<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> nexts(vector<int>& heights){
        int n = heights.size();
        vector<int> ans(n);
        stack<int> st;
        st.push(-1);
        for(int i=n-1; i>=0; i--){
            while(st.top()!=-1 && heights[st.top()]>=heights[i]){
                st.pop();
            }
            ans[i]=st.top();
            st.push(i);
        }
        return ans;
    }
    vector<int> prevs(vector<int>& heights){
        int n = heights.size();
        vector<int> ans(n);
        stack<int> st;
        st.push(-1);
        for(int i=0; i<n; i++){
            while(st.top()!=-1 && heights[st.top()]>=heights[i]){
                st.pop();
            }
            ans[i]=st.top();
            st.push(i);
        }
        return ans;
    }
    int largestRectangleArea(vector<int>& heights) {
        int n = heights.size();
        vector<int> ps(n,-1);
        ps = prevs(heights);
        vector<int> ns(n,-1);
        ns = nexts(heights);

        int ans = 0;
        for(int i=0; i<n; i++){
            int psidx = ps[i];
            int nextidx = ns[i];
            if(nextidx==-1){
                nextidx = n;
            }
            ans = max(ans, (nextidx-psidx-1)*heights[i]);
        }
        return ans;
    }
    int maximalRectangle(vector<vector<char>>& matrix) {
        int ans = INT_MIN;
        vector<int> heights(matrix[0].size(), 0);
        for(int i=0; i<matrix.size(); i++){
            for(int j=0; j<matrix[0].size(); j++){
                if(matrix[i][j]=='1'){
                    heights[j]++;
                }
                else{
                    heights[j]=0;
                }
            }
            ans = max(ans, largestRectangleArea(heights));
        }
        return ans;
    }
};
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