<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 {
    vector<int> leftSmaller(vector<int> &arr)
    {
        int n = arr.size();
        stack<int> st;
        vector<int> ans;
        st.push(1);
        ans.push_back(0);
        for (int i = 1; i < n; i++) {
            while (!st.empty() && arr[i] <= arr[st.top()-1]) {
                st.pop();
            }
            if (st.empty()) ans.push_back(0);
            else ans.push_back(st.top());
            st.push(i+1);
        }
        return ans;
    }

    int rightSmaller(vector<int> &arr, vector<int> &leftArr)
    {
        int n = arr.size();
        stack<int> st;
        int back = n;
        int maxArea = (back-leftArr[n-1])*arr[n-1];
        st.push(n-1);
        for (int i = arr.size()-2; i >= 0; i--) {
            while (!st.empty() && arr[i] <= arr[st.top()]) {
                st.pop();
            }
            if (st.empty()) back = n;
            else back = st.top();
            st.push(i);
            maxArea = max(maxArea, arr[i]*(back - leftArr[i]));
        }
        return maxArea;
    }

    int maxAreaRectangle(vector<int>& hist) {
        vector<int> arr;
        stack<int> st;
        int maxArea = INT32_MIN, n = hist.size();
        st.push(0);
        arr.push_back(0);
        for (int i = 1; i < n; i++) {
            while (!st.empty() && hist[i] <= hist[st.top()]) {
                arr[st.top()] = hist[st.top()]*(i - arr[st.top()]);
                maxArea = max(arr[st.top()], maxArea);
                st.pop();
            }
            if (st.empty()) arr.push_back(0);
            else arr.push_back(st.top()+1);
            st.push(i);
        }
        while (!st.empty()) {
            arr[st.top()] = hist[st.top()]*(n - arr[st.top()]);
            maxArea = max(arr[st.top()], maxArea);
            st.pop();
        }
        return maxArea;
    }

    int largestSquare(vector<int>& histograms) {
        // vector<int> leftArr = leftSmaller(histograms);
        // return rightSmaller(histograms, leftArr);
        return maxAreaRectangle(histograms);
    }
public:
    int maximalRectangle(vector<vector<char>>& arr) {
        int n = arr[0].size();
        vector<int> histograms(n, 0);
        int ans = INT32_MIN;
        for (int i = 0; i < arr.size(); i++) {
            int localMax = INT32_MIN;
            for (int j = 0; j < arr[0].size(); j++) {
                if (arr[i][j] == '1') histograms[j]++;
                else histograms[j] = 0;
            }
            ans = max(ans, largestSquare(histograms));
        }
        return ans;
    }
};
20208