
        return area;
    }
    int maximalRectangle(vector<vector<char>>& matrix) {
        int n = matrix.size();
        int m = matrix[0].size();
        int area = 0;
        // first make the histogram row wise
        vector<int> rowHistogram(m, 0);
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; ++j) {
                if (matrix[i][j] == '1') {
                    rowHistogram[j]++;
                } else {
                    rowHistogram[j] = 0;
                }
            }
            // now I have created the histogram for the particular level
            // now find the largest area in that level
            area = max(area, findLargestAreaInHistogram(m, rowHistogram));
        }
        return area;
    }
};
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