<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> arr, int n) {
        stack<int> s;
        s.push(-1);
        vector<int> ans(n, n);

        for (int i = n - 1; i >= 0; i--) {
            int curr = arr[i];
            while (s.top() != -1 && arr[s.top()] >= curr) {
                s.pop();
            }
            // ans is stack ka top
            ans[i] = s.top();
            s.push(i);
        }
        return ans;
    }

    vector<int> prevSmallerElement(vector<int> arr, int n) {
        stack<int> s;
        s.push(-1);
        vector<int> ans(n);

        for (int i = 0; i < n; i++) {
            int curr = arr[i];
            while (s.top() != -1 && arr[s.top()] >= curr) {
                s.pop();
            }
            // ans is stack ka top
            ans[i] = s.top();
            s.push(i);
        }
        return ans;
    }

    int largestRectangleArea(vector<int>& heights) {
        int n = heights.size();
        
        vector<int> next(n);
        next = nextSmallerElement(heights, n);
        
        vector<int> prev(n);
        prev = prevSmallerElement(heights, n);
        
        int area = INT_MIN;
        for (int i = 0; i < n; i++) {
            int l = heights[i];
            
            if (next[i] == -1) {
                next[i] = n;
            }
            int b = next[i] - prev[i] - 1;
            int newArea = l * b;
            area = max(area, newArea);
        }
        return area;
    }

    int maximalRectangle(vector<vector<char>>& matrix) {
        int r = matrix.size();
        if (r == 0) return 0;
        int c = matrix[0].size();

       
        vector<vector<int>> PrefixSum(r, vector<int>(c, 0)); // store sum of each level

        
        for (int i = 0; i < r; i++) {
            for (int j = 0; j < c; j++) {
                if (matrix[i][j] == '1') {
                    if(i==0) 
                     PrefixSum[i][j] = 1;
                    else
                        PrefixSum[i][j] = PrefixSum[i - 1][j] + 1;  // Add from previous row
                    
                }
            }
        }

        int ans = 0;
        for (int i = 0; i < r; i++) {
            ans = max(ans, largestRectangleArea(PrefixSum[i]));
        }

        return ans;
    }
};

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